Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 3 - Current Electricity

Answer:

Current density is basically electric current per unit area i.e. $j=\frac{I}{A}$. 
However, electric field and current density are also inter related by a formula,
$\overrightarrow{j}=\sigma \overrightarrow{E}$
This means that current density changes due to the electric field produced by charges get collected on the surface of wire.
Hence, the correct answer is option (b).
 

Answer:

For parallel combination, equivalent emf
$$\begin{gathered} \frac{{\epsilon }_{eq}}{R_{eq}}=\frac{{\epsilon }_1}{r_1}+\frac{{\epsilon }_2}{r_2} \\ \Rightarrow {\epsilon }_{eq}=\left(\frac{{\epsilon }_1}{r_1}+\frac{{\epsilon }_2}{r_2}\right)\times R_{eq} \\ \Rightarrow {\epsilon }_{eq}=\left(\frac{{\epsilon }_1}{r_1}+\frac{{\epsilon }_2}{r_2}\right)\times \frac{r_1{r}_2}{r_1+r_2} \left(\because R_{eq}=\frac{r_1{r}_2}{r_1+r_2} \right) \\ \Rightarrow {\epsilon }_{eq}=\frac{{\epsilon }_1{r}_2+{\epsilon }_2{r}_1}{r_1+r_2} \end{gathered}$$

It is given that,

Hence, the correct answer is option (a).

Answer:

Given, standard resistance, S = 100 Ω
l₁ = 2.9 cm
When meter bridge is balanced_,

Answer:

In case of potentiometer experiment, emf of a cell can be measured, if along the potentiometer wire potential drop is more than the emf of the cell. Here, emf of two cells are 5V and 10 V, so potential drop across potentiometer wire should be more than 10 V.
Hence, the correct answer is option (b).

Answer:

Resistance of the wire is given as,
R = $\rho \frac{l}{A}$
Here, ρ is resistivity
         l is length
        A is area of cross-section
For greater value of R, l must be maximum and A should be minimum and this is only possible when battery is connected across
 $\left(1\mathrm{cm}\times \frac{1}{2} \mathrm{cm}\right)$ faces. 
Hence, the correct answer is option (a).

Answer:

Current in a conductor is given as
I = n A_e v_d
Here, I is current
         n is number of electrons per unit volume
         A is area of cross-section
         v_d is drift velocity
         e is charge of electron.
⇒ I $\propto$ v_d 
Hence, the correct answer is option (a).

Answer:

Kirchhoff's junction law states that the total current into junction will be equal to the total current going out of the junction
I_in – I_out = 0
Kirchhoff's junction rule is based on the law of conservation of charge and the fact that there is no charge accumulation at a junction.
Hence, the correct options are (b) and (d).

Answer:

Internal resistance = r
Variable resistance = R'
Here, r << R << R'
Applying Ohm's law, V = IR

$\Rightarrow I=\frac{V}{\left(r+\frac{RR\text{'}}{R+R\text{'}}\right)}$ 

Since, r << R << R'

$\Rightarrow I=\frac{V}{r+{\displaystyle \frac{RR\text{'}}{R+R\text{'}}}}\ge \frac{V}{r+R}$

Potential difference across A and B, V_AB = $I\times \frac{RR\text{'}}{R+R\text{'}}$
$\Rightarrow V_{\mathrm{AB}}=\frac{V}{r+\left({\displaystyle \frac{RR\text{'}}{R+R\text{'}}}\right)}\times \frac{RR\text{'}}{R+R\text{'}}$

$=\frac{V \left(R+R\text{'}\right)}{r\left(R+R\text{'}\right)+RR\text{'}}\times \frac{RR\text{'}}{R+R\text{'}}$

$=\frac{VR}{r+R}$

Therefore, V_AB  is constant if R' is varied.
Hence, the correct  options are (a) and (d).

Answer:

Resistivity $\rho \left(T\right)=\frac{m}{n{e}^2\tau }$
When number of charge carriers changes with temperature T, then the resistivity will also change. Also, here τ is the relaxation time which is the time interval between two successive collisions, which decrease with increase in temperature.
Hence, the correct options are (a) and (b).

Answer:

Given for first student,
R₁ = 5 Ω
R₂ = 10 Ω
R₃ = 5 Ω
For second student,
R₁ = 500 Ω
R₂ = 1000 Ω
R₃ = 5 Ω

Applying wheat-stone bridge rule,

$\frac{R_2}{R}=\frac{R_1}{R_3}$

$\Rightarrow R=R_3\times \frac{R_2}{R_1}$
For student 1,
 $R=\cancel{5}\times \frac{10}{\cancel{5}}=10 \mathrm{\Omega }$

For student 2,
 $R=\cancel{5}\times \frac{\cancel{1000}}{\cancel{500}}=10 \mathrm{\Omega }$  
So, wheat-stone bridge is most sensitive and accurate if resistances are of same value.
When the values of R₁ and R₂ becomes large current through the arms become weak which can make the determination of null point accurately more difficult.
Hence, the correct options are (b) and (c).

Answer:

When jockey is at a neutral point D, current will not flow through galvanometer which means potentials at B and D are equal. However, potentials at different point will be different. So, the point D is unique to get null point. Now, when jockey is shifted in right of D on wire, potential in wire towards the right becomes smaller. i.e., V_B > V_O.
So, current flows from B to D in the wire. Now, when R is increased, potential drop across R increases which results in in-case in potential at B to get null point.
​Hence, the correct options are (a) and (c).

Answer:

In an electric circuit, when an electron approaches a junction, in addition to the uniform electric field E that it normally faces (which keep the drift velocity dv fixed), there are accumulation of charges on the surface of wires at the junction. Due to which an electric field is produced Thus these electric fields alter direction of momentum.
Hence, when charge crosses a junction in an electric circuit the momentum is not conserved at the junction.

Answer:

Relaxation time,
 $\tau =\frac{m{v}_d}{eE}$

$\Rightarrow \tau \alpha \frac{v_d}{E}$
This means relaxation time depends upon velocity of electrons and ions and electric field. The applied electric field affects the velocities of electron at order of 1 mm/s which can be neglected and this also supports Ohm's law.
However, change in temperature affects velocities of electrons and ions at order 10² m/s, which is significant. 
​
Resistivity, $\rho =\frac{m}{n{e}^2\tau }$

Thus, τ can be affected to an extent and hence the resistivity.

Answer:

The necessary condition for wheatstone bridge is given as $\frac{P}{Q}=\frac{R}{S}$
Here, P and Q are ratio arms, R is known resistance and S is unknown resistance-advantage of the null-point method:-

1. Resistance of galvanometer does not affect the balance point. So, no need for determination of current in resistances and internal resistance of a galvanometer.
2. By calculating least count and readings of ammeter and voltmeter, R unknown resistance can be calculated by applying kirchhoff's law and Ohm's law.

Answer:

In potentiometer, thick metallic strips are used, as they will introduce least resistance while joining the wires. Being metallic strips of least resistance, they can be ignored while measuring the length of the wire which is used for determination of null point.

Answer:

For wiring in the home, Cu or Al wires are used the main considerations involved are:
(i) Good conductivity of metal.
(ii) Cost of metal used

Answer:

Alloys are used for making standard resistance coils, this is because they have:
(i) Low value of temperature coefficient of resistivity.
(ii) High resistivity.

Answer:

Power delivered,
P = Voltage × Current = VI

$\Rightarrow I=\frac{P}{V}$

If R_c is the resistance of transmission lines then power wasted:
P_c = I²R_c

$$\begin{gathered} \Rightarrow p_c=\frac{p^2}{V^2} R_c \\ \Rightarrow p_c \alpha \frac{1}{V^2} \end{gathered}$$

Thus, in order to reduce P_{c ,} V should be increased.
Hence, the power wastage can be reduced by transmitting power of high voltage.

Answer:

When R is increased, the current through the wire will decrease and hence the potential gradient will also decrease. Hence, the balance length. will increase. Thus point J will shift towards B to obtain the balance point.

Answer:

(i) Current in auxiliary circuit decreases whereas potential difference across A and jockey increases. This is possible when positive terminal of E₁ is connected at X and E₁ > E.
(ii) Current in auxiliary circuit increases whereas potential difference across A and jockey increases. This is possible when E₁ negative terminal is connected at X.

Answer:

Consider a circuit in which battery of emf E is connected to resistance R and having internal resistances r.
Let V be the potential drop across resistance and I be the current flowing through the circuit.
Applying, Kirchhoff's voltage law,
E – IR – Ir = 0
⇒ E = I(R + r) 
Potential drop across the battery, V = IR
⇒ V = $\frac{ER}{R+r}$
As R becomes large R + r becomes approximately equal to R and V ≈ E.

 

Answer:

When n resistances each of resistance R are connected in series and then in parallel.
For series combination:
Rs = R₁ + R₂ + . . . . + R_n = R + R + . . . . .n times
⇒ R_s = n R
For parallel combination

$$\begin{gathered} \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+. . . . +\frac{1}{R_n}=\frac{1}{R}+\frac{1}{R}+ . . . . +n \mathrm{time} \\ \Rightarrow R_{\mathit{p}}\mathit{=}\frac{\mathit{R}}{\mathit{n}} \end{gathered}$$
In series combination, current, I = $\frac{E}{R+nR}$

In parallel combination, current, 10 I = $\frac{E}{R+{\displaystyle \frac{R}{n}}}$
$\Rightarrow \frac{E}{R+{\displaystyle \frac{R}{n}}}=\frac{10E}{R+nR}$
$$\begin{gathered} \Rightarrow \frac{1}{R\left(1+{\displaystyle \frac{1}{n}}\right)}=\frac{10}{R\left(1+n\right)} \\ \Rightarrow 10\left(1+\frac{1}{n}\right)=1+n \\ \Rightarrow 10+\frac{10}{n}-1-n=0 \\ \Rightarrow n^2-9n-10=0 \\ \Rightarrow \left(n+1\right)\left(n-10\right)=0 \end{gathered}$$

Here n = –1 (not possible)
⇒ n = 10
Hence, there are 10 resistance in combination.

Answer:

Let R_min and R_max be the minimum and maximium resistance among R₁, R₂, . . . . ., R_n. When resistance are connected in parallel,

$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+. . . . +\frac{1}{R_n}$

On multiplying both sides by R_min , we get

$\frac{R_{min}}{R_p}=\frac{R_{min}}{R_1}+\frac{R_{min}}{R_2}+. . . . +\frac{R_{min}}{R_n}$

Here, R₁, R₂ . . . , R_n there must be a resistance which is minimum.
This means there must be a term R_min in R.H.S. which is equal to 1 and the R_min other terms are positive.

$$\begin{gathered} \Rightarrow \frac{R_{min}}{R_p}=\frac{R_{min}}{R_1}+\frac{R_{min}}{R_2}+. . . . +\frac{R_{min}}{R_n}>1 \\ \Rightarrow R_{min}>R_p \end{gathered}$$

So, in parallel combination, equivalent resistance R_p is always less than any smallest resistance in the combination.



When resistance are connected in series, then equivalent resistance
R_s = R₁ + R₂ + . . . + R_n
In R.H.S. there must be a term R_max which has maximum value among R₁. R₂ . . , R_n.
⇒ R_s = R₁ + R₂ + . . + R_max + . . + R_n = R₊₊
⇒ R_s > R_max
So, in series combination equivalent resistance is always greater than the maximum resistance available in combination of resistance.​

Answer:

Given,
emf of cell E₁ = 6 V
Internal resistance, r₁ = 2 Ω
emf of cell E₂ = 4 V
Internal resistance, r₂ = 8 Ω
Effective resistance, R = 2 + 8 = 10 Ω
Effective emf, E = 6 – 4 = 2 V
Using Ohm's law,  V = IR
$\Rightarrow I=\frac{6-4}{2+8}=\frac{1}{5} \mathrm{A}$ 



The direction of flow of current is from higher potential to lower potential.
Since, $V_B>V_A$
$$\begin{gathered} \Rightarrow V_{AB}=V_B-V_A \\ =4+\frac{1}{5}\times 8=4+1.6 \\ =5.6 \mathrm{V} \end{gathered}$$
Hence, the potential difference between A and B is 5.6 V.

Answer:

Emf of two cells = E
​Internal resistance = r₁ and r₂
External resitance = R
Terminal voltage across first cell = 0
Now effective emf,
$E_{effective}=E-E=0$

Electric current, I = $\frac{E-E}{R+r_1+r_2}=0$

Disclaimer: According to the circuit shown in the question, current through the circuit will be zero. So, the question/image seems incorrect.

Answer:

Diameter of solid wire A, d₁ = 1 mm.
Diameter of outer solid hollow tube D₁ = 2 mm
Diameter of inner solid hollow tube, D₂ = 1 mm
$\mathrm{Resistance}, R=\frac{\rho l}{A}$
Resistance of first conductor, R_A = $\frac{\rho l}{\pi {\left({\displaystyle \frac{d_1}{2}}\right)}^2}$

Resistance of second conductor, $R_B=\frac{\rho l}{\pi \left[{\left({\displaystyle \frac{D_1}{2}}\right)}^2-{\left({\displaystyle \frac{D_2}{2}}\right)}^2\right]}$

Ratio of two resistors, $\frac{R_A}{R_B}=\frac{{\left({10}^{-3}\right)}^2-{\left(0.5\times {10}^{-3}\right)}^2}{{\left(0.5\times {10}^{-3}\right)}^2}$

$$\begin{gathered} \Rightarrow \frac{R_A}{R_B}=\frac{1-0.25}{0.25} \\ \Rightarrow \frac{R_A}{R_B}=\frac{0.75}{0.25} \\ \Rightarrow \frac{R_A}{R_B}=\frac{3}{1} \end{gathered}$$

Hence, the ratio of resistance R_A to R_B is 3 : 1.

Answer:

Let n be the number of batteries in series of the circuit and R_eff be the effective internal resistance of n resistors ∊_eff be the effective emf of n batteries in series and R be the external resistance in the circuit.


Current through R, I = $\frac{{\epsilon }_{eff}}{R_{eff}+R}$

On increasing voltage and resistances n-fold

$$\begin{gathered} {\epsilon }_{eff}^{new}=n {\epsilon }_{eff} \\ {R}_{eff}^{new}=n R_{eff} \end{gathered}$$

$$\begin{gathered} R^{new}=nR \\ \Rightarrow I^{new}=\frac{{\epsilon }_{eff}^{new}}{R_{eff}^{new}+R^{new}} \\ =\frac{n{\epsilon }_{eff}}{n{R}_{eff}+nR}=\frac{n{\epsilon }_{eff}}{n\left(R_{eff}+R\right)}=\frac{{\epsilon }_{eff}}{R_{eff}+R}=I \\ \Rightarrow I^{new}=I \end{gathered}$$
Thus, the current remains unaltered when all voltage and all resistances are increased to n-folds.

Answer:

Voltage of cells, V₁ = 10 V and V₂ = 2 V
   But, V = IR
Applying Kirchhoff's junction rule:
I₂ = I₁ − I
In bigger outer loop having 10 V cell and resistance R:
applying Ohm's law,
10 = IR + 10I_{1                                    .....}(1)
Now, for outer loop having 2 V cell and resistance R:
2 = 5I₂ – RI
⇒ 2 = 5(I₁ – I) – RI                     .....(2)
Solving equation (1) and (2) we get:
6 = 3RI + 10I
⇒ 2 = $I\left(R+\frac{10}{3}\right)$                        .....(3)
Effective potential difference due to both batteries is,
V_eff = I (R + R_eff)                       .....(4)
On comparing equation (3) and (4), we get:
V_eff = 2 V and R_eff = $\frac{10}{3} \mathrm{\Omega }$.

Answer:

Energy consumed per day, E = 10 kWh
Time for using energy per day to 5 h

Power consumed, P = $\frac{10 \mathrm{kWh}}{5 \mathrm{h}}=2 \mathrm{kW}=2000 \frac{\mathrm{J}}{\mathrm{S}}$

Current supplied,
I = $\frac{P}{V}=\frac{200\cancel{0}}{22\cancel{0}}\approx 9 \mathrm{A}$

Power loss in copper wire,

P = $I^{2}R=I^2\times \frac{\rho l}{\pi r^2} \left(\mathrm{as}, R=\rho \frac{l}{A}=\frac{\rho l}{\pi r^2} \right)$

$\Rightarrow P_{\mathrm{cu}}=9^2\times \left[\frac{1.7\times {10}^{-8}\times 10}{{\displaystyle \frac{22}{7}\times {\left({10}^{-3}\right)}^2}}\right]\approx 4 \mathrm{J}/\mathrm{s}.$

Fractional loss of power in Cu wire, 
$=\frac{4}{2000}\times 100=0.2\%$

Similarly power loss in Al wire,
P_Al = $I^2\frac{{\rho }_{Al}\times l}{\pi r^2}$

$\Rightarrow P_{\mathrm{Al}}=9^2\times \frac{\left(2.7\times {10}^{-8}\right)\times 10}{{\displaystyle \frac{22}{7}}\times {\left({10}^{-3}\right)}^2}=6.9 \mathrm{J}/\mathrm{s}$


Fractional loss of power in Al wire, 
= $\frac{6.9}{2000}\times 100=0.34\%.$

Answer:

Variable resistance, R = 50 Ω
Let R' be the potentiometer wire resistance.
Current supplying in primary circuit be I which have E_B = 10 V
Here, I = $\frac{V_{\mathrm{B}}}{R+R\text{'}} \left[\mathrm{Using} \mathrm{Ohm}\text{'}\mathrm{s} \mathrm{law}\right]$

$\Rightarrow I=\frac{10}{50+R}$

Potential difference across wire of potentiometer, V = IR'

$\Rightarrow V\text{'}=\frac{10 R\text{'}}{R+50}$

Null deflection can not be obtained by 8 V.
it is possible only when:

$$\begin{gathered} \Rightarrow 10 R\text{'}<8 R+400 \\ \Rightarrow 2 R\text{'}<400 \\ \Rightarrow R\text{'}<200 \mathrm{\Omega } \end{gathered}$$

Similarly, for R = 10 Ω, null point is possible when,

$\frac{10 R\text{'}}{10+R\text{'}}>8$

$$\begin{gathered} \Rightarrow 2 R\text{'}>80 \\ \Rightarrow R\text{'}>40 \mathrm{\Omega } \end{gathered}$$

Null point is obtained on 4th segment or at $\frac{3}{4}$ of the total length.
$$\begin{gathered} \Rightarrow \frac{10\times {\displaystyle \frac{3}{4}}R\text{'}}{10+R\text{'}}<8 \left(\mathrm{At} \mathrm{balance} \mathrm{point}\right) \\ \Rightarrow 7.5 R\text{'}<80+8 R\text{'} \\ \Rightarrow –R\text{'}<160 \\ \Rightarrow R\text{'}>160 \end{gathered}$$

⇒ 160 < R' < 200
Any R' between 160 Ω and 200 Ω can achieve a null point null pint is on the 4th segment of potentiometer wire.
So, 400 cm of wire > 8 V is the potential drop.
Potential gradient (k),
k × 400 cm > 8 V
 ⇒ k × 4 m > 8 V
⇒ k > 2 V/m         .....(1)
Similarly, as balance point is at 4th segment of the wire, so, no balance point at 3 m.
k × 300 cm < 8 V = k × 3 m < 8 V
⇒ k < $\frac{8}{3}$  V/m      .....(2)
By using equation (1) and (2), we get:

​$\frac{8}{3} \mathrm{V}/\mathrm{m} > k > 2 \mathrm{V}/\mathrm{m}$

Answer:

Given.
Area of cross-section, A = 1 mm² = 10^{– 6} m²
Resistance, R = 6 Ω
Voltage, V = 6 V

n = $\frac{{10}^{29}}{m^2}$

Applying Ohm's law, V = IR
                              ⇒ 6 = I (6)
                              ⇒ I = 1 A
a. Current, I = Ane V_d
           ⇒ V_d = $\frac{I}{\mathrm{Ane}}$
           ⇒ V_{d = $\frac{1}{{10}^{-6}\times {10}^{29}\times 1.6\times {10}^{-19}}$}
           
          ⇒ V_{d = $\frac{{10}^{-4}}{1.6} \mathrm{m}/\mathrm{s}$}
Number off electrons in wire = n (volume of the wire)
                                               = n × Al
Here, energy is absorbed in the form of kinetic energy.

$\Rightarrow E=\frac{1}{2} m{v}_d^2 \mathrm{A}nl$

      $=\frac{1}{2}\times 9.1\times {10}^{-31}\times \frac{{10}^{-4}\times {10}^{-4}}{1.6\times 1.6}\times {10}^{29}\times {10}^{-6}\times {10}^{-1}$

      $=\frac{9.1}{2\times 2.56}\times {10}^{-17}=1.78\times {10}^{-17}\approx 2\times {10}^{-17}$

b. Power loss, P = I²R = (1)² × 6 = 6 j/s
Energy, E = P × t
         $\Rightarrow t=\frac{E}{P}=\frac{\cancel{2}\times {10}^{-17}}{{\cancel{6}}_3}=0.33\times {10}^{-17} \sec$
Hence, the time scale is approximately 3 × 10^–18 sec.