Hc Verma II Solutions for Class 12 Science Physics Chapter 32 Electric Current In Conductors are provided here with simple step-by-step explanations. These solutions for Electric Current In Conductors are extremely popular among Class 12 Science students for Physics Electric Current In Conductors Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma II Book of Class 12 Science Physics Chapter 32 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Hc Verma II Solutions. All Hc Verma II Solutions for class Class 12 Science Physics are prepared by experts and are 100% accurate.

#### Page No 196:

#### Answer:

(a) When the three resistors are connected in series:

The resultant resistance, *R _{eq} = R + R + R *= 90 $\Omega $

(b) When the three resistors are connected in parallel:

The resultant resistance of the combination,

$\frac{1}{{R}_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R}=\frac{3}{30}=\frac{1}{10}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow {R}_{eq}=10\Omega $

(c) When two of the resistors are connected in parallel and this combination is connected in series with the third resistor:

Let

*R'*be the resultant resistance of the two resistors connected in parallel to each other. Therefore,

$\frac{1}{R\text{'}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}=\frac{2}{30}=\frac{1}{15}$

$\Rightarrow R\text{'}=15\Omega $

Now, the net resistance of the combination of the resistors,

*R*=

_{eq}*R' + R =*15 + 30 = 45 $\Omega $

(d) When two of the resistors are connected in series and the combination is connected to the third resistor in parallel:

Let

*R'*be the resultant resistance of the series in combination. Therefore,

*R' = R + R =*30 + 30 = 60 $\Omega $

Now, let the net resultant of the combination be

*R*. So,

_{eq}$\frac{1}{{R}_{eq}}=\frac{1}{R\text{'}}+\frac{1}{R}=\frac{1}{60}+\frac{1}{30}=\frac{3}{60}=\frac{1}{20}$

$\Rightarrow {R}_{eq}=20\Omega $

#### Page No 196:

#### Answer:

Yes, there is an electric current in the direction of the proton beam. Conventionally, the direction of electric current is same as that of the positive charge flow.

#### Page No 196:

#### Answer:

Yes, there is a net current that flows from left to right. By convention, the direction of current is along the flow of positive charge. As the positive ions move from left to right, current also moves from left to right. Also, since the negative ions move from right to left, current moves from left to right. Thus, the movement of net current is from left to right.

#### Page No 196:

#### Answer:

Conventionally, the direction of current is along the direction of flow of positive charge and opposite the direction of flow of negative charge. In a TV tube, current will flow from the front to the rear, as the direction of current is opposite the direction in which electrons are moving.

#### Page No 196:

#### Answer:

An electron drifts under the influence of an external electric field. During the course of this motion, an electron follows a very random path. So, we have to take the average for a very long time. Hence, we cannot define drift speed as the limit of Δ*l*/Δ*t** *as Δ*t *→ 0 because this is a very short interval of time that is not enough to get the desired result.

#### Page No 196:

#### Answer:

You have studied in the previous chapters that when a static charge is given to a metal, the charge resides on its surface and there is no electric field inside it. But this is not the case when a charge is in motion. Current can flow through a conductor only when an electric field is established inside it. An electric field exists inside the conductor, as it is connected to the battery. Thus, current flows through the conductor.

#### Page No 196:

#### Answer:

No, the number of electrons in the wire remains constant. The electrons that drift and move to the positive terminal of the battery, under the influence of the external electric field, are replaced by the battery in the circuit. As a result, the total number of free electrons in the wire is always constant.

#### Page No 196:

#### Answer:

A copper wire has higher conductance than an aluminium wire. So, a copper wire offer less resistance to the current flow than an aluminium wire. Thus, there is more heat dissipation in an aluminium wire than in a copper wire. This is why a fan with copper winding in its motor consumes less power compared to a similar fan with aluminium winding.

#### Page No 196:

#### Answer:

In the expression *U = Vit,* voltage *V* and current *i* are variables for a given time interval. So, based on this expression we cannot say that *U* is proportional to *i*. In the expression *U* = *i*^{2}*Rt**,* the resistance *R* is fixed for a circuit for a given time interval. So, based on this expression, we can say that *U* is proportional to *i*^{2} and not* i*.

#### Page No 196:

#### Answer:

Yes, the "work done by the battery" and "the thermal energy developed" represent two names of the same physical quantity. The work done by the battery on the resistor is dissipated by the resistor in the form of thermal energy. Hence the " work done by the battery " and " the thermal energy developed " represent two names of the same physical quantity.

However, a non-ideal battery, in this case, would have to do extra work in order to overcome the internal resistance of the battery. Hence, the work done by a non-ideal battery will not be equal to the thermal energy developed.

#### Page No 196:

#### Answer:

No, the work done by a battery is not always equal to the thermal energy developed in the electrical circuit. In case of a non-ideal battery, the work done by the battery is the sum of the thermal energy developed in the electric circuit and the thermal energy developed in the internal resistance of the battery. In case of a capacitor, the work done by the battery is equal to *C V ^{2}*. An amount of energy equal to $\frac{1}{2}$

*C V*is stored in it when it is fully charged, which is a form of electrical energy and not a form of thermal energy. During the charging of the capacitor, $\frac{1}{2}$

^{2}*C*

*V*

^{2}

*of energy is lost in the form of heat and electromagnetic radiation.*

^{ }#### Page No 196:

#### Answer:

No, the work done by a non-ideal battery is not equal to the thermal energy developed in the resistor, as energy is spent to overcome the internal resistance of the battery and the resistance of the wire that connects the circuit elements/resistor to the battery. However, the resistance of the wire is generally negligible.

Yes, the answer will change if the battery is ideal. An ideal battery has no internal resistance. Hence, the work done by an ideal battery will be equal to the thermal energy developed in the resistor, assuming that the resistance of the wires used for connection is negligible.

#### Page No 196:

#### Answer:

Yes, the given statement is correct. When charge flows through a conductor, its electric potential energy decreases. This loss in electric potential energy appears as increased heat energy of the resistor. Thus, heat energy is developed in a resistor when there is an electric current in it.

#### Page No 196:

#### Answer:

When there is a transfer of charge through a wire, current is said to be flowing through it. It is the electron/charge that drifts through the wire. The assertions "charge is going" and "current is going through the wire" are correct, as they signify the same thing, that is, flow of charge.

#### Page No 196:

#### Answer:

A potentiometer is preferred to measure the emf of a battery, as it gives a more accurate result. This is because a potentiometer uses the null method to measure emf and it hardly draws any current from the primary circuit.

When a voltmeter is used in the circuit, its equivalent resistance is connected parallel to some element of the circuit. This changes the overall current in the circuit and, hence, the potential difference to be measured also changes. The error can be minimised if the equivalent resistance of the voltmeter is increased. However, we also need to keep in mind the heat dissipated due to high resistance while deciding the value of resistance of the voltmeter.

Hence a potentiometer is preferred.

#### Page No 196:

#### Answer:

No, a conductor does not become charged when a current is passed through it. The free electrons present in the valence shell in a circuit drift from a lower potential to a higher potential and, thus, current is produced. A battery does not provide any extra electrons or charge to the circuit. It just provides a potential difference across two points, which helps in creating an electric field. This further helps in moving the electrons along the conductor.

#### Page No 196:

#### Answer:

The potential difference across a battery cannot be greater than its emf. Basically, emf is the maximum potential difference between the terminals of a battery when the terminals are not connected externally to an electric circuit. When the same battery is connected to an electric circuit, current flows in the closed circuit. When current flows, the potential difference across the terminals of the battery is decreased as some potential drop due to its internal resistance.

Due to the internal resistance in the battery, the potential difference across it is less than its emf. However, for an ideal battery, potential difference and emf are equal.

#### Page No 196:

#### Answer:

(a) increase

If the number of collisions of the free electrons with the lattice is decreased, then the drift velocity of the electrons increases.

Current *i* is directly proportional to the drift velocity '*V*_{d}' and is given by the following relation:

*i = neAV _{d}_{ ,}* where '

*n*' is the number density of electrons and '

*A'*is the area of the cross-section of the conductor.

So, we can easily see that current increases with increase in drift velocity.

#### Page No 196:

#### Answer:

(d) The information is not sufficient to find the relation between ρ_{A}_{ }and ρ_{B}.

The resistance *R* of a conductor depends on its resistivity *ρ*, length *l* and cross-sectional area *A*. Thus,

$R=\rho \frac{l}{A}$

From the given comparison of resistances, we cannot derive the correct relation between the resistances. We also need to know the cross-sectional areas and the lengths of both the conductors before concluding about their resistivities. Only then can the relation between the resistivities be found.

The information is not sufficient to find the relation between ρ_{A}_{ }and ρ_{B}.

Hence, the correct option is (d)

#### Page No 196:

#### Answer:

(d) None of these

The relation between the resistivity *ρ* and conductivity *σ* of a material is given by

$\rho =\frac{1}{\sigma}\phantom{\rule{0ex}{0ex}}\Rightarrow \rho \times \sigma =1$

The product of conductivity and resistivity is unity. So, this product does not depend on any of the given quantities.

#### Page No 196:

#### Answer:

(d) may increase or decrease

The product of resistivity and conductivity is independent of temperature. As the temperature of a metallic resistor is increased, the resistivity increases and conductivity decreases. Hence, both the conductivity and resistivity of the metallic resistor nullify the effect of the change in temperature.

#### Page No 196:

#### Answer:

(b) may go from the positive terminal to the negative terminal

In the study of electric current, the direction opposite the flow of electrons is regarded as the direction of flow of positive charge. In a battery, positive charge flows from the negative terminal to the positive terminal when it is discharging (connected to external circuit). But when the battery is charged, the positive charge flows from the positive terminal to the negative terminal.

#### Page No 196:

#### Answer:

(a) increase

As the resistance is connected to an ideal battery, it provides a constant potential difference across the two terminals. The internal resistance of the battery is also zero.

The power dissipated in the resistor,

*P* = $\frac{{V}^{2}}{R}$

*V* is constant; hence V $\propto \frac{1}{{R}^{}}$

Thus, if the value of the resistance is decreased, the power dissipated in the resistor will increase.

#### Page No 196:

#### Answer:

(c) *K*_{1} > *K*_{2}_{ }

The metal ions are bound at their positions and vibrate due to collisions with electrons and due to thermal energy. The conduction electrons are free to move. They get energy from the electric field set inside the conductor by connection with a battery and due to thermal motion. The velocity of the electrons is high.

Thus, the kinetic energy of the electrons is greater than the kinetic energy of the metal ions.

#### Page No 196:

#### Answer:

(a) 1 : 2

Thermal energy developed across a resistor,

*U = i ^{2}Rt* ,

where

*i*is the current flowing through the resistor of resistance

*R*for time

*t.*Since the resistors are connected in series, the current flowing through both the resistors is same and the time for which the current flows is also same.

Thus, the ratio of the thermal energy developed in

*R*and

*2 R*is 1 : 2.

#### Page No 196:

#### Answer:

(b) 2 : 1

Thermal energy developed in the resistances,

*H *=$\frac{{V}^{2}}{R}t$

Heat developed in the resistance *R*, *H*_{1}=$\frac{{V}^{2}}{R}t$

Heat developed in the resistance 2*R**,* *H*_{2} = $\frac{{V}^{2}}{2R}t$

Thus, Heat developed in the resistance *R *and 2*R* are in ratio 2:1.

#### Page No 196:

#### Answer:

(a) 2 Ω

Resistance of a wire is directly proportional to its length.

So, when we cut the wire into 5 equal parts, the resistance of each part will be 10 Ω.

On connecting these wires in parallel, the net resistance will be

$\frac{1}{R}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow R=2\mathrm{\Omega}$

#### Page No 197:

#### Answer:

(a) *A *and *B* are correct.

According to Kirchhoff's junction law, the net charge coming towards a point should be equal to the net charge going away from that point in the same time. It follows from the principle of conservation of charge.

The loop law follows from the fact that electrostatic force is a conservative force and the work done by it in any closed path is zero.

#### Page No 197:

#### Answer:

(b)* A* is correct but *B* is wrong.

Let the emfs of the batteries be *e*_{1} and *e*_{2}, and their respective resistances be *r*_{1} and *r*_{2}.

Since the batteries are connected in series, the equivalent emf will be the sum of the emf of the two batteries ( *e = **e*_{1} + *e*_{2}) .

Thus*, e > **e*_{1} and *e > **e*_{2}

Thus, the equivalent emf is larger than either of the two emfs. Hence, statement A is correct.

Since the batteries are connected in series, the equivalent internal resistance (*r*) of the combination will be the sum of the internal resistance of the two batteries ( *r = **r*_{1} + *r*_{2}).

*r > **r*_{1} and *r >* *r*_{2}

Thus, the equivalent internal resistance is greater that either of the two resistances. Hence, statement B is wrong.

#### Page No 197:

#### Answer:

(c) *B* is correct but *A *is wrong.

The equivalent emf ${\epsilon}_{0}$ , when two non-ideal batteries of emfs ${\epsilon}_{1}$ and ${\epsilon}_{2}$ and internal resistances *r _{1}* and

*r*respectively are connected in parallel to each other, is given by:

_{2}${\epsilon}_{0}=\frac{{\epsilon}_{1}{r}_{1}+{\epsilon}_{2}{r}_{2}}{{r}_{1}+{r}_{2}}$

Thus, the equivalent emf is greater than either of the emfs. Thus, statement A is wrong.

For the parallel combination of the batteries of internal resistance

*r*and

_{1}*r*, the equivalent internal resistance

_{2}*r*is given as

$r=\frac{{r}_{1}{r}_{2}}{{r}_{1}+{r}_{2}}$

Thus, the value of the resultant resistance is even smaller than either resistance.

#### Page No 197:

#### Answer:

(d) it does not appreciably change the current to be measured

The ammeter is connected in series with the circuit to measure the net amount of current flowing through the circuit. If the net resistance of the ammeter is high, then the amount of current that the circuit draws will have error. Hence, we cannot accurately measure the amount of current that the circuit draws from the voltage source. Hence, the net resistance or the equivalent resistance of an ammeter should be small enough to ensure that it does not appreciably change the current to be measured.

#### Page No 197:

#### Answer:

(d) it does not appreciably change the potential difference to be measured

To measure potential difference across any circuit element, the voltmeter is connected in parallel to that circuit element. Let *R _{eq}* be the equivalent resistance of the voltmeter and

*V*be the potential difference across the voltmeter.

Then, the current through the voltmeter,

*i*= $\frac{V}{{R}_{eq}}$

Hence, the deflection in the voltmeter is proportional to the current

*i*and, hence, proportional to

*V*. However, when the voltmeter is used in a circuit, its resistance

*R*

_{eq}is connected in parallel to some circuit element. This might change the overall resistance of the circuit and, hence, the current. Consequently, the potential difference to be measured is also changed. To minimise the error due to this, the equivalent resistance

*R*of the voltmeter should be large. When a large resistance is connected in parallel to a small resistance, the equivalent resistance is only slightly less than the smaller one.

_{eq}#### Page No 197:

#### Answer:

(b) *Q*_{1} > *Q*_{2}, *t*_{1} <* **t*_{2}

The charge *Q* on the plates of a capacitor at time *t* after connecting it in a charging circuit,

$Q=\epsilon C(1-{e}^{-t/RC})$,

where

ε = emf of the battery connected in the charging circuit

*C* = capacitance of the given capacitor

*R* = resistance of the resistor connected in series with the capacitor

The charge developed on the plates of the capacitor in first 10 mili seconds is given by

${Q}_{1}=\epsilon C(1-{e}^{-10\times {10}^{-3}/RC})$

The charge developed on the plates of the capacitor in first 20 mili seconds is given by

$Q\text{'}=\epsilon C(1-{e}^{-20\times {10}^{-3}/RC})$

The charge developed at the plates of the capacitor in the interval *t* = 10 mili seconds to 20 mili seconds is given by

${Q}_{2}=Q\text{'}-{Q}_{1}\phantom{\rule{0ex}{0ex}}{Q}_{2}=\left[\epsilon C(1-{e}^{-20\times {10}^{-3}/RC})\right]-\left[\epsilon C(1-{e}^{-10\times {10}^{-3}/RC})\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {Q}_{2}=\left[\epsilon C({e}^{-10\times {10}^{-3}/RC}-{e}^{-20\times {10}^{-3}/RC})\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {Q}_{2}=\left[\epsilon C\left({e}^{-10\times {10}^{-3}/RC}\right)(1-{e}^{-10\times {10}^{-3}/RC})\right]\phantom{\rule{0ex}{0ex}}$

Comparing *Q*_{1}_{ }with *Q*_{2}_{ }

${Q}_{1}=\epsilon C(1-{e}^{-10\times {10}^{-3}/RC}),{Q}_{2}=\epsilon C{e}^{-10\times {10}^{-3}/RC}(1-{e}^{-10\times {10}^{-3}/RC})\phantom{\rule{0ex}{0ex}}\frac{{Q}_{1}}{{Q}_{2}}=\frac{1}{{e}^{-10\times {10}^{-3}/RC}}\phantom{\rule{0ex}{0ex}}{e}^{-10\times {10}^{-3}/RC}1\phantom{\rule{0ex}{0ex}}\therefore {Q}_{1}{Q}_{2}$

For second part of the question

10 μC charge be deposited in a time interval *t*_{1} and the next 10 μC charge is deposited in the next time interval* **t*_{2}.

The time taken for 10 μC charge to develop on the plates of the capacitor is *t*_{1}

$10\mathrm{\mu C}=\epsilon C(1-{e}^{-{t}_{1}/RC})$ ...(1)

The time taken for 20 μC charge to develop on the plates of the capacitor is *t'*

$20\mathrm{\mu C}=\epsilon C(1-{e}^{-{t}_{2}/RC})$ ....(2)

Dividing (2) by (1)

$\frac{20\mathrm{\mu C}}{10\mathrm{\mu C}}=\frac{\epsilon C(1-{e}^{-{t}_{2}/RC})}{\epsilon C(1-{e}^{-{t}_{1}/RC})}\phantom{\rule{0ex}{0ex}}2=\frac{(1-{e}^{-{t}_{2}/RC})}{(1-{e}^{-{t}_{1}/RC})}\phantom{\rule{0ex}{0ex}}2(1-{e}^{-{t}_{1}/RC})=(1-{e}^{-{t}_{2}/RC})\phantom{\rule{0ex}{0ex}}{e}^{-{t}_{2}/RC}=2{e}^{-{t}_{1}/RC}-1$

Taking natural log both side,

$\frac{{t}_{2}}{RC}=\mathrm{ln}\left(2\right)+\frac{{t}_{1}}{RC}$

$\Rightarrow $*t*_{1} <* **t*_{2}

#### Page No 197:

#### Answer:

(a) The speed of the electrons is more at *B* than at *A*.

Let the potentials at A and B be *V*_{A} and *V*_{B}.

As potential, $E=-\frac{dV}{dr}$ ,

potential increases in the direction opposite to the direction of the electric field.

Thus, *V*_{A} < *V*_{B}

Potential energy of the electrons at points A and B:

*U*_{A}_{ }= $-$*e**V*_{A}

*U*_{B}_{ }= $-$*e**V*_{B}

Thus, *U*_{A}_{ }> *U*_{B}

Let the kinetic energy of an electron at points A and B be *K*_{A }and *K*_{B} respectively.

Applying the principle of conservation of mechanical energy, we get:

*U*_{A}_{ }+ *K*_{A} = *U*_{B}_{ }*+ K*_{B}

As, *U*_{A}_{ }> *U*_{B},

*K*_{A} < *K*_{B}

Therefore, the speed of the electrons is more at *B* than at *A*.

#### Page No 197:

#### Answer:

(b) There is no current through *B*.

(c) There is a current through *A *as long as the charging is not complete.

As the capacitor is connected to the battery at *t* = 0, current flows through the wire up to the time the capacitor is charged. As the capacitor is completely charged, the potential difference across the capacitor is equal to the terminal potential of the battery. So, the current stops flowing and does not pass through the point between the plates of the capacitor, as there is no medium for the flow of charge.

#### Page No 197:

#### Answer:

(c) the average velocity of a free electron over a large period of time is zero

(d) the average of the velocities of all the free electrons at an instant is zero

When no current is passed through a conductor, it means that there is no net charge transfer through it. The free electrons are in random motion. The net charge transfer through the cross-section of the conductor is zero. As there is no net charge transfer at any instant, it means that the average of the velocities of all the free electrons at an instant is zero. If we see the motion of each electron as its motion is random and there is no net transfer of charge. Thus, it can be concluded that the average velocity of a free electron over a large period of time is zero.

#### Page No 197:

#### Answer:

(d) Number of free electrons

As the resistor connected to the battery is heated, the thermal energy of the electrons increases. Thus, the relaxation time of the electrons will decrease and so will the drift velocity. Resistivity is inversely proportional to the relaxation time; thus, resistivity will increase with decrease in relaxation time. Therefore, resistance will also change. The number of electrons in a resistor will remain same.

#### Page No 197:

#### Answer:

(a) increases

Resistivity (*ρ*) of a conductor is the reciprocal of the its conductivity(*σ*). Thus,

$\rho =\frac{1}{\sigma}$

The ratio of the resistivity and the conductivity,

$\frac{\rho}{\sigma}=\frac{\rho}{1/\rho}={\rho}^{2}$

As resistivity increases with temperature, the square of resistivity will also increase. Hence, the ratio of resistivity and conductivity will increase with increase in temperature.

#### Page No 197:

#### Answer:

(a) The charge crossing in a given time interval

(d) Free-electron density

Drift speed and current density are inversely proportional to the area of cross-section of a wire. Thus, they are dependent on the cross-section. The charge crossing in a given time interval is independent of the area of cross-section of the wire. Free electron density is the total number of free electrons per unit volume of the wire. The density of free electrons depends on the distribution of the free electrons throughout the volume of the wire. It does not depend on the cross-section of the wire.

#### Page No 197:

#### Answer:

(a) An ammeter should have small resistance.

(d) A voltmeter should have large resistance.

The ammeter is connected in series in the circuit whose current is to be measured. If the net resistance of the ammeter is high, then the amount of current that the circuit draws will have error . So, we won't be able to accurately measure the amount of current that the circuit draws from the voltage source. Thus, it should have a small resistance.

To measure the potential difference across any circuit element, the voltmeter is connected in parallel to that circuit element. If the resistance of the voltmeter is large, then maximum voltage drop occurs across the voltmeter and it will measure the correct value of the potential.

#### Page No 198:

#### Answer:

(a) 5 s

(b) 50 s

(c) 500 s

(d) 500 s

The charge (*Q*) on the capacitor at any instant *t*,

$Q=CV(1-{e}^{-t/RC})$,

where

*C* = capacitance of the given capacitance

*R* = resistance of the resistor connected in series with the capacitor

*RC* = (10 × 10^{3}) × (500 × 10^{$-$6}) = 5 s

The charge on the capacitor in the first 5 seconds,

${Q}_{0}=CV(1-{e}^{-5/5})=CV\times 0.632\phantom{\rule{0ex}{0ex}}$

The charge on the capacitor in the first 10 seconds,

${Q}_{1}=CV(1-{e}^{-10/5})\phantom{\rule{0ex}{0ex}}{Q}_{1}=CV(1-{e}^{-2})=0.864\times CV$

Charge developed in the next 5 seconds,

*Q*' = *Q*_{1} $-$ *Q*_{0}

*Q*' = *CV*(0.864 $-$ 0.632) = 0.232 *CV*

The charge on the capacitor in the first 55 seconds,

${Q}_{2}=CV(1-{e}^{-55/5})\phantom{\rule{0ex}{0ex}}{Q}_{2}=CV(1-{e}^{-11})=0.99\times CV$

Charge developed in the next 50 seconds,

*Q*' = *Q*_{2} $-$ *Q*_{0}

*Q*' = *CV*(0.99 $-$ 0.632) = 0.358 *CV*

Charge developed in the first 505 seconds,

${Q}_{3}=CV(1-{e}^{-500/5})=CV(1-{e}^{-100})\approx CV$

Charge developed in the next 500 seconds,

*Q*' = *CV* (1$-$ 0.632) = 0.368 *CV*

Thus, the charge developed on the capacitor in the first 5 seconds is greater than the charge developed in the next 5,50, 500 seconds.

Disclaimer : Out of the four given options, two options are same.

#### Page No 198:

#### Answer:

(b) The currents in the two discharging circuits at *t* = 0 are equal but not zero.

(d) *C*_{1} loses 50% of its initial charge sooner than *C*_{2} loses 50% of its initial charge.

Let the voltage of the battery connected to the capacitors be *V*. Both the capacitors will charge up to the same potential (*V*).

The charge on the capacitors *C*_{1}_{ }is *C*_{1}*V* = (1 μF)×*V*

The charge on the capacitors *C*_{2}_{ }is *C*_{2}*V** =* (2 μF)×*V*

The charge on the discharging circuit at an instant *t*,

$Q=CV{e}^{-t/RC}$

The current through the discharging circuit,

$\frac{\mathrm{d}Q}{\mathrm{d}t}=-\frac{CV}{RC}{\mathrm{e}}^{-t/RC}=\frac{\mathit{V}}{\mathit{R}}{\mathrm{e}}^{-\mathrm{t}/\mathrm{R}\mathrm{C}}$

At *t* = 0, the current through the discharging circuit will be $\frac{V}{R}$ for both the capacitors.

Let the time taken by the capacitor C_{1} to lose 50% of the charge be *t*_{1}.

${Q}_{1}=\frac{{C}_{1}V}{2}\phantom{\rule{0ex}{0ex}}\frac{{C}_{1}V}{2}={C}_{1}V{e}^{-{t}_{1}/RC}\phantom{\rule{0ex}{0ex}}\frac{1}{2}={e}^{-{t}_{1}/RC}$

Taking natural log on both sides, we get:

$-\mathrm{ln}2=-\frac{{t}_{1}}{R{C}_{1}}\phantom{\rule{0ex}{0ex}}{t}_{1}=R{C}_{1}\mathrm{ln}2$

Similarly,

Time taken for capacitor *C*_{2}: ${t}_{2}=R{C}_{2}\mathrm{ln}2$

As, *C*_{1} < *C*_{2}, *t*_{1} < *t*_{2}

Thus, we can say that *C*_{1} loses 50% of its initial charge sooner than *C*_{2} loses 50% of its initial charge.

#### Page No 198:

#### Answer:

(a) Amount of charge,

*Q*(*t*) =* **At*^{2} + *Bt* +* **C*

We can only add the terms with the same dimensions. So, all the individual terms will have dimensions equal to the dimensions of the charge.

Comparing the dimensions of each term separately, we get:

$A{t}^{2}=Q\phantom{\rule{0ex}{0ex}}\Rightarrow A=\frac{Q}{{t}^{2}}\phantom{\rule{0ex}{0ex}}Q=It\phantom{\rule{0ex}{0ex}}\Rightarrow A=\frac{I}{t}=\left[{\mathrm{AT}}^{-1}\right],$

where *I* = current through the wire

Now,

$Bt=Q\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{Q}{t}=I=\left[\mathrm{A}\right]$

Also,

$C=Q\phantom{\rule{0ex}{0ex}}\Rightarrow C=\left[\mathrm{AT}\right]$

(b) Current *I* = rate of flow of charge

$\Rightarrow I=\frac{\mathrm{d}Q}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(A{t}^{2}+Bt+C\right)\phantom{\rule{0ex}{0ex}}\Rightarrow I=2At+B\phantom{\rule{0ex}{0ex}}A=5,B=3\mathrm{and}t=5\mathrm{s}\phantom{\rule{0ex}{0ex}}\Rightarrow I=2\times 5\times 5+3=53\mathrm{A}$

#### Page No 198:

#### Answer:

Rate of emission of electrons = 2 ×10^{16} per second

Total charge passing through the gun = 2 ×10^{16} ×1.6 × 10^{–19 }= 3.2 × 10^{–3} per second

We know:

Current*, I* = rate of flow of charge

∴ *I* = 3.2 × 10^{–3} A

#### Page No 198:

#### Answer:

Let *Q* amount of charge be transferred across the cross-section of the tube.

Given:

Current through the discharge tube, *I* = 2 μA = 2 × 10^{–6} A

Time taken to transfer charge, *t *= 5 mins = 300 s

We know:

*Q* = *It*

∴ *Q *= 2 × 10^{–6} × 300 = 6× 10^{–4} C

#### Page No 198:

#### Answer:

Current through the wire varies with time as

*i** *= *i*_{0}_{ }+ *α t*

Time interval, *t* = 10 s

*i*_{0} = 10 A

*α* = 4 A/sec

We know that for variable current,

$q={\int}_{0}^{t}i\mathrm{d}t\phantom{\rule{0ex}{0ex}}\Rightarrow q={\int}_{0}^{t}\left({i}_{0}+\alpha t\right)\mathrm{d}t\phantom{\rule{0ex}{0ex}}\Rightarrow q={i}_{0}t+\frac{\alpha {t}^{2}}{2}\phantom{\rule{0ex}{0ex}}\therefore q=10\times 10+4\times \frac{{\left(10\right)}^{2}}{2}\phantom{\rule{0ex}{0ex}}=100+200\phantom{\rule{0ex}{0ex}}=300\mathrm{C}$

#### Page No 198:

#### Answer:

Given:

Current*, i *= 1 A

Area of cross-section, *A* = 1 mm^{2} = 1 × 10^{–6} m^{2}

Density of copper, $\rho $ = 9000 kg/m^{3}

Length of the conductor = *l*

Also,

Mass of copper wire = Volume × density

$\Rightarrow m\mathit{=}A\times l\times \rho \phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}m\mathit{=}A\times l\times 9000\mathrm{kg}$

We know that the number of atoms in molecular mass *M* = N_{A}

∴ Number of atoms in mass *m*, *N* = $\left(\frac{{\mathrm{N}}_{\mathrm{A}}}{M}\right)m$,

where N_{A }is known as Avagadro's number and is equal to 6 × 10^{23} atoms.

*$\Rightarrow N=\left(\frac{{\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}}\right)m\phantom{\rule{0ex}{0ex}}\Rightarrow N=\left(\frac{{\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}}\right)\times A\times l\times 9000\phantom{\rule{0ex}{0ex}}$*

Also, it is given that

No. of free electrons = No. of atoms

Let *n* be the number of free electrons per unit volume

$n=\frac{\mathrm{Number}\mathrm{of}\mathrm{electrons}}{\mathrm{Volume}}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{N}}_{\mathrm{A}}\times A\times l\times 9000}{M\times A\times l}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{N}}_{\mathrm{A}}\times 9000}{M}\phantom{\rule{0ex}{0ex}}=\frac{6\times {10}^{23}\times 9000}{63.5\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}\therefore i\mathit{=}{V}_{\mathrm{d}}nAe\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{d}=\frac{1}{{\displaystyle \frac{6\times {10}^{23}\times 9000}{63.5\times {10}^{-3}}\times {10}^{-6}\times 1.6\times {10}^{-19}}}\phantom{\rule{0ex}{0ex}}=\frac{63.5\times {10}^{-3}}{6\times {10}^{23}\times 9000\times {10}^{-6}\times 1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}=\frac{63.5\times {10}^{-3}}{6\times {10}^{26}\times 9\times {10}^{-6}\times 1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}=\frac{63.5\times {10}^{-3}}{6\times 9\times 16}\phantom{\rule{0ex}{0ex}}=0.073\times {10}^{-3}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}=0.073\mathrm{mm}/\mathrm{s}$

#### Page No 198:

#### Answer:

Let *ρ* be the resistivity of the wire.

Given:

Length*, l* = 1 m

Radius*, r *= 0.1 × 10^{–3} m

Area, *A* = $\mathrm{\pi}{r}^{2}$

Resistance, *R* = 1000 Ω

We know:

$R=\rho \frac{l}{A}\phantom{\rule{0ex}{0ex}}\Rightarrow \rho =\frac{RA}{l}\phantom{\rule{0ex}{0ex}}=\frac{1000\times \mathrm{\pi}\times 0.1\times 0.1\times {10}^{-6}}{1}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\times {10}^{-6}\mathrm{\Omega m}$

#### Page No 198:

#### Answer:

Let

Resistivity of the wire = *ρ*

Original length of the wire = *l*

New length of the wire = *l'*

Original area of the wire = *A*

New area of the wire = *A' *

Original resistance of the wire = *R* = 100 Ω

New resistance of the wire = *R'*

Given:

*l'* = 2*l*

The volume of the wire remains constant. So,

$Al=A\text{'}l\text{'}\phantom{\rule{0ex}{0ex}}\Rightarrow A\text{'}=\frac{A}{2}$

We know:

$R=\frac{\rho l}{A}\phantom{\rule{0ex}{0ex}}\Rightarrow R\text{'}=\frac{\rho l\text{'}}{A\text{'}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{R\text{'}}{R}=\frac{l\text{'}A}{lA\text{'}}\phantom{\rule{0ex}{0ex}}\because l\text{'}=2l,A\text{'}=\frac{A}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{R\text{'}}{R}=\frac{\left(2l\right)A}{l\left({\displaystyle \frac{A}{2}}\right)}=4\phantom{\rule{0ex}{0ex}}\Rightarrow R\text{'}=4R=400\mathrm{\Omega}$

#### Page No 198:

#### Answer:

Given:

Current through the wire, *i* = 2 A

Length of the wire*, l *= 4 m

Area of cross section, *A* = 1 mm^{2} = 1 × 10^{–6} m^{2}

Number of electrons per unit volume, *n* = 10^{29}

We know:

$i=nA{V}_{d}e\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{d}=\frac{i}{nAe}\phantom{\rule{0ex}{0ex}}=\frac{2}{{10}^{29}\times {10}^{-6}\times 1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{\mathrm{d}}=\frac{1}{8000}\mathrm{m}/\mathrm{s},$

where *V _{d}* is the drift speed.

Let

*t*be the time taken by an electron to cross the length of the wire.

$\Rightarrow t=\frac{\text{L}\mathrm{ength}\mathrm{of}\text{the}\mathrm{wire}}{\text{D}\mathrm{rift}\mathrm{speed}}\phantom{\rule{0ex}{0ex}}=\frac{l}{{V}_{\mathrm{d}}}\phantom{\rule{0ex}{0ex}}\therefore t=\frac{4}{{\displaystyle \frac{1}{8000}}}\phantom{\rule{0ex}{0ex}}=32\times {10}^{3}\mathrm{s}$

#### Page No 198:

#### Answer:

Given:

Resistivity of copper, *ρ*_{cu}* _{ }*= 1.7 × 10

^{–8}Ωm

Area of cross-section,

*A*= 0.01 mm

^{2}

^{ }= 0.01 × 10

^{–6}m

^{2}

Required resistance

*, R*= 1 kΩ = 10

^{3}Ω

Let

*l*be the required length of the copper wire.

We know:

$R=\frac{\rho l}{A}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\frac{RA}{{\rho}_{\mathrm{cu}}}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\frac{{10}^{3}\times 0.01\times {10}^{-6}}{1.7\times {10}^{-8}}\phantom{\rule{0ex}{0ex}}\therefore l=0.58\times {10}^{3}\mathrm{m}=0.6\mathrm{km}$

#### Page No 198:

#### Answer:

Let us consider a small element strip of length *dx* at a distance *x* from one end, as shown below.

Let the resistance of the small element strip be *dR*. Let the radius at that point be *c* .

Then, the resistance of this small strip,

$\mathrm{d}R=\frac{\rho \mathrm{d}x}{\mathrm{\pi}{c}^{2}}...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{tan\theta}=\frac{c-\mathrm{a}}{x}=\frac{\mathrm{b}-\mathrm{a}}{L}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{c-\mathrm{a}}{x}=\frac{\mathrm{b}-\mathrm{a}}{L}\phantom{\rule{0ex}{0ex}}\Rightarrow L\times \left(c-\mathrm{a}\right)=x\times \left(\mathrm{b}-\mathrm{a}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow Lc-\mathrm{La}=xb-xa\phantom{\rule{0ex}{0ex}}$

Differentiating w.r.t to *x*, we get:

$\mathrm{L}\frac{\mathrm{d}c}{\mathrm{d}x}-0=\mathrm{b}-\mathrm{a}\phantom{\rule{0ex}{0ex}}dx=\frac{Ldc}{\left(b\mathit{-}a\right)}...\left(\mathrm{ii}\right)$

Substituting the value of *dx *in equation (i), we get:

$\mathrm{d}R=\frac{\rho L\mathrm{d}c}{\mathrm{\pi}{c}^{2}\left(b-a\right)}\phantom{\rule{0ex}{0ex}}\mathrm{d}R=\frac{\rho L}{\mathrm{\pi}\left(b-a\right)}\xad\xb7\frac{\mathrm{d}c}{{c}^{2}}\phantom{\rule{0ex}{0ex}}$

Integrating *dR* from *a* to *b*, we get:

${\int}_{0}^{R}dR=\frac{\rho L}{\mathrm{\pi}\left(\mathit{b}\mathit{-}\mathit{a}\right)}{\int}_{a}^{b}\frac{\mathrm{d}c}{{c}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{\rho L}{\mathrm{\pi}\left(\mathit{b}\mathit{-}\mathit{a}\right)}{\left[\frac{-1}{c}\right]}_{a}^{b}\phantom{\rule{0ex}{0ex}}=\frac{\rho L}{\mathrm{\pi}\left(\mathit{b}\mathit{-}\mathit{a}\right)}\left(\frac{-1}{\mathrm{b}}-\frac{-1}{\mathrm{a}}\right)\phantom{\rule{0ex}{0ex}}=\frac{\rho L}{\mathrm{\pi ab}}$

#### Page No 198:

#### Answer:

Given:

Radius of the wire, *r = *0.1 mm *= *10^{-4} m

Resistance, *R* = 1 kΩ = 10^{3} Ω

Voltage across the ends of the wire,* V* = 20* *V

(a) Let *q* be the charge transferred per second and *n* be the number of electrons transferred per second.

We know:

$i=\frac{V}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow i=\frac{20\mathrm{V}}{{10}^{3}\mathrm{\Omega}}\phantom{\rule{0ex}{0ex}}\Rightarrow i=20\times {10}^{-3}=2\times {10}^{-2}\mathrm{A}\phantom{\rule{0ex}{0ex}}q\mathit{=}it\phantom{\rule{0ex}{0ex}}\Rightarrow q=2\times {10}^{-2}\times 1\phantom{\rule{0ex}{0ex}}\Rightarrow q=2\times {10}^{-2}\mathrm{C}$

Also,

*q* = *ne*

$\Rightarrow n=\frac{q}{e}=\frac{2\times {10}^{-2}}{1.6\times {10}^{-19}}\phantom{\rule{0ex}{0ex}}\Rightarrow n=1.25\times {10}^{17}$

(b) Current density of a wire,

$j=\frac{i}{A}\phantom{\rule{0ex}{0ex}}\Rightarrow j=\frac{2\times {10}^{-2}}{3.14\times {10}^{-8}}\phantom{\rule{0ex}{0ex}}\Rightarrow j=6.37\times {10}^{5}\mathrm{A}/{\mathrm{m}}^{2}$

#### Page No 198:

#### Answer:

Given:

Area of cross-section*, A* = 2 × 10^{–6 }m^{2}

Current* *through the wire*, i* = 1 A

Resistivity of copper, *ρ *= 1.7 × 10^{–8} Ωm

Resistance of a wire,

*$R=\rho \frac{l}{A}$*

Also from Ohm's Law, voltage across a wire,

$V=iR=\frac{i\rho l}{A}$

The electric field of the wire,

$E=\frac{V}{l}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{i\rho l}{Al}=\frac{i\rho}{A}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{1\times 1.7\times {10}^{-8}}{2\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=8.5\mathrm{V}/\mathrm{m}$

#### Page No 198:

#### Answer:

Given:

Length of the wire, *l *= 2 m

Resistance of the wire, *R* = 5 Ω

Current through the wire, *i* = 10 A

From Ohm's Law, the potential difference across the wire,

*V* = *iR** *

*∴ V*= 10 × 5 = 50 V

Electric field,

$\mathrm{E}=\frac{\mathrm{V}}{l}\phantom{\rule{0ex}{0ex}}=\frac{50}{2}\phantom{\rule{0ex}{0ex}}=25\mathrm{V}/\mathrm{m}$

#### Page No 198:

#### Answer:

Given:

Resistance *R*_{Fe} of the iron wire at 20°C = 3.9 Ω

Resistance *R*_{Cu} of the copper wire at 20°C = 4.1 Ω

Temperature coefficient *α*_{Fe} for iron = 5.0 × 10^{–3} K^{–1}

Temperature coefficient *α*_{Cu} for copper = 4.0 × 10^{–3} K^{–1}

Let

The temperature at which the resistance be equal = T

Resistance of iron wire at T °C = *R*_{Fe}'

Resistance of copper wire at T °C = *R*_{Cu}'

We know:

$R={R}_{0}(1+\alpha \u2206T)$

Here, Δ*T** *= *T - 20*

$\Rightarrow {R}_{\mathrm{Fe}}\text{'}={R}_{\mathrm{Fe}}\left[1+{\alpha}_{\mathrm{Fe}}\left(T-20\right)\right]\phantom{\rule{0ex}{0ex}}{R}_{\mathrm{Cu}}\text{'}={R}_{\mathrm{Cu}}\left[1+{\alpha}_{\mathrm{Cu}}\left(T-20\right)\right]\phantom{\rule{0ex}{0ex}}{R}_{\mathrm{Fe}}\mathit{\text{'}}\mathit{=}{R}_{\mathrm{Cu}}\text{'}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}{R}_{\mathrm{Fe}}\left[1+{\alpha}_{\mathrm{Fe}}\left(T-20\right)\right]={R}_{\mathrm{Cu}}\left[1+{\alpha}_{\mathrm{Cu}}\left(T-20\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 3.9\left[1+5\times {10}^{-3}\left(T-20\right)\right]=4.1\left[1+4\times {10}^{-3}\left(T-20\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 3.9+3.9\times 5\times {10}^{-3}\left(T-20\right)=4.1+4.1\times 4\times {10}^{-3}\left(T-20\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 4.1\times 4\times {10}^{-3}\left(T-20\right)-3.9\times 5\times {10}^{-3}\left(T-20\right)=-4.1+3.9\phantom{\rule{0ex}{0ex}}\Rightarrow 16.4\left(T-20\right)-19.5\left(T-20\right)=-0.2\times {10}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(T-20\right)\left(-3.1\right)=-0.2\times {10}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow T-20=64.5\phantom{\rule{0ex}{0ex}}\Rightarrow T=84.5\xb0\mathrm{C}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 198:

#### Answer:

Let the magnitude of zero error in the voltmeter reading be *V.*

We need to subtract the zero error from the readings obtained under the two given conditions to obtain the true value of potential difference.

Under both the conditions ,the resistance of the wire will not change.

$\Rightarrow {R}_{1}={R}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{V}_{1}}{{I}_{1}}=\frac{{V}_{2}}{{I}_{2}}$

$\frac{{I}_{1}R}{{I}_{2}R}=\frac{{V}_{1}}{{V}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1.75}{2.75}=\frac{14.4-V}{22.4-V}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{0.35}{0.55}=\frac{14.4-V}{22.4-V}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{7}{11}=\frac{14.4-V}{22.4-V}\phantom{\rule{0ex}{0ex}}\Rightarrow 7\times \left(22.4-V\right)=11\left(14.4-V\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 156.8-7V=158.4-11V\phantom{\rule{0ex}{0ex}}\Rightarrow \left(7-11\right)V=156.8-158.4\phantom{\rule{0ex}{0ex}}\Rightarrow -4V=-1.6\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{V}=0.4\mathrm{V}$

Magnitude of zero error, V = 0.4 V, which can either be positive or negative. Positive or negative zero error just indicates that the needle of the voltmeter is to the right or left of the zero marked on the device if zero voltage is applied across the voltmeter.

#### Page No 198:

#### Answer:

(a)

When the switch *S* is opened, loop 2 will be open and no current will pass through the ammeter. Loop 1 will be closed; so, a current will flow through it. But since the voltmeter has very high resistance, compared to the internal resistance of the battery, the voltage-drop across the internal resistance can be ignored, compared to the voltage drop across the voltmeter. So, the voltage appearing across the voltmeter will be almost equal to the emf of the battery.

∴ *ε* = 1.52 V

(b) When the switch is closed, current will pass through the circuit in loop 2. In this case, there will be a voltage drop across *r* due to current *i* flowing through it.

Applying the loop rule, we get:

*ε – **ir* = 1.45

⇒ 1.52 *– **ir* = 1.45

⇒ *ir* = 0.07

⇒ 1.r = 0.07

r = 0.07 Ω

#### Page No 198:

#### Answer:

Given:

Emf of the battery, *E* = 6 V

Internal resistance, *r* = 1 Ω

Potential difference*, V* = 5.8 V

Let *R* be the resistance of the external resistor.

Applying KVL in the above circuit, we get:

*$E\mathit{-}i\mathit{(}R\mathit{+}r\mathit{)}=0\phantom{\rule{0ex}{0ex}}\Rightarrow i\mathit{=}\frac{E}{R\mathit{+}r}\phantom{\rule{0ex}{0ex}}\mathit{=}\frac{6}{R\mathit{+}1}$*

Also,

$V\mathit{=}E\mathit{-}ir\phantom{\rule{0ex}{0ex}}\Rightarrow 5.8=6-\frac{6}{R+1}\times 1\phantom{\rule{0ex}{0ex}}\frac{6}{R+1}=0.2\phantom{\rule{0ex}{0ex}}R+1=\frac{6}{0.2}=30\phantom{\rule{0ex}{0ex}}R=29\mathrm{\Omega}$

#### Page No 199:

#### Answer:

Given:

Potential difference across the terminals of the battery, *V* = 7.2 V

Emf of the battery, *E* = 6 V

Current flowing through the circuit,* i* = 2 A

As the battery is getting charged, the voltage drop across the terminals of the battery will be equal to the emf of the battery plus the voltage drop across the internal resistance of the battery.

So,* V* = *E* + *ir*

⇒ 7.2 = 6 + 2 × *r*

⇒ 1.2 = 2*r*

⇒ *r* = 0.6 Ω

#### Page No 199:

#### Answer:

(a) **When the battery is being charged:**

Emf of the battery, *E* = 6 V

Internal resistance of the battery, *r* = 10 Ω

Potential difference, *V* = 9 V

Net e.m.f. across the resistance while charging = 9 – 6 = 3 V

$\therefore \mathrm{Curre}\text{nt}=\frac{3}{10}=0.3\mathrm{A}$

(b) **When the battery is completely charged: **

Internal resistance, *r*' = 1 Ω

$\therefore \mathrm{Current}=\frac{3}{1}=3\mathrm{A}$

#### Page No 199:

#### Answer:

(a) For *R* = 0.1 Ω:

Applying KVL in the given circuit, we get:

$0.1{i}_{1}+1{i}_{1}-6+1{i}_{1}-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow 0.1{i}_{1}+1{i}_{1}+1{i}_{1}=12\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{1}=\frac{12}{\left(2.1\right)}=5.71\mathrm{A}\phantom{\rule{0ex}{0ex}}$

Now, consider the given circuit.

Applying KVL in the loop ABCDA, we get:

$0.1{i}_{2}+1i-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow 0.1{i}_{2}+i=6\phantom{\rule{0ex}{0ex}}\Rightarrow i=6-0.1{i}_{2}$

Applying KVL in ADEFA, we get:

$i-6+6-\left({i}_{2}-i\right)1=0\phantom{\rule{0ex}{0ex}}\Rightarrow i-{i}_{2}+i=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\mathrm{i}-{i}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left[6-0.1{i}_{2}\right]-{i}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{2}=10\mathrm{A}$

$\therefore \frac{{i}_{1}}{{i}_{2}}=0.571$

(b) For *R* = 1 Ω:

Applying KVL in the circuit given in figure 1, we get:

$1{i}_{1}+1.{i}_{1}-6+{i}_{1}-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3{i}_{1}=12\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{1}=4\phantom{\rule{0ex}{0ex}}$

Now, for figure 2:

Applying KVL in ABCDA, we get:

${i}_{2}+i-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{2}+i=6$

Applying KVL in ADEFA, we get:

$i-6+6-\left({i}_{2}-i\right)1=0\phantom{\rule{0ex}{0ex}}\Rightarrow i-{i}_{2}+i=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2i-{i}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left[6-{i}_{2}\right]-{i}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 12-3{i}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{2}=4\mathrm{A}\phantom{\rule{0ex}{0ex}}\therefore \frac{{i}_{1}}{{i}_{2}}=1$

(c) For R = 10 Ω:

Applying KVL in the circuit given in figure 1, we get:

$10{i}_{1}+1{i}_{1}-6+1{i}_{1}-6=0$

$\Rightarrow 12{i}_{1}=12\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{1}=1\phantom{\rule{0ex}{0ex}}$

Now, for figure 2:

Applying KVL in ABCDA, we get:

$10{i}_{2}+i-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow i=6-10{i}_{2}$

Applying KVL in ADEFA, we get:

$i-6+6-\left({i}_{2}-i\right)1=0\phantom{\rule{0ex}{0ex}}\Rightarrow i-{i}_{2}+i=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2i-{i}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left[6-10{i}_{2}\right]-{i}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 12-21{i}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{2}=0.57\mathrm{A}\phantom{\rule{0ex}{0ex}}\therefore \frac{{i}_{1}}{{i}_{2}}=1.75$

#### Page No 199:

#### Answer:

(a)

Given:

Emf of one cell = *E*

∴ Total e.m.f. of *n*_{1} cells in one row = *n*_{1}*E*

Total emf of one row will be equal to the net emf across all the *n _{2}* rows because of parallel connection.

Total resistance in one row =

*n*

_{1}

*r*

Total resistance of

*n*rows in parallel $=\frac{{n}_{1}r}{{n}_{2}}$

_{2}_{ }Net resistance of the circuit =

*R*+ $\frac{{n}_{1}r}{{n}_{2}}$

$\therefore \mathrm{Current},I=\frac{{n}_{1}E}{R+{\displaystyle \frac{{n}_{1}r}{{n}_{2}}}}=\frac{{n}_{1}{n}_{2}\mathrm{E}}{{n}_{2}\mathrm{R}+{n}_{1}r}$

(b) From (a),

$I=\frac{{n}_{1}{n}_{2}\mathrm{E}}{{n}_{2}\mathrm{R}+{n}_{1}r}$

For

*I*to be

*maximum,*(

*n*

_{1}

*r*

*+*

*n*

_{2}R) should be minimum

$\Rightarrow {\left(\sqrt{{n}_{1}r}-\sqrt{{n}_{2}\mathrm{R}}\right)}^{2}+2\sqrt{{n}_{1}\mathrm{R}{n}_{2}r}=\mathrm{min}$

It is minimum when

$\sqrt{{n}_{1}r}=\sqrt{{n}_{2}\mathrm{R}}\phantom{\rule{0ex}{0ex}}{n}_{1}r={n}_{2}\mathrm{R}$

*∴ I*is maximum when

*n*

_{1}

*r*

*=*

*n*

_{2}R .

#### Page No 199:

#### Answer:

Given:

Emf of the battery, *E* = 100 volt

Resistance in series with battery, *R' *= 10 kΩ = 10000 Ω

External resistance, *R* = (1-100) Ω

When no external resistor is present (*R* = 0), current through the circuit,

$i=\frac{\mathrm{E}}{\mathrm{R}\mathit{\text{'}}}=\frac{100}{1000}=0.01\mathrm{Amp}.\phantom{\rule{0ex}{0ex}}$

When *R* = 1 Ω,

$i=\frac{100}{10000+1}=\frac{100}{10001}\phantom{\rule{0ex}{0ex}}=0.009999\mathrm{A}$

When *R* = 100 Ω,

$i=\frac{100}{10000+100}\phantom{\rule{0ex}{0ex}}=\frac{100}{10100}=0.009900\mathrm{A}$

We can see that up to *R* = 100 Ω, the current does not change up to two significant digits.

#### Page No 199:

#### Answer:

Given*: i _{1}* = 2.4 A

Since the 20 Ω and 30 Ω resistors are connected in parallel, the voltage across them will be the same.

From the figure,

$20{i}_{1}=30\left(i-{i}_{1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 50{i}_{1}=30i\phantom{\rule{0ex}{0ex}}\Rightarrow i=\frac{50{i}_{1}}{30}=3.99=4\mathrm{A}\phantom{\rule{0ex}{0ex}}$

∴ The reading of ammeter A3 =

*i*= 4 A

The reading of ammeter A2 =

*i*-

*i*= 1.6 A

_{1}#### Page No 199:

#### Answer:

Current will be minimum when the rheostat will have maximum resistance, i.e 30 Ω.

In this case, total resistance of the circuit = $\frac{20\times 10}{30}+30=36.67\mathrm{\Omega}$

The minimum current through the circuit,

${i}_{\mathrm{min}}=\frac{5.5}{36.67}=0.15\mathrm{A}$

Current will be maximum when the rheostat will have minimum resistance, i.e. 0 Ω.

In this case, total resistance of the circuit = $\frac{20\times 10}{30}=6.67\mathrm{\Omega}$

The maximum current through the circuit,

${i}_{\mathrm{max}}=\frac{5.5}{6.67}=0.825\mathrm{A}$

#### Page No 199:

#### Answer:

(a) When all the bulbs are switched on, we have a combination of three bulbs in parallel.

The effective resistance of the circuit,

${R}_{\mathrm{eff}}=\frac{180}{3}=60\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

The current delivered by the battery,

$i=\frac{60}{60}=1\mathrm{A}$

(b) When two of the bulbs are switched on, they are parallel to each other.

The effective resistance of the circuit,

${R}_{\mathrm{eff}}=\frac{180}{2}=90\mathrm{\Omega}$

The current delivered by the battery,

$i=\frac{60}{90}=0.67\mathrm{A}$

(c) When only one bulb is switched on, the effective resistance of the circuit,

${R}_{\mathrm{eff}}=180\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

The current delivered by the battery,

$i=\frac{60}{180}=0.33\mathrm{A}$

#### Page No 199:

#### Answer:

We can obtain maximum resistance when all the resistors are connected in series.

So, maximum resistance,

${R}_{\mathrm{max}}=\left(20+50+100\right)\mathrm{\Omega}=170\mathrm{\Omega}$

The minimum resistance will be obtained when all the resistors are connected in parallel

So, minimum resistance,

${R}_{\mathrm{min}}=\frac{1}{{\displaystyle \frac{1}{20}}+{\displaystyle \frac{1}{50}}+{\displaystyle \frac{1}{100}}}=\frac{100}{8}=12.5\mathrm{\Omega}$

#### Page No 199:

#### Answer:

We know:

$R=\frac{{V}^{2}}{P}$,

where R is the resistance, *V* is the voltage drop and *P* is the power on which the bulb is operated.

We can calculate the resistance of the bulb for the same *V* and different *P*.

The resistances of the bulb for three different powers are:

${R}_{1}=\frac{{\left(15\right)}^{2}}{5}=45\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}{R}_{2}=\frac{{\left(15\right)}^{2}}{10}=22.5\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}{R}_{3}=\frac{{\left(15\right)}^{2}}{15}=15\mathrm{\Omega}$

When the two resistance are used in parallel, the equivalent resistance will be less than their individual resistance.

∴ The two resistance are 45 Ω and 22.5 Ω.

#### Page No 199:

#### Answer:

Let the current flowing through the resistors be marked as shown in the figure.

Given*: i* = 12 mA

The resistors of 20 kΩ and 10 kΩ are connected in parallel. So, the potential difference across their ends will be same, i.e.

$20{i}_{1}=10\left(i-{i}_{1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 30{i}_{1}=10i\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{1}=\frac{i}{3}=\frac{12}{3}=4\mathrm{mA}$

∴ The current flowing through the 20 kΩ resistor = *i _{1}* = 4 mA

The current flowing through the 10 kΩ resistor =

*i*-

*i*= 8 mA

_{1}From the figure,

The current flowing through the 100 kΩ resistor =

*i*= 12 mA

The equivalent resistance between the points A and B,

${R}_{\mathrm{net}}=\left(5+\frac{20\times 10}{20+10}+100\right)\mathrm{k\Omega}=111.67\mathrm{k\Omega}$

The potential difference

*V*between the points A and B,

$V=i{R}_{\mathrm{net}}\phantom{\rule{0ex}{0ex}}\Rightarrow V=12\times {10}^{-3}\times 111.67\times {10}^{3}=1340\mathrm{V}$

#### Page No 199:

#### Answer:

Let the resistance of the first resistor be *R*.

If *V* is the potential difference across *R,* then the current through it,

$i=5=\frac{V}{R}$

Now, the other resistor of 10 Ω is connected in parallel with *R*.

It is given that the new value of current through the circuit,* i'* = 6 A

The effective resistance of the circuit, *R'* = $\frac{10R}{10+R}$

Since the potential difference is constant,

$iR=i\text{'}R\text{'}\phantom{\rule{0ex}{0ex}}\Rightarrow 5\times R=6\times \frac{10R}{10+R}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(10+R\right)5=60\phantom{\rule{0ex}{0ex}}\Rightarrow 5R=10\phantom{\rule{0ex}{0ex}}\Rightarrow R=2\mathrm{\Omega}$

#### Page No 199:

#### Answer:

The simplified circuit can be drawn as shown below.

Equivalent resistance $=\frac{1}{{\displaystyle \frac{1}{r}}+{\displaystyle \frac{1}{r}}+{\displaystyle \frac{1}{r}}}=\frac{r}{3}$

#### Page No 199:

#### Answer:

(a)

From the figure, it can be seen that between points A and B, the resistance of the first side of the hexagon will be in parallel with the total resistance of the other five sides.

The resistance of the first side can be calculated as shown below.

Resistance of the first leg=$\frac{\text{L}\mathrm{ength}\mathrm{of}\mathrm{the}\mathrm{that}\mathrm{portion}}{\text{T}\mathrm{otal}\mathrm{length}}\times R=\frac{{\displaystyle \frac{1}{6}}}{{\displaystyle \frac{5}{6}}+{\displaystyle \frac{1}{6}}}\times 15=15\times \frac{1}{6}$

Total resistance of the 5 legs=$\frac{{\displaystyle \frac{5}{6}}}{{\displaystyle \frac{1}{6}}+{\displaystyle \frac{5}{6}}}\times 15=\frac{5}{6}\times 15$

∴ The effective resistance between the points A and B,

${R}_{\mathrm{eff}}=\frac{{\displaystyle \frac{15\times 5}{6}}\times {\displaystyle \frac{15}{6}}}{{\displaystyle \frac{15\times 5}{6}}+{\displaystyle \frac{15}{6}}}=\frac{{\displaystyle \frac{15\times 5\times 15}{6\times 6}}}{{\displaystyle \frac{75+15}{6}}}\phantom{\rule{0ex}{0ex}}=\frac{15\times 5\times 15}{6\times 90}=\frac{25}{12}\phantom{\rule{0ex}{0ex}}=2.08\mathrm{\Omega}$

(b) From the figure, it can be seen that between points A and C, the resistance of the first two sides of the hexagon will be in parallel with the total resistance of the other four sides.

Resistance of the two legs =$\frac{\text{L}\mathrm{ength}\mathrm{of}\mathrm{the}\mathrm{that}\mathrm{portion}}{\text{T}\mathrm{otal}\mathrm{length}}\times R=\frac{{\displaystyle \frac{2}{6}}}{{\displaystyle \frac{4}{6}}+{\displaystyle \frac{2}{6}}}\times 15=15\times \frac{2}{6}$

Total resistance of the four legs=$\frac{{\displaystyle \frac{4}{6}}}{{\displaystyle \frac{2}{6}}+{\displaystyle \frac{4}{6}}}\times 15=\frac{4}{6}\times 15$

∴ The effective resistance between the points A and C,

${\mathrm{R}}_{eff}=\frac{{\displaystyle \frac{15\times 4}{6}}\times {\displaystyle \frac{15\times 2}{6}}}{{\displaystyle \frac{15\times 4}{6}}+{\displaystyle \frac{15\times 2}{6}}}=\frac{{\displaystyle \frac{15\times 4\times 2\times 15}{6\times 6}}}{{\displaystyle \frac{60+30}{6}}}\phantom{\rule{0ex}{0ex}}=\frac{15\times 2\times 4\times 15}{6\times 90}\phantom{\rule{0ex}{0ex}}=\frac{10}{3}=3.33\mathrm{\Omega}$

(c) From the figure, it can be seen that between points A and D, the resistance of the first three sides of the hexagon will be in parallel with the total resistance of the other three sides.

Resistance of the three legs = $\frac{\text{L}\mathrm{ength}\mathrm{of}\mathrm{the}\mathrm{that}\mathrm{portion}}{\text{T}\mathrm{otal}\mathrm{length}}\times R=\frac{{\displaystyle \frac{3}{6}}}{{\displaystyle \frac{3}{6}}+{\displaystyle \frac{3}{6}}}\times 15=15\times \frac{3}{6}$

∴ The effective resistance between the points A and D,

${\mathrm{R}}_{\mathrm{eff}}=\frac{{\displaystyle \frac{15\times 3}{6}}\times {\displaystyle \frac{15\times 3}{6}}}{{\displaystyle \frac{15\times 3}{6}}+{\displaystyle \frac{15\times 3}{6}}}=\frac{{\displaystyle \frac{15\times 3\times 3\times 15}{6\times 6}}}{{\displaystyle \frac{90}{6}}}\phantom{\rule{0ex}{0ex}}=\frac{15\times 3\times 3\times 15}{6\times 90}=\frac{15}{4}=3.75\mathrm{\Omega}$

#### Page No 199:

#### Answer:

(a) When *S* is open, the resistors will be connected in series, as the current will flow through the 20 Ω resistor. The effective resistance,

${\mathrm{R}}_{\mathrm{eff}}=\left(10+20\right)\mathrm{\Omega}=30\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

The current flowing through the circuit,

$i=\left(\frac{3}{30}\right)\mathrm{A}=0.1\mathrm{A}$

(b) When *S* is closed, current will flow through the switch that offers the least resistive path. So, the 20 Ω resistance will be ineffective in this case. The effective resistance in this case,

${R}_{\mathrm{eff}}=10\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

The current flowing through the circuit,

$i=\left(\frac{3}{10}\right)\mathrm{A}=0.3\mathrm{A}$

#### Page No 200:

#### Answer:

From the figure, we can see that current will flow only in loop 1 because current follows the least resistive path. All the current will pass through the wire connected in parallel to the 4 Ω resistor in loop 2.

∴ There will be no current through the 4 Ω resistor in loop 2.

For loop1, applying KVL, we get:

$6i+4i-4+2=0\phantom{\rule{0ex}{0ex}}\Rightarrow 10i=2\phantom{\rule{0ex}{0ex}}\Rightarrow i=0.2\mathrm{A}\phantom{\rule{0ex}{0ex}}$

∴ The current through the 4 Ω and 6 Ω resistors, *i* = 0.2 A

#### Page No 200:

#### Answer:

Let the potential at the point *o* be *X* volts.

From the figure,

${i}_{1}=\frac{{V}_{\mathrm{a}}-{V}_{\mathrm{o}}}{10}\phantom{\rule{0ex}{0ex}}{V}_{\mathrm{a}}=30\mathrm{V}\mathrm{and}{V}_{\mathrm{o}}=X\phantom{\rule{0ex}{0ex}}So,{i}_{1}=\frac{30-X}{10}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\phantom{\rule{0ex}{0ex}}{i}_{2}=\frac{{V}_{\mathrm{o}}-{V}_{\mathrm{b}}}{20}\phantom{\rule{0ex}{0ex}}=\frac{X-12}{20}\phantom{\rule{0ex}{0ex}}\mathrm{And}\phantom{\rule{0ex}{0ex}}{i}_{3}=\frac{{V}_{\mathrm{o}}-{V}_{\mathrm{c}}}{30}\phantom{\rule{0ex}{0ex}}=\frac{X-2}{30}$

Also, from kirchoff's junction law we have:

*i*_{1} = *i*_{2} + *i*_{3}

$\Rightarrow \frac{30-X}{10}=\frac{X-12}{20}+\frac{X-2}{30}\phantom{\rule{0ex}{0ex}}\Rightarrow 30-X=\frac{X-12}{2}+\frac{X-2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 30-X=\frac{3X-36+2X-4}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow 180-6X=5X-40\phantom{\rule{0ex}{0ex}}\Rightarrow 11X=220\phantom{\rule{0ex}{0ex}}\Rightarrow X=\frac{220}{11}=20\mathrm{V}$

Thus, the currents through the three resistors are:

${i}_{1}=\frac{30-20}{10}=1\mathrm{A}\phantom{\rule{0ex}{0ex}}{i}_{2}=\frac{20-12}{20}=\frac{8}{20}=0.4\mathrm{A}\phantom{\rule{0ex}{0ex}}{i}_{3}=\frac{20-2}{30}=\frac{18}{30}=0.6\mathrm{A}$

#### Page No 200:

#### Answer:

(*a*)

From the figure,

Potential difference between the two ends of resistor *a* = 10 V

Resistance of resistor *a* =10 Ω

∴ Current through *a* = $\frac{10}{10}=1\mathrm{A}$

Since the two batteries are connected opposite each other,

potential difference between the terminals of *b* = 10 – 10 = 0 V

∴ Current through *b* = 0 A

(b)

From the figure,

Potential difference between the two ends of resistor *a* = 10 V.

Resistance of resistor *a* =10 Ω

∴ Current through *a* = $\frac{10}{10}=1\mathrm{A}$

Since the two batteries are again connected opposite to each other,

Potential difference between terminal of *b* = 10 – 10 = 0 V

∴ Current through *b* = 0 A

#### Page No 200:

#### Answer:

Applying KVL in loop 1, we get:

${i}_{1}{R}_{2}-{E}_{2}+\left({i}_{1}+{i}_{2}\right){R}_{3}=0\phantom{\rule{0ex}{0ex}}({R}_{2}+{R}_{3}){i}_{1}+{R}_{3}{i}_{2}={E}_{2}...\left(1\right)$

Applying KVL in loop 2, we get:

${i}_{2}{R}_{1}-{E}_{1}+\left({i}_{1}+{i}_{2}\right){R}_{3}=0\phantom{\rule{0ex}{0ex}}\left({R}_{1}+{R}_{3}\right){i}_{2}+{R}_{3}{i}_{1}={E}_{1}...\left(2\right)$

Multiplying equation (1) by (*R _{1}*+

*R*) and (2) by

_{3}*R*and then subtracting (2) from (1), we get:

_{3}${i}_{1}=\frac{{E}_{2}\left({R}_{1}+{R}_{3}\right)-{E}_{1}{R}_{3}}{\left({R}_{1}{R}_{2}+{R}_{2}{R}_{3}+{R}_{3}{R}_{1}\right)}$

Similarly, multiplying equation (1) by

*R*and (2) by (

_{3}*R*+

_{1}*R*), and then subtracting (2) from (1), we get:

_{3}${i}_{2}=\frac{{E}_{1}\left({R}_{2}+{R}_{3}\right)-{E}_{2}{R}_{3}}{\left({R}_{1}{R}_{2}+{R}_{2}{R}_{3}+{R}_{3}{R}_{1}\right)}$

From the figure,

${V}_{\mathrm{a}}-{V}_{\mathrm{b}}=\left({i}_{1}+{i}_{2}\right){R}_{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{\mathrm{a}}-{V}_{\mathrm{b}}=\left[\frac{{E}_{1}{R}_{2}+{E}_{2}{R}_{1}}{\left({R}_{1}{R}_{2}+{R}_{2}{R}_{3}+{R}_{3}{R}_{1}\right)}\right]{R}_{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{\mathrm{a}}-{V}_{\mathrm{b}}=\frac{{\displaystyle \frac{{E}_{1}}{{R}_{1}}}+{\displaystyle \frac{{E}_{2}}{{R}_{2}}}}{{\displaystyle \frac{1}{{R}_{1}}}+{\displaystyle \frac{1}{{R}_{2}}}+{\displaystyle \frac{1}{{R}_{3}}}}$

(

*b*)

*The circuit in figure b can be redrawn as shown below:*

We can see that it is similar to the circuit in figure a and, hence, the answer obtained will be same.

#### Page No 200:

#### Answer:

Applying KVL in loop 1, we get:

${i}_{1}+\left({i}_{1}-{i}_{2}\right)-1+2=0\phantom{\rule{0ex}{0ex}}2{i}_{1}-{i}_{2}=-1...\left(1\right)\phantom{\rule{0ex}{0ex}}$

Similarly, for loop 2:

${i}_{2}-\left({i}_{1}-{i}_{2}\right)-2+3=0\phantom{\rule{0ex}{0ex}}2{i}_{2}-{i}_{1}=-1...\left(2\right)$

Solving (1) and (2), we get:

${i}_{1}={i}_{2}=1\mathrm{A}\phantom{\rule{0ex}{0ex}}$

Potential difference between A and B:

${V}_{\mathrm{A}}-{V}_{\mathrm{B}}={E}_{2}-\left({i}_{1}-{i}_{2}\right)\times 1=2-\left(1-1\right)=2\mathrm{V}$

Current through the top branch = *i _{2}* = 1 A

Current through the middle branch =

*i*-

_{1}*i*= 0 A

_{2}Current through the bottom branch =

*i*= 1 A

_{1}#### Page No 200:

#### Answer:

Applying KVL in loop 1, we get:

3*i** *+ 6*i*_{1} = 4.5 ...(1)

Applying KVL in loop 2, we get:

$\left(i-{i}_{1}\right)10+3-6{i}_{1}=0\phantom{\rule{0ex}{0ex}}10i-16{i}_{1}=-3...\left(2\right)\phantom{\rule{0ex}{0ex}}$

Multiplying equation (1) by 10 and (2) by 3 and then, subtracting (2) from (1), we get:

$-108{i}_{1}=-54\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{1}=\frac{54}{108}=\frac{1}{2}=0.5\phantom{\rule{0ex}{0ex}}$

Substituting the value of* i _{1}* in (1), we get:

$3i+6\times \frac{1}{2}-4.5=0\phantom{\rule{0ex}{0ex}}3i-1.5=0\phantom{\rule{0ex}{0ex}}\Rightarrow i=\frac{1.5}{3}=0.5$

So, current flowing through the 10 Ω resistor =

*i*-

*i*= 0.5 - 0.5 = 0 A

_{1}#### Page No 200:

#### Answer:

Applying KVL in loop 1, we get:

$2+\left({i}_{1}-{i}_{2}\right)\times 1-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{1}={i}_{2}$

Applying KVL in loop 2, we get:

$2+\left({i}_{2}-{i}_{3}\right)\times 1-2-\left({i}_{1}-{i}_{2}\right)\times 1=0\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{2}-{i}_{3}-{i}_{1}+{i}_{2}=0\phantom{\rule{0ex}{0ex}}{i}_{1}={i}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{2}-{i}_{3}-{i}_{2}+{i}_{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{2}={i}_{3}$

Applying KVL in loop 3, we get:

$2+{i}_{3}-2-\left({i}_{2}-{i}_{3}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{3}=0\phantom{\rule{0ex}{0ex}}{i}_{1}={i}_{2}={i}_{3}\phantom{\rule{0ex}{0ex}}\therefore {i}_{1}={i}_{2}={i}_{3}=0$

#### Page No 200:

#### Answer:

For current to be zero in the middle branch, the circuit should be balanced. According to the Wheatstone bridge principle, balancing a circuit does not depend on the value of resistance R in the middle branch. The circuit in this case is balanced. Hence, any value of R is acceptable.

#### Page No 200:

#### Answer:

(a) The simplified circuit can be drawn as shown below.

Here cdeo forms a balanced Wheatstone bridge; therefore, branch od will become become ineffective.

The simplified circuit will then be as shown below.

The equivalent resistance between points c and e,

${R}_{cd}=\frac{2r\times 2r}{4r}=r$

The equivalent resistance between a and b,

${R}_{ab}=\frac{r\times r}{2r}=\frac{r}{2}$

(b)

Let R_{eff} be the effective resistance of the circuit.Then, from the symmetry of the circuit, we can assume that the current moving along CO enters OB and the current moving along EO enters OD.

Current on CO = current on OB

Current on EO = current on OD

So, the circuit can be simplified as shown below.

From the simplified circuit diagram, effective resistance of the upper half of the circuit will be

$\left[\left[r+\left(\frac{2{r}^{2}}{3r}\right)+r\right]=2r+\left(\frac{2r}{3}\right)=\frac{8}{3}r\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{R}}_{\mathrm{eff}}=\frac{{\displaystyle \frac{8r}{6}}\times 2r}{{\displaystyle \frac{8r}{6}}+2r}\phantom{\rule{0ex}{0ex}}=8{r}^{2}\times \frac{2}{20r}=\frac{8r}{10}=\frac{4r}{5}$

#### Page No 200:

#### Answer:

From the symmetry of the circuit,we can see that the potential at point C will be equal to the potential at point F. This implies that no current will flow in this branch. Hence, this arm of the circuit can be removed.

Similarly, the potential at point B will be equal to the potential at point G. So, this arm of the circuit can also be removed.

The simplified circuit diagram will be as shown below.

(b)

From the figure, the effective resistance of the circuit,

${R}_{\mathrm{eff}}=\frac{30\times 30}{30+30}=15\mathrm{\Omega}$

The current through the circuit,

$i=\frac{6}{15}=0.4\mathrm{A}$

So, the ammeter reading will be 0.4 A.

#### Page No 201:

#### Answer:

(a)

Applying KVL in the above loop, we get:

$10i+6+5i+12=0\phantom{\rule{0ex}{0ex}}\Rightarrow 10i+5i=-18\phantom{\rule{0ex}{0ex}}\Rightarrow 15i=-18\phantom{\rule{0ex}{0ex}}\Rightarrow i=-\frac{18}{15}=-\frac{6}{5}=-1.2\mathrm{A}$

The negative sign indicates that current is flowing in the direction opposite to our assumed direction.

(b) Potential drop across the 5 Ω resistor= 5*i** *= 5×(-1.2 ) = -6 V

(c) Potential drop across the 10 Ω resistor = 10*i* = (-1.2) × 10 = 12 V

(d)

Applying KVL in the above loop, we get:

$10i+5i+6+12=0\phantom{\rule{0ex}{0ex}}\Rightarrow 15i=-18\phantom{\rule{0ex}{0ex}}\Rightarrow i=-1.2\mathrm{A}$

* *

Potential drop across the 5Ω register = -6 V

Potential drop across the 10Ω register = -12 V

#### Page No 201:

#### Answer:

Let *V* be the potential difference between the points a and f. Let current *i *enter a and leave from f. The distribution of current in various branches is shown in figure.

To calculate the potential difference between a and f, consider the path abcf and apply Kirchofff's Law:

$\frac{i}{3}r+\frac{i}{6}r+\frac{i}{3}r=V$

$\Rightarrow \left(\frac{2ir}{3}+\frac{ir}{6}\right)=V\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{5ir}{6}\right)=V\phantom{\rule{0ex}{0ex}}$

The effective resistance between a and f,

${R}_{\mathrm{eff}}\mathit{}=\frac{V}{i}=\frac{5}{6}r$

#### Page No 201:

#### Answer:

(a) The circuit can be simplified stepwise, as shown below.

The effective resistance between the points a and b,

${\mathrm{R}}_{\mathrm{eff}}=\frac{{\displaystyle \frac{5r}{3}}\times r}{{\displaystyle \frac{5r}{3}}+r}=\frac{5r}{8}\phantom{\rule{0ex}{0ex}}$

(b) The circuit can be simplified, as shown below.

The effective resistance between the points a and b,

${\mathrm{R}}_{\mathrm{eff}}=\left(\frac{r}{3}\right)+r=\frac{4r}{3}\phantom{\rule{0ex}{0ex}}$

(c)

From the figure, it can be seen that axbya is a balanced Wheatstone bridge. The resistors in branch xy will, thus, become ineffective. The circuit can be simplified as under

The effective resistance between the points a and b,

${\mathrm{R}}_{\mathrm{eff}}=\left(\frac{2r\times 2r}{2r+2r}\right)=r$

(d) The circuit can be simplified as shown below.

The effective resistance between the points a and b,

${\mathrm{R}}_{\mathrm{eff}}=\frac{r}{4}$

(e) The circuit can be redrawn as shown below.

Now, we can see that the circuit is a balanced Wheatstone bridge. So, the branch xy will become ineffective. Thus, the simplified circuit will become as shown below.

The effective resistance between the points a and b,

${\mathrm{R}}_{\mathrm{eff}}=\left(\frac{2r\times 2r}{2r+2r}\right)=r$

#### Page No 201:

#### Answer:

(a) Let the effective resistance of the combination be *R*. The circuit can be redrawn as shown below.

From the figure,

$\frac{2R}{R+2}+1=R\phantom{\rule{0ex}{0ex}}\Rightarrow 3R+2={R}^{2}+2R\phantom{\rule{0ex}{0ex}}\Rightarrow {R}^{2}-R-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{+1+\sqrt{1+4\times 1\times 2}}{2\times 1}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{+1+\sqrt{9}}{2.1}=2\mathrm{\Omega}$

(b) Total current sent by the battery $=\frac{6}{R}=\frac{6}{2}=3\mathrm{A}$

Applying Kirchoff's Law in loop 1, we get:

$3\times 1+2i=6\phantom{\rule{0ex}{0ex}}\Rightarrow 2i=3\phantom{\rule{0ex}{0ex}}\Rightarrow i=\frac{3}{2}=1.5\mathrm{A}$

#### Page No 201:

#### Answer:

(a)

The 50 Ω and 200 Ω resistances are in parallel. Their equivalent resistance,

${R}_{\mathrm{eqv}}=\frac{50\times 200}{50+200}=40\mathrm{\Omega}$

This equivalent resistance and the 2 Ω and 1 Ω resistances are connected in series. The effective resistance of the circuit,

${R}_{\mathrm{eff}}=\left(40+2+1\right)=43\mathrm{\Omega}$

In this case, the ammeter will read the total current of the circuit. The current through the ammeter,

$i=\frac{4.3}{43}=0.1\mathrm{A}$

This current will be distributed in the inverse ratio of resistance between the resistances 50 Ω and 200 Ω. The current through the voltmeter,

$i\text{'}=\frac{50}{\left(50+200\right)}i\phantom{\rule{0ex}{0ex}}\Rightarrow i\text{'}=0.02\mathrm{A}$

Reading of the voltmeter = 0.02 × 200 = 4 V

(b)

b

The two branches AB and CD are in parallel. Their equivalent resistance,

${R}_{\mathrm{eqv}}=\frac{52\times 200}{52+200}=41.27\mathrm{A}$

This equivalent resistance is in series with the 1 Ω resistance. The effective resistance of the circuit,

${R}_{\mathrm{eff}}=\left(41.27+1\right)\mathrm{\Omega}=42.27\mathrm{\Omega}$

The total current through the circuit,

$i=\frac{4.3}{42.27}=0.1\mathrm{A}$

In this case, the ammeter will read the current flowing through the 50 Ω resistance, which is *i*_{1}, as shown. The currents in the two parallel branches will distribute in the inverse ratio of the resistances.

$\therefore {i}_{1}=\left(\frac{200}{200+52}\right)i\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{i}=0.08\mathrm{A}$

The current through the voltmeter = *i *- *i _{1}* = 0.02 A

The reading of the voltmeter = 0.02 × 200 = 4 V

#### Page No 201:

#### Answer:

(a) The effective resistance of the circuit,

${\mathrm{R}}_{\mathrm{eff}}=\frac{100\times 400}{500}+200=280\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

The current through the circuit,

$i=\frac{84}{280}=0.3\mathrm{A}$

Since 100 Ω resistor and 400 Ω resistor are connected in parallel, the potential difference will be same across their ends. Let the current through 100 Ω resistor be *i _{1}* ; then, the current through 400 Ω resistor will be

*i*-

*i*.

_{1}$100{i}_{1}=400\left(i-{i}_{1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 500{i}_{1}=400i\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{i}=\frac{4}{5}i=0.24\mathrm{A}$

The reading of the voltmeter = 100 × 0.24 = 24 V

(b) Before the voltmeter is connected, the two resistors 100 Ω resistor and 200 Ω resistor are in series.

The effective resistance of the circuit,

${R}_{\mathrm{eff}}=\left(200+100\right)\phantom{\rule{0ex}{0ex}}=300\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

The current through the circuit,

$i=\frac{84}{300}=0.28\mathrm{A}$

∴ Voltage across the 100 Ω resistor = (0.28 × 100) = 28 V

#### Page No 201:

#### Answer:

Let *R *be the resistance of the voltmeter. For the given circuit, 50 Ω and *R *are in parallel. The equivalent resistance of these two resistors,

${R}_{\mathrm{eqv}}=\frac{50R}{50+R}$

This equivalent resistance is now in series with the 24 Ω resistor. The effective resistance of the circuit,

${R}_{\mathrm{eff}}=\left(\frac{50R}{50+R}+24\right)\mathrm{\Omega}$

Current *t*hrough the circuit,

$i=\frac{30}{{R}_{\mathrm{eff}}}$

Voltage across the 24 Ω resistor, *V _{1}* = $i\times 24=\frac{30}{{R}_{\mathrm{eff}}}\times 24\mathrm{V}$

Voltage across the 50 Ω resistor,

*V*

_{2}_{ }= $30-{V}_{1}=30-\frac{30}{{R}_{\mathrm{eff}}}\times 24\mathrm{V}(\because \mathit{}{V}_{1}+{V}_{2}=30)$

It is given that

*V*= 18 V

_{2}$\Rightarrow 18=30-\frac{30}{{R}_{eff}}\times 24\phantom{\rule{0ex}{0ex}}\Rightarrow 18=30-\frac{30}{\left({\displaystyle \frac{50R}{50+R}}+24\right)}\times 24\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{30}{\left({\displaystyle \frac{74R+1200}{50+R}}\right)}\times 24=30-18\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{30}{\left({\displaystyle \frac{74R+1200}{50+R}}\right)}=\frac{12}{24}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 60\left(50+R\right)=\left(74R+1200\right)\phantom{\rule{0ex}{0ex}}\Rightarrow R\approx 130\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

#### Page No 201:

#### Answer:

It is given that for maximum current, *i* = 10 mA, the potential drop across the voltmeter will be maximum.

The effective resistance of the circuit,

${R}_{\mathrm{eff}}=\left(575+25\right)\mathrm{\Omega}=600\mathrm{\Omega}$

The maximum value of potential difference measured,

*V* = R_{eff} × *i*

= 600 × 10 × 10^{−3} = 6 V

#### Page No 201:

#### Answer:

Let *R* be the resistance of the shunt used.

Coil resistance*, g *= 25 Ω

Current through the coil, *i*_{g} = 1 mA

Total current *i* through the circuit in which the ammeter is connected = 2 A

Since *R* and *g* are connected in parallel,

$g\times {i}_{\mathrm{g}}=R\left(i-{i}_{\mathrm{g}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{g\times {i}_{\mathrm{g}}}{i-{i}_{\mathrm{g}}}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{25\times 1\times {10}^{-3}}{2-{10}^{-3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{R}=1.25\times {10}^{-2}\mathrm{\Omega}$

#### Page No 201:

#### Answer:

The effective resistance for the voltmeter,

*R*_{eff}_{ }= (1150 + 50) Ω = 1200 Ω

The maximum current *i _{g}_{ }*through the coil for maximum deflection of 12 V,

${i}_{\mathrm{g}}=\left(\frac{12}{1200}\right)\mathrm{A}=0.01\mathrm{A}$

Let

*R*be the resistance of the shunt used for ammeter.

Resistance of the coil, g = 50 Ω

Maximum deflection current,

*i,*through the ammeter = 2 A

Since

*R*and

*g*are connected in parallel,

$g\times {i}_{\mathrm{g}}=R\times \left(i-{i}_{\mathrm{g}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{g\times {i}_{\mathrm{g}}}{\left(i-{i}_{\mathrm{g}}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{50\times 0.01}{2-0.01}=0.251\mathrm{\Omega}$

#### Page No 201:

#### Answer:

Let D be the null-point of the potentiometer. Since the bridge is balanced at this point,

$\frac{{R}_{\mathrm{AD}}}{{R}_{\mathrm{DB}}}=\frac{8}{12}$

According to the principle of a potentiometer,

$\frac{{R}_{\mathrm{AD}}}{{R}_{\mathrm{DB}}}=\frac{{l}_{\mathrm{AD}}}{{l}_{\mathrm{DB}}}=\frac{8}{12}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {l}_{\mathrm{AD}}=\frac{2}{3}{l}_{\mathrm{DB}}\phantom{\rule{0ex}{0ex}}{l}_{\mathrm{AD}}+{l}_{\mathrm{DB}}=40\mathrm{cm}\phantom{\rule{0ex}{0ex}}\Rightarrow {l}_{\mathrm{DB}}\frac{2}{3}+{l}_{\mathrm{DB}}=40\mathrm{cm}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5}{3}{l}_{\mathrm{DB}}=40\mathrm{cm}\phantom{\rule{0ex}{0ex}}\Rightarrow {l}_{\mathrm{DB}}=40\times \frac{3}{5}=24\mathrm{cm}$

Hence, the null-point is obtained 24 cm from B.

#### Page No 202:

#### Answer:

It is given that at point D, there is no deflection in the galvanometer, i.e. the bridge is balanced

$\Rightarrow \frac{{R}_{\mathrm{AD}}}{{R}_{\mathrm{DB}}}=\frac{6}{R}$

According to the principle of a potentiometer,

$\frac{{R}_{\mathrm{AD}}}{{R}_{\mathrm{DB}}}=\frac{{l}_{\mathrm{AD}}}{{l}_{\mathrm{DB}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{6}{R}=\frac{30}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{20\times 6}{30}=4\mathrm{\Omega}$

#### Page No 202:

#### Answer:

(a) Potential difference across AB = Potential at A - Potential at B

Potential at B = 0 V

â‡› Potential at point A = Potential difference across AB = 6 V

Potential difference across AC = Potential at A - Potential at B

â‡› 4 = 6 - Potential at C

â‡› Potential at C = 2 V

(b) Given:

Potential across AD = Potential across AC = 4 V

â‡› Potential across DB = 2 V

$\frac{{V}_{\mathrm{AD}}}{{V}_{\mathrm{BD}}}=\frac{{l}_{\mathrm{AD}}}{{l}_{\mathrm{DB}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4}{2}=\frac{{l}_{\mathrm{AD}}}{100-{l}_{\mathrm{AD}}}\phantom{\rule{0ex}{0ex}}\Rightarrow 4\left(100-{l}_{\mathrm{AD}}\right)=2{l}_{\mathrm{AD}}\phantom{\rule{0ex}{0ex}}\Rightarrow 6{l}_{\mathrm{AD}}=400\phantom{\rule{0ex}{0ex}}\Rightarrow {l}_{\mathrm{AD}}=\frac{400}{6}=66.7\mathrm{cm}$

(c) When the points C and D are connected by a wire, current flowing through the wire will be zero because the points are at the same potential.

(d) Potential difference across AC = Potential at A - Potential at C

â‡› 7.5 = 6 - Potential at C

â‡› Potential at C = −1.5 V

Since the potential at C is negative now, this point will go beyond point B, which is at 0 V. Hence, no such point D will exist between the points A and B.

#### Page No 202:

#### Answer:

Let X be the null point on the wire at a distance *x* cm from point A, as shown.

Given:

Total resistance of the wire AB = 15*r*

Resistance per unit cm = $\frac{15r}{600}$

Resistance of *x* cm of the wire = $\frac{15rx}{600}$

Resistance of (600 - *x *) cm of the wire = $\frac{15r\left(600-x\right)}{600}$

(a) Applying KVL in loop 1, we get:

$\left({i}_{1}+{i}_{2}\right)\frac{15}{600}rx+\frac{15}{600}r\left(600-x\right){i}_{1}+{i}_{1}r=\epsilon ...\left(\mathrm{i}\right)$

Applying KVL in loop 2, we get:

${i}_{2}r+\frac{15}{600}rx\left({i}_{1}+{i}_{2}\right)=\frac{\epsilon}{2}...\left(\mathrm{ii}\right)$

For zero deflection in the galvanometer, *i*_{2} = 0. From equation (ii),

$\frac{15}{600}rx\left({i}_{1}\right)=\frac{\epsilon}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{1}=\frac{20\epsilon}{rx}$

Substituting the values of *i _{1}* and

*i*in equation (i), we get:

_{2}*x*= 320 cm

(b) Putting

*x*= 560 cm and solving equations (i) and (ii), we get:

${i}_{2}=\frac{3\epsilon}{22r}$

#### Page No 202:

#### Answer:

In steady state, the capacitor is fully charged and then, it offers infinite resistance to the direct current flow. So, no current can flow through the capacitor in steady state.

The effective resistance of the circuit,

R_{eff}_{ }= 10 + 20 = 30 Ω

The current *i *through the circuit,

$i=\frac{2}{30}=\frac{1}{15}\mathrm{A}$

Voltage drop across the 10 Ω resistor,

V = *i *×* r*

$=\frac{1}{15}\times 10\phantom{\rule{0ex}{0ex}}=\frac{10}{15}=\frac{2}{3}\mathrm{V}$

Since the potential drops across the capacitor and the 10 Ω resistor are the same,

the charge stored on the capacitor,

*Q* = *CV*

$=6\times {10}^{-6}\times \frac{2}{3}\phantom{\rule{0ex}{0ex}}=4\times {10}^{-6}\mathrm{C}=4\mathrm{mC}$

#### Page No 202:

#### Answer:

(a) Applying Kirchoff's voltage law in loop 1, we get:

In the circuit ABEDA,

10*i*_{1} + 20 (*i*_{1} + *i _{2}*) − 5 = 0

⇒ 30

*i*

_{1}+ 20

*i*= 5 ...(i)

_{2}Applying Kirchoff's voltage law in loop 2, we get:

20 (

*i*

_{1}+

*i*) − 5 + 10

_{2}*i*

_{2}= 0

⇒ 20

*i*

_{1}*+*30

*i*

_{2}= 5 ...(ii)

Multiplying equation (i) by 20 and (ii) by 30 and subtracting (ii) from (i), we get:

*i*= 0.1 A

_{2}and

*i*= 0.1 A

_{1}∴ Current through the 20 Ω resistor =

*i*+

_{1}_{ }*i*= 0.1 + 0.1 = 0.2 A

_{2}(b) Potential drop across across AB is,

${V}_{\mathrm{AB}}=0.2\times 20=4\mathrm{V}$

Electrostatic energy stored in the capacitor is given by,

$U=\frac{1}{2}C{{V}_{\mathrm{AB}}}^{2}\phantom{\rule{0ex}{0ex}}U=\frac{1}{2}\times 4\times {10}^{-6}\times (0.2\times 20{)}^{2}\phantom{\rule{0ex}{0ex}}U=32\times {10}^{-6}\mathrm{J}$

#### Page No 202:

#### Answer:

When the capacitors are fully charged, they attain steady state and no current flows through them. Then, equivalent resistance of the circuit,

${R}_{\mathrm{eff}}=\frac{3\times 6}{3+6}=2\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

Current through the circuit,

$i=\frac{6}{2}=3\mathrm{A}$

Current *i *is divided in the inverse ratio of the resistance in each branch. One branch has resistance of 3 Ω and the other branch has resistance of 6 Ω.

Current *i'* through the 3 Ω branch,

$i\text{'}=\frac{6}{9}i=\frac{2}{3}\times 3\mathrm{A}=2\mathrm{A}\phantom{\rule{0ex}{0ex}}$

Current* i'' *through the 6 Ω branch,

$i\text{'}\text{'}=i-i\text{'}=1\mathrm{A}$

Voltage across the 1 Ω resistor = 2 A × 1 Ω = 2 V

Charge on the 1 μF capacitor = 2 × 1 μF = 2 μC

Voltage across the 2 Ω resistor = 2 Ω × 2 A = 4 V

Charge on the 2 μF capacitor = 4V × 2 μF = 8 μC

Voltage across each 3 Ω resistor = 3 Ω × 1 A = 3 V

Charge on the 4 μF capacitor = 3 × 4 μC = 12 μC

Charge on the 3 μF capacitor = 3 × 3 μC = 9 μC

#### Page No 202:

#### Answer:

Equivalent capacitance of the circuit can be calculatesd as,

C_{eq}:

$1\mathrm{\mu F}\mathrm{is}\mathrm{in}\mathrm{parallel}\mathrm{with},C=\frac{1}{{\displaystyle \frac{1}{(3+3)}}+{\displaystyle \frac{1}{(1+1)}}}=\frac{3}{2}\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}\mathrm{So},{C}_{\mathrm{eq}}=\frac{3}{2}+1=\frac{5}{2}\mathrm{\mu F}$

*V* = 100 V

Total charge in the circuits is,

$Q={C}_{\mathrm{eq}}V=\frac{5}{2}\times 100=250\mathrm{\mu}c$

As the volatge across 1μ*f** *is 100 V, therfore charge stored on 1 μ*f* capacitors = 100 μC

Charge flowing from A to B = (250 $-$ 100) = 150 μC

*C*_{eq} between* *A and B is 6 μ*f*.

Potential drop across AB,

$V=\frac{Q}{C}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{\mathrm{AB}}=\frac{150}{6}=25\mathrm{V}$

Potential drop across BC = (100 $-$ 25) = 75 V

#### Page No 202:

#### Answer:

(a) When the charge on the capacitor is zero, it acts as short circuit.

Thus, maximum value of potential difference across the resistor = *ε *(at *t* = 0)â€‹

(b) Maximum value of current in the circuit$=\frac{\mathrm{\epsilon}}{r}$ (at *t* = 0)

(c) Maximum value of potential difference across the capacitor = $\epsilon $

(at *t* = ∞, when the capacitor is fully charged and acts as a open circuit)

(d) Maximum energy stored in the capacitor$=\frac{1}{2}\mathrm{C}{\epsilon}^{2}$ (at *t* = ∞)

(e) Maximum power delivered by the battery$=\frac{{\mathrm{\epsilon}}^{2}}{r}$

(f) Maximum power converted to heat$=\frac{{\mathrm{\epsilon}}^{2}}{r}$

#### Page No 202:

#### Answer:

Given:

Area of the plates*, A* = 20 cm^{2} = 20 × 10^{−4} m^{2}

Separation between the plates*, d* = 1 mm = 1 × 10^{−3} m,

Resistance of the circuit, *R* = 10 kΩ

The capacitance of a parallel-plate capacitor,

$C=\frac{{\mathit{\in}}_{\mathit{0}}\mathit{}A}{d}\phantom{\rule{0ex}{0ex}}=\frac{8.85\times {10}^{-12}\times 20\times {10}^{-4}}{1\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=\frac{8.85\times {10}^{-12}\times 2\times {10}^{-4}}{{10}^{-3}}\phantom{\rule{0ex}{0ex}}=17.7\times {10}^{-12}\mathrm{F}$

Time constant = *CR*

= 17.7 × 10^{−12} × 10 × 10^{3}

= 17.7 × 10^{−8} s

= 0.177 × 10^{−6} s

= 0.18 μs

#### Page No 202:

#### Answer:

Given:

Capacitance of the capacitor, *C* = 10 μF = 10^{−5} F

Emf of the battery*, E*= 2 V

Time taken to charge the capacitor completely, *t* = 50 ms = 5 × 10^{−2} s

The charge growth across a capacitor,

*q* = $Q{e}^{-\frac{t}{RC}}$

*Q* = *CE* = 10^{−5} × 2 C

and* q* = 12.6 × 10^{−6} C

$\Rightarrow 12.6\times {10}^{-6}=2\times {10}^{-5}{\mathrm{e}}^{-\frac{5\times {10}^{-2}}{R\times {10}^{-5}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{12.6\times {10}^{-6}}{2\times {10}^{-5}}={\mathrm{e}}^{-\frac{5\times {10}^{3}}{R}}\phantom{\rule{0ex}{0ex}}\Rightarrow 0.63={\mathrm{e}}^{-\frac{5\times {10}^{3}}{R}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{log}0.63=-\frac{5\times {10}^{3}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{-5\times {10}^{3}}{-0.2}=2.5\mathrm{k\Omega}$

#### Page No 202:

#### Answer:

The growth of charge across a capacitor,

$Q={Q}_{0}.{\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{RC}}}\phantom{\rule{0ex}{0ex}}{Q}_{0}=\mathit{}CV=20\times 6\times {10}^{-6}\mathrm{C}\phantom{\rule{0ex}{0ex}}\Rightarrow Q=120\times {10}^{-6}.{\mathrm{e}}^{-\frac{2\times {10}^{-3}}{{10}^{2}.20\times {10}^{-6}}}\phantom{\rule{0ex}{0ex}}\Rightarrow Q=120\times {10}^{-6}.{\mathrm{e}}^{-1}=76\mathrm{\mu C}$

#### Page No 202:

#### Answer:

Given:

Capacitance of the capacitor, *C* = 10 μF

Initial charge on capacitor, *Q* = 60 μC

Resistance of the circuit, *R* = 10 Ω

(a)Decay of charge on the capacitor,

*$q=Q{\mathrm{e}}^{-\frac{t}{RC}}$*

At* t *=* 0,
q* =

*Q*= 60 μC

(b) At

*t*= 30 μs,

*$q=Q.{\mathrm{e}}^{-\frac{t}{RC}}\phantom{\rule{0ex}{0ex}}\Rightarrow q=Q.{\mathrm{e}}^{-\frac{30\times {10}^{-6}}{10\times 10\times {10}^{-6}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{q}=60.{\mathrm{e}}^{-0.3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{q}=44\mathrm{\mu C}$*

(c) At

*t*= 120 μs,

*$q=Q.{\mathrm{e}}^{-\frac{t}{RC}}\phantom{\rule{0ex}{0ex}}\Rightarrow q=Q.{\mathrm{e}}^{-\frac{120\times {10}^{-6}}{10\times 10\times {10}^{-6}}}\phantom{\rule{0ex}{0ex}}\Rightarrow q=60.{\mathrm{e}}^{-1.2}\phantom{\rule{0ex}{0ex}}\Rightarrow q=18\mathrm{\mu C}$*

(d) At

*t*= 1 ms,

*$q=Q.{\mathrm{e}}^{-\frac{t}{RC}}\phantom{\rule{0ex}{0ex}}\Rightarrow q=Q.{\mathrm{e}}^{-\frac{1\times {10}^{-3}}{10\times 10\times {10}^{-6}}}\phantom{\rule{0ex}{0ex}}\Rightarrow q=60.{\mathrm{e}}^{-10}\phantom{\rule{0ex}{0ex}}\Rightarrow q=0.003\mathrm{\mu C}$*

#### Page No 202:

#### Answer:

Given:

Capacitance, *C* = 8 μF

Emf of the battery, *V*= 6 V

Resistance, *R* = 24

(a) Just after the connections are made, there will be no charge on the capacitor and, hence, it will act as a short circuit. Current through the circuit,

$i=\frac{V}{R}=\frac{6}{24}=0.25\mathrm{A}$

(b) The charge growth on the capacitor,

$q=Q\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)$

One time constant = *RC* = 8 × 24 = 192 × 10^{-6} s

For* t* = *RC*, we have:

$q=Q.\left(1-{\mathrm{e}}^{\frac{-RC}{RC}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow q=CV\left(1-{\mathrm{e}}^{-1}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow q=8\times {10}^{-6}\times 6\times 0.632\phantom{\rule{0ex}{0ex}}=3.036\times {10}^{-5}\mathrm{C}$

$V=\frac{Q}{C}=\frac{3.036\times {10}^{-5}}{8\times {10}^{-6}}=3.792\mathrm{V}$

Applying KVL in the circuit, we get:

*E *= *V *+ *iR*

⇒ 6 = 3.792 + 24*i*

⇒ *i* = 0.09 A

#### Page No 203:

#### Answer:

Given:

Area of plates*, A* = 40 cm^{2} = 40 × 10^{−4} m^{2}

Separation between the plates, *d* = 0.1 mm = 1 × 10^{−4} m

Resistance*, R* = 16 Ω

Emf of the battery,${V}_{0}$= 2 V

The capacitance *C* of a parallel plate capacitor,

$C=\frac{{\mathit{\in}}_{0}A}{d}\phantom{\rule{0ex}{0ex}}=\frac{8.85\times {10}^{-12}\times 40\times {10}^{-4}}{1\times {10}^{-4}}\phantom{\rule{0ex}{0ex}}=35.4\times {10}^{-11}\mathrm{F}$

So, the electric field,

$E=\frac{V}{d}=\frac{Q}{Cd}=\frac{{Q}_{0}}{A{\mathit{\in}}_{0}}\left(1-{\mathrm{e}}^{\mathit{-}\frac{t}{RC}}\right)\phantom{\rule{0ex}{0ex}}=\frac{C{V}_{0}}{A{\mathit{\in}}_{0}}\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\phantom{\rule{0ex}{0ex}}=\frac{35.4\times {10}^{-11}\times 2}{8.85\times {10}^{-12}\times 40\times {10}^{-4}}\left(1-{\mathrm{e}}^{-1.76}\right)\phantom{\rule{0ex}{0ex}}=1.655\times {10}^{4}\phantom{\rule{0ex}{0ex}}=1.7\times {10}^{4}\mathrm{V}/\mathrm{m}$

#### Page No 203:

#### Answer:

Given:

Area of the plates, *A* = 20 cm^{2}

Separation between the plates*, d *= 1 mm

Dielectric constant, *k* = 5

Emf of the battery, *E *= 6 V

Resistance of the circuit, *R* = 100 × 10^{3} Ω

The capacitance of a parallel-plate capacitor,

$C=\frac{K{\mathit{\in}}_{0}A}{d}\phantom{\rule{0ex}{0ex}}=\frac{5\times 8.85\times {10}^{-12}\times 20\times {10}^{-4}}{1\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=\frac{10\times 8.85\times {10}^{-12}\times 20\times {10}^{-4}}{1\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=88.5\times {10}^{-12}\mathrm{C}$

After the connections are made, growth of charge through the capacitor,

*Q* = *EC *(1 − ${\mathrm{e}}^{-\frac{t}{RC}}$)

= 6 × 88.5 × 10^{−12} (1 − ${e}^{-\frac{8.9}{8.85}}$)

= 335.6 × 10^{−12} C

Thus, energy stored in the capacitor,

$U=\frac{1}{2}\frac{{Q}^{2}}{C}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \frac{335.6\times 335.6\times {10}^{-24}}{88.5\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}=\frac{335.6\times 335.6}{88.5\times 2}\times {10}^{-12}\phantom{\rule{0ex}{0ex}}=6.3\times {10}^{-10}\mathrm{J}$

#### Page No 203:

#### Answer:

Time constant of the circuit,

*τ* = *R*C

= 1 × 10^{6} × 100 × 10^{−6}

= 100 s

The growth of charge through a capacitor,

*q* = *VC* (1 − ${\mathrm{e}}^{-\frac{\mathit{t}}{\mathit{R}\mathit{C}}}$)

The current through the circuit,

$i=\frac{dq}{dt}\phantom{\rule{0ex}{0ex}}=\frac{VC}{RC}\xb7{e}^{-\frac{t}{RC}}\phantom{\rule{0ex}{0ex}}=\frac{V}{R}.{e}^{-\frac{t}{RC}}\phantom{\rule{0ex}{0ex}}=24\times {10}^{-6}\xb7{e}^{-\frac{t}{100}}$

For* t* = 10 min = 600 s

*q* = 24 × 10^{−4} × (1 − *e*^{−6})

= 23.99 × 10^{−4}

$i=\frac{24}{{10}^{6}}.\frac{1}{e}=5.9\times {10}^{-8}\mathrm{A}$

(a) The plot between current and time for the first 10 minutes is shown below.

(b) The plot between charge and time for the first 10 minutes is shown below.

#### Page No 203:

#### Answer:

The growth of charge across a capacitor,

$q=Q\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\phantom{\rule{0ex}{0ex}}q=\frac{Q}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{Q}{2}=Q\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {e}^{-\frac{t}{RC}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{t}{R\mathrm{C}}=\mathrm{ln}2\phantom{\rule{0ex}{0ex}}\text{L}\mathrm{et}t=nRC\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{nRC}{RC}=0.69\phantom{\rule{0ex}{0ex}}\Rightarrow n=0.69$

The decay of charge across a capacitor,

$q=Q{\mathrm{e}}^{-\frac{t}{RC}}\phantom{\rule{0ex}{0ex}}q=\frac{Q}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{Q}{2}=Q{\mathrm{e}}^{-\frac{t}{RC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}={\mathrm{e}}^{-\frac{t}{RC}}\phantom{\rule{0ex}{0ex}}\text{L}\mathrm{et}\mathit{}t\mathit{=}nRC\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{nRC}{RC}=\mathrm{ln}2\phantom{\rule{0ex}{0ex}}\Rightarrow n=0.69$

#### Page No 203:

#### Answer:

The decay of charge across a capacitor,

*q* = $Q{\mathrm{e}}^{\mathit{-}\frac{\mathit{t}}{\mathit{R}\mathit{C}}}$

Here,* q* = 0.1 % and Q = 1 × 10^{−3} Q

⇒ 1 × 10^{−3} Q = $Q{\mathrm{e}}^{\mathit{-}\frac{\mathit{t}}{\mathit{R}\mathit{C}}}$

⇒ ${\mathrm{e}}^{-\frac{t}{RC}}$= 10^{−3}

$\Rightarrow \frac{t}{r\mathrm{C}}=-\left(-6.9\right)=6.9$

Let *t *= *nRC*

$\Rightarrow \frac{nRC}{RC}=6.9\phantom{\rule{0ex}{0ex}}\Rightarrow n=6.9$

#### Page No 203:

#### Answer:

The equilibrium value of energy in a capacitor,

$U=\frac{1}{2}\frac{{Q}^{2}}{C}$, where *Q *is the steady state charge.

Let *q* be the charge for which energy reaches half its equilibrium. Then,

$\frac{1}{2}\frac{{q}^{2}}{C}=\frac{1}{2}U\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\frac{{q}^{2}}{C}=\frac{1}{2}\left(\frac{1}{2}\frac{{Q}^{2}}{C}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow q=\sqrt{\frac{{Q}^{2}}{2}}$

The growth of charge in a capacitor,

$q=Q\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\phantom{\rule{0ex}{0ex}}\because q=\sqrt{\frac{{Q}^{2}}{2},}\phantom{\rule{0ex}{0ex}}\sqrt{\frac{{Q}^{2}}{2}}=Q\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{Q}{\sqrt{2}}=Q\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{e}}^{-\frac{t}{RC}}=\left(1-\frac{1}{\sqrt{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{t}{RC}=\mathrm{ln}\left(1-\frac{1}{\sqrt{2}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Let}t\mathit{=}nRC\phantom{\rule{0ex}{0ex}}\text{So},-\frac{nRC}{RC}=\mathrm{ln}\left(1-\frac{1}{\sqrt{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow n=1.23$

#### Page No 203:

#### Answer:

Power = CV^{2} = *q* × V

$\mathrm{Now},\frac{q\mathrm{V}}{2}=q\mathrm{V}\times {e}^{-t/rc}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}={e}^{-t/rc}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{t}{rc}=-\mathrm{ln}\left(0.5\right)\phantom{\rule{0ex}{0ex}}=-\left(-0.69\right)=0.69$

#### Page No 203:

#### Answer:

The rate of growth of charge for the capacitor,

*q* = ε*C* (1 − ${\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{RC}}}$)

Let *E* be the energy stored inside the capacitor. Then,

$E=\frac{{q}^{2}}{2C}=\frac{{\epsilon}^{2}{C}^{2}}{2C}{\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{{\epsilon}^{2}C}{2}{\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)}^{2}$

Let *r* be the rate of energy stored inside the capacitor. Then,

$r=\frac{dE}{dt}=\frac{2{\epsilon}^{2}C}{2}\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-\frac{1}{RC}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{{\epsilon}^{2}}{R}\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left({\mathrm{e}}^{-\frac{t}{RC}}\right)$

$\frac{dr}{dt}=\frac{{\epsilon}^{2}}{R}\left[\left(-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-\frac{1}{RC}\right)\left({\mathrm{e}}^{-\frac{t}{RC}}\right)+\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left({\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-\frac{1}{RC}\right)\right]$

For *r *to be maximum, $\frac{dr}{dt}=0$

$\Rightarrow \frac{{\epsilon}^{2}}{R}\left[\left(-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-\frac{1}{RC}\right)\left({\mathrm{e}}^{-\frac{t}{RC}}\right)+\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left({\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-\frac{1}{RC}\right)\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left[\frac{{\mathrm{e}}^{-{\displaystyle \frac{2t}{RC}}}}{RC}+\frac{{\mathrm{e}}^{-{\displaystyle \frac{2t}{RC}}}}{RC}-\frac{{\mathrm{e}}^{{\displaystyle \frac{-t}{RC}}}}{RC}\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{\mathrm{e}}^{-\frac{2t}{RC}}={\mathrm{e}}^{-\frac{t}{RC}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{e}}^{-\frac{t}{RC}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{t}{RC}=-\mathrm{ln}2\phantom{\rule{0ex}{0ex}}\Rightarrow t=RC\mathrm{ln}2$

#### Page No 203:

#### Answer:

Given,

Capacitance of capacitor, *C*= 12.0 μF = 12 × 10^{−6} F

Emf of battery,* **V*_{0} = 6.00 V

Internal resistance of battery,* R* = 1 Ω

Time interval,* t * = 12 μs

(a)

Charging current in the circuit is given as,

*i *= *i*_{0}*e*^{−t}^{/RC}

Current at,* t* = 12.0 μs

$i=\frac{{V}_{0}}{R}{e}^{-t/RC}\phantom{\rule{0ex}{0ex}}i=\frac{6}{1}\times {e}^{-1}\phantom{\rule{0ex}{0ex}}i=2.207=2.21\mathrm{A}$

(b)

During charging, charge on the capacitor at any time ''*t*'' is given as

$Q=C{V}_{0}(1-{e}^{-\frac{t}{RC}})$

Work done by battery in in time delivering this charge is,

*W = **QV*_{0}

Power deliver by the battery in time ''*t*'' is,

$P=\frac{C{{V}_{0}}^{2}(1-{e}^{-{\displaystyle \frac{t}{RC}}})}{t}$

Putting,* t* = 12 μs

$P=\frac{12\times {10}^{-6}{{V}_{0}}^{2}(1-{e}^{-{\displaystyle 1}})}{12\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow P=13.25W$

(c)

Energy stroed in the capacitor at any instant of time is given as,

$U=\frac{1}{2}\frac{{Q}^{2}}{C}\phantom{\rule{0ex}{0ex}}\Rightarrow U=\frac{1}{2}\frac{{C}^{2}{{V}_{0}}^{2}(1-{e}^{-{\displaystyle \frac{t}{RC}}}{)}^{2}}{C}\phantom{\rule{0ex}{0ex}}\Rightarrow U=\frac{1}{2}C{{V}_{0}}^{2}(1-{e}^{-\frac{t}{RC}}{)}^{2}$

Rate at which the energy stored in the capacitor is,

$\frac{dU}{dt}=\frac{1}{2}C{{V}_{0}}^{2}\times 2(1-{e}^{-\frac{t}{RC}})\times ({e}^{-\frac{t}{RC}})\times \frac{1}{RC}$

$\Rightarrow \frac{dU}{dt}=\frac{{{V}_{0}}^{2}}{R}({e}^{-\frac{t}{RC}}-{e}^{-\frac{2t}{RC}})\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dU}{dt}=\frac{6\times 6}{1}({e}^{-1}-{e}^{-2})\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dU}{dt}=8.37\mathrm{W}\phantom{\rule{0ex}{0ex}}$

So, the power dissipated in heat = $P-\frac{dU}{dt}$

= 13.25 $-$ 8.37 = 4.87 W

(d)

Rate at which the energy stored in the capacitor is increasing = $\Rightarrow \frac{dU}{dt}=8.37\mathrm{W}$

#### Page No 203:

#### Answer:

Let *Q _{0}* be the initial charge on the capacitor. Then,

*Q*=

_{0}*CV*

The charge on the capacitor at time

*t*after the connections are made,

$Q={Q}_{0}{e}^{-\frac{t}{RC}}$

$i=\frac{dQ}{dt}=-\left(\frac{{Q}_{0}}{RC}\right){\mathrm{e}}^{-\frac{t}{RC}}$

Heat dissipated during time

*t*to

_{1}*t*,

_{2}$U={\int}_{{t}_{1}}^{{t}_{2}}{i}^{2}Rdt\phantom{\rule{0ex}{0ex}}=\frac{{{Q}_{0}}^{2}}{2C}\left({\mathrm{e}}^{-\frac{2{t}_{1}}{RC}}-{\mathrm{e}}^{-\frac{2{t}_{2}}{RC}}\right)$

Time constant =

*RC*

Putting

*t*= 0 and

_{1}*t*=

_{2}*RC*, we get:

$U=\frac{{{Q}_{0}}^{2}}{2C}\left({\mathrm{e}}^{-0}-{\mathrm{e}}^{-2}\right)\phantom{\rule{0ex}{0ex}}\because {Q}_{0}=CV,\phantom{\rule{0ex}{0ex}}U=\frac{CV}{2}\left(1-\frac{1}{{\mathrm{e}}^{2}}\right)$

**Alternative method:**

Heat dissipated at any time = Energy stored at time 0 − Energy stored at time

*t*

$\Rightarrow U=\frac{1}{2}C{V}^{2}-\frac{1}{2}C{V}^{2}{\mathrm{e}}^{-\frac{2t}{RC}}\phantom{\rule{0ex}{0ex}}\because t=RC,\phantom{\rule{0ex}{0ex}}U=\frac{1}{2}C{V}^{2}-\frac{1}{2}C{V}^{2}{\mathrm{e}}^{-2}\phantom{\rule{0ex}{0ex}}\Rightarrow U=\frac{1}{2}C{V}^{2}\left(1-\frac{1}{{\mathrm{e}}^{2}}\right)$

#### Page No 203:

#### Answer:

The growth of charge on the capacitor at time* t *,

$Q={Q}_{0}\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)$

$i=\frac{dQ}{dt}=\left(\frac{{Q}_{0}}{RC}\right){\mathrm{e}}^{-\frac{t}{RC}}$

Heat dissipated during time *t _{1}* to

*t*,

_{2}$U={\int}_{{t}_{1}}^{{t}_{2}}{i}^{2}Rdt\phantom{\rule{0ex}{0ex}}=\frac{{{Q}_{0}}^{2}}{2C}\left({\mathrm{e}}^{-\frac{2{t}_{1}}{RC}}-{\mathrm{e}}^{-\frac{2{t}_{2}}{RC}}\right)\phantom{\rule{0ex}{0ex}}\because {Q}_{0}=C{V}_{0,}\phantom{\rule{0ex}{0ex}}U=\frac{1}{2}C{{V}_{0}}^{2}\left({\mathrm{e}}^{-\frac{2{t}_{1}}{RC}}-{\mathrm{e}}^{-\frac{2{t}_{2}}{RC}}\right)$

The potential difference across a capacitor at any time

*t*,

$V={V}_{0}\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)$

The energy stored in the capacitor at any time

*t,*

$E=\frac{1}{2}C{V}^{2}=\frac{1}{2}C{{V}_{0}}^{2}{\left(1-{\mathrm{e}}^{-\frac{2t}{RC}}\right)}^{2}$

∴ The energy stored in the capacitor from

*t*

_{1}to

*t*

_{2},

$E=\frac{1}{2}C{{V}_{0}}^{2}\left({\mathrm{e}}^{-\frac{2{t}_{1}}{RC}}\right)-\frac{1}{2}C{{V}_{0}}^{2}\left(1-{\mathrm{e}}^{-\frac{2{t}_{2}}{RC}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{1}{2}C{{V}_{0}}^{2}\left({\mathrm{e}}^{-\frac{2{t}_{1}}{RC}}-{\mathrm{e}}^{-\frac{2{t}_{2}}{RC}}\right)$

#### Page No 203:

#### Answer:

The capacitance of a parallel plate capacitor,

$C=\frac{K{\epsilon}_{0}A}{d}$

The resistance of dielectric material,

$R=\frac{\mathrm{\rho}d}{A}$

Time constant,

$\tau =RC=\mathrm{\rho}K{\epsilon}_{0}$,

which is independent of the plate area or separation between the plates.

#### Page No 203:

#### Answer:

The equivalent capacitance of the circuit,

*${C}_{\mathrm{eqv}}={C}_{1}+{C}_{2}=2+2=4\mathrm{\mu F}$*

The growth of charge through the capacitor,

*q* = *q*_{0}(1 − *e*^{−t}^{/RC})

${q}_{0}=CV=4\times {10}^{-6}\times 6=24\times {10}^{-6}\mathrm{C}\phantom{\rule{0ex}{0ex}}\frac{\mathit{t}}{\mathit{R}\mathit{C}}=\frac{0.20\times {10}^{-3}}{25\times 4\times {10}^{-6}}=2$

⇒ *q* = 24 × 10^{−6} (1 − *e*^{−2})

= 18.4 × 10^{−6} C

This is the total charge on both capacitors. As the capacitors are in parallel, the total charge will be shared between them. Also, both the capacitors are of same capacitance; so, they will share equal amount of charge.

∴ Charge on each capacitor$=\frac{18.4}{2}\mathrm{\mu C}=9.2\mathrm{\mu C}$

#### Page No 203:

#### Answer:

Initially, the switch S was closed; so, the capacitor was getting charged. Also, the two resistors are connected in parallel. The equivalent resistance of the circuit,

$R=\frac{1}{{\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{10}}}=5\mathrm{\Omega}$

Initially, the switch was closed and the capacitor was getting charged. So, the two resistances were in parallel connection. Hence, their effective resistance will be 5 Ω.

Potential difference across the 10 Ω resistance, $V=\frac{12}{5}\times 10=24\mathrm{V}$

When the switch is opened, the decay of charge through the capacitor,

$Q={Q}_{0}{\mathrm{e}}^{-\frac{t}{RC}}=V\times C{\mathrm{e}}^{-\frac{t}{RC}}\phantom{\rule{0ex}{0ex}}\frac{t}{RC}=\frac{1\times {10}^{-3}}{10\times 25\times {10}^{-6}}=\frac{1000}{250}=4\phantom{\rule{0ex}{0ex}}\Rightarrow Q=24\times 25\times {10}^{-6}\times {\mathrm{e}}^{-4}\phantom{\rule{0ex}{0ex}}=24\times 25\times {10}^{-6}\times 0.0183\phantom{\rule{0ex}{0ex}}=10.9\times {10}^{-6}\mathrm{C}$

Current in the 10 Ω resistor,

$i=\frac{Q}{t}=\frac{10.9\times {10}^{-6}\mathrm{C}}{1\times {10}^{-3}\mathrm{sec}}=11\mathrm{mA}$

#### Page No 203:

#### Answer:

Given:

Capacitance of capacitor* ,C* = 100 μF

Emf of battery *,E *= 6V

Resistance* ,R *= 20 kΩ

Time for charging , *t*_{1} = 4 s

Time for discharging ,*t*_{2} = 4 s

During charging of the capacitor, the growth of charge across it,

$\mathrm{Q}=CE\left(1-{\mathrm{e}}^{-\frac{{t}_{1}}{RC}}\right)\phantom{\rule{0ex}{0ex}}\frac{{t}_{1}}{RC}=\frac{4}{20\times {10}^{3}\times 100\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}=2\phantom{\rule{0ex}{0ex}}\Rightarrow Q=6\times {10}^{-4}\left(1-{\mathrm{e}}^{-2}\right)\phantom{\rule{0ex}{0ex}}=5.187\times {10}^{-4}\mathrm{C}$

This is the amount of charge developed on the capacitor after 4s.

During discharging of the capacitor, the decay of charge across it,

$Q\text{'}=Q\left({\mathrm{e}}^{-\frac{t}{RC}}\right)\phantom{\rule{0ex}{0ex}}=5.184\times {10}^{-4}\times {\mathrm{e}}^{-2}\phantom{\rule{0ex}{0ex}}=0.7\times {10}^{-4}\mathrm{C}=70\mathrm{\mu C}$

#### Page No 203:

#### Answer:

The two capacitors are connected in series. Their equivalent capacitance,

${C}_{eqv}=\frac{{C}_{\mathit{1}}{C}_{\mathit{2}}}{{C}_{\mathit{1}}+{C}_{\mathit{2}}}$

The growth of charge in the capacitors,

$q={C}_{\mathrm{eqv}}\epsilon \left(1-{\mathrm{e}}^{-\frac{1}{r{C}_{\mathrm{eqv}}}}\right)$

#### Page No 203:

#### Answer:

Given:

Initial charge on first capacitor = *Q*

Let *q* be the charge on the second capacitor after time *t*.

According to the principle of conservation of charge, charge on the first capacitor after time *t* = *Q *- *q*.

Let* V _{1}* be the potential difference across the first capacitor and

*V*be the potential difference across the second capacitor after time

_{2}*t*. Then,

${V}_{\mathit{1}}=\frac{Q-q}{C}\phantom{\rule{0ex}{0ex}}{V}_{\mathit{2}}=\frac{q}{C}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{V}}_{1}-{\mathrm{V}}_{2}=\frac{Q-q}{C}-\frac{q}{C}\phantom{\rule{0ex}{0ex}}=\frac{Q-2q}{C}$

The current through the circuit after time

*t*,

$i=\frac{{V}_{1}-{V}_{2}}{R}=\frac{dq}{dt}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{Q-2q}{CR}=\frac{dq}{dt}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dq}{Q-2q}=\frac{1}{RC}dt\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dq}{Q-2q}=\frac{1}{RC}dt\phantom{\rule{0ex}{0ex}}$

Integrating both sides within the limits time =0 to t and charge on the second capacitor varying from q=0 to q, we get:

$\frac{1}{2}\left[\mathrm{ln}\left(Q-2q\right)-\mathrm{ln}Q\right]=\frac{-1t}{RC}\phantom{\rule{0ex}{0ex}}\mathrm{ln}\frac{Q-2q}{Q}=\frac{-2t}{RC}\phantom{\rule{0ex}{0ex}}Q-2q=Q{\mathrm{e}}^{-\frac{2t}{RC}}\phantom{\rule{0ex}{0ex}}2q=Q\left(1-{\mathrm{e}}^{-\frac{2t}{RC}}\right)\phantom{\rule{0ex}{0ex}}q=\frac{Q}{2}\left\{1-{\mathrm{e}}^{-\frac{2t}{RC}}\right\}$

#### Page No 203:

#### Answer:

Given:

Initial charge given to the capacitor = *Q*

When the capacitor is connected to a battery, it will charge through the battery. The initial charge will also decay.

The growth of charge across the capacitor due to the battery of emf *E,*

$q=CE\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)$

The decay of charge through the capacitor,

$q\text{'}=Q{\mathrm{e}}^{-\frac{t}{RC}}$

The net charge on the capacitor at any time *t*,

${q}_{\mathrm{net}}=q+q\text{'}\phantom{\rule{0ex}{0ex}}=CE\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)+Q{\mathrm{e}}^{-\frac{t}{RC}}$

View NCERT Solutions for all chapters of Class 12