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Page No 5:

The unit joule/coulomb is called by another name, volt. Volt is the SI unit of potential difference.
Potential difference = Work done/Quantity of charge moved

Page No 5:

(b) A volt is a joule per coulomb.
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to another.
1 volt = 1 joule / 1 coulomb

Page No 5:

(a) The acronym p.d. stands for the potential difference between two points in an electric circuit.
(b) Potential difference (p.d.) is measured by a device called voltmeter.

Page No 5:

When we say that the electric potential at a point is 1 volt, it means that 1 joule of work has been done in moving 1 coulomb of positive charge from infinity to that point.

Page No 5:

Charge given, Q = 1 C
Potential difference, V = 1 V
Thus, the amount of work done, W in moving the charge is:

W = V × Q
= 1 V × 1 C
= 1 J

Page No 5:

The SI unit of potential difference is volt and is denoted by the letter, V.

Page No 5:

The amount of charge, Q that flows between two points at a potential difference, V (= 12 V) is 2 C.
Therefore, the amount of work done, W in moving the charge is
given by:

W = V × Q
= 12 V × 2 C
= 24 J

Page No 5:

The SI unit of electric charge is coulomb and is denoted by the letter, C.

Page No 5:

One coulomb is the quantity of electric charge which exerts a force of $9×{10}^{9}\mathrm{N}$ on an equal charge placed at a distance of one metre from it.

Page No 5:

(a) Potential difference is measured in volts by using a voltmeter placed in parallel across a component.
(b) Copper is a good conductor. Plastic is an insulator.

Page No 5:

The substances through which electricity can flow are called conductors. Two examples are copper and aluminium.

The substances through which electricity cannot flow are called insulators. Two examples are glass and paper.

Page No 5:

Conductors: silver, copper, aluminium, nichrome, graphite, manganin, mercury
Insulators: cotton, air, paper, porcelain, mica, bakelite, polythene, sulphur

Page No 5:

The electric potential (or potential) at a point in an electric field is defined as the work done in moving a unit positive charge from infinity to that point. It is denoted by the symbol, V and its unit is volt.

Page No 5:

(a) The potential difference between two points in an electric circuit is equal to the amount of work done in moving a unit charge from one point to another point.

(b) Here:
Potential difference, V = 230 − 220 = 10 V
Charge moved, Q = 4 C
So, using the relation V = W/Q, we have:
W = V $×$ Q
= 10 $×$ 4 = 40 J

Page No 5:

(a) The potential difference across a conductor is measured by an instrument called voltmeter.
(b) The term 'each coulomb' means 'every 1 coulomb'.
So, charge, Q = 1 C
Potential difference, V = 12 V
Now:
VW/Q
12 = W/1
So, work done, W = 12 x 1 = 12 J
Since the work done on each coulomb of charge is 12 J, the energy transferred to each coulomb of charge is also 12 J.

Page No 6:

(a) The potential difference between two points in an electric circuit is equal to the amount of work done in moving a unit charge from one point to another point.

The SI unit of potential difference is volt.

(b) The potential difference between two points is said to be one volt if one joule of work is done in moving one coulomb of electric charge from one point to the other.
Thus:
1 volt = 1 joule/1 coulomb

(c) Here:
Work, W = 250 J
Charge, Q = 20 C
Using the relation, V = W/Q:
Potential difference, V = 250/20 = 12.5 V

(d) A voltmeter is an instrument that is used to measure the potential difference across a conductor.
A voltmeter is always connected in parallel across the points where the potential difference is to be measured.

Suppose we have a conductor AB, such as a resistance wire, and we want to measure the potential difference across its ends.
So one end of the voltmeter is connected to point A and the other end to the point B. We can read the value of potential difference in volts on the dial of the voltmeter.

(e) A voltmeter has a high resistance so that it takes a negligible current from the circuit.
In a circuit, a voltmeter is connected in parallel across the element of whose potential difference it has to measure. Since we don't want the current diverted through the voltmeter, the voltmeter has a high internal resistance so that the circuit maintains the same current as if the voltmeter was not present.

Page No 6:

(b) potential difference
The work done in moving a unit charge across two points in an electric circuit is a measure of potential difference.

Page No 6:

d) voltmeter
It is connected in parallel across the points where the potential difference is to be measured.

Page No 6:

(d) coulomb
It is the SI unit of electric charge.

Page No 6:

(c) volt
It is the SI unit of potential difference.

Page No 6:

(d) 6.25 × 1018 electrons
Since the charge of an electron is $1.6×{10}^{-19}\mathrm{C}$, 6.25 × 1018 electrons taken together constitute (6.25 × 1018 × 1.6 × 1019 C) or 1 coulomb of charge.

Page No 6:

(a) The potential difference across each cell is 2 V. So, the potential difference at the terminals of the battery is 3 x 2 V = 6 V.

(b)
(i) For one cell:
V = 2 V
Q = 1 C
And also:
Electrical energy gained = Work done
So using the relation, VW/Q or W = V x Q
= 2 x 1 = 2 J
(ii) For three cells:
V = 6 V
Q = 1 C
So, W = V x Q = 6 x 1 = 6 J
So 2 joules and 6 joules of electrical energy is gained for the above two cases (i) one cell and (ii) all three cells, respectively.

Page No 6:

In a copper atom, the electrons in the outermost orbit are lightly bound to the nucleus of the atom and so they move freely. This free flow of electrons makes copper a good conductor of electricity. In rubber, the electrons in the outermost orbits are tightly bound to the nucleus. So, they do not flow freely and hence, rubber is not a good conductor of electricity.

Page No 11:

(a) A cell or a battery helps to maintain potential difference across a conductor.
(b) Here:
Potential difference, V = 10 V
Current, I = 2 A
Time, t = 1 minute = (1 x 60) s = 60 s
We know that the amount of energy transferred is equal to the work done.
Now:
VW/Q
Or W = V x Q
= V x I x t (since Q = It)
= 10 x 2 x 60
= 1200 J
Thus, the amount of energy transferred is 1200 J.

Page No 11:

(a) An electric current is the flow of electric charge called electrons in a conductor and its magnitude is the amount of electric charge passing through a given point of the conductor in one second. The potential difference between the ends of the wire makes the current flow through the wire.

(b) The SI unit of electric current is ampere.
When one coulomb of charge flows through any cross-section of a conductor in one second, the electric current flowing through it is said to be one ampere.
1 ampere=

Page No 11:

An ammeter is an instrument that is used to measure electric current. It is always connected in series with the circuit in which the current is to be measured.

Suppose we want to find out the current flowing through a conductor AB. Here, we should connect the ammeter in series with the conductor AB as above.

Page No 11:

(a) If a charge of Q coulombs flows through a conductor in time t seconds, the magnitude, I of the electric current flowing through it is given by:
$I\mathit{=}\frac{\mathit{Q}}{\mathit{t}}$
(b) Given:
Current, I = 0.36 A
Time, t = 15 minutes = (15 $×$ 60) s = 900 s
Using the formula, $I\mathit{=}\frac{\mathit{Q}}{\mathit{t}}$:
or Q = I $×$ t = 0.36 $×$ 900 = 324 C
Thus, the amount of electric charge that flows through the circuit is 324 C.

Page No 11:

(a) An ammeter should have a very low resistance so that it will not change the value of the current flowing in the circuit.
(b) A voltmeter should have a high resistance so that it can take only a negligible amount of current from the circuit.

Page No 11:

(a) Fixed resistance

(b) Variable resistance

(c) A cell

(d) A battery of three cells

(e) An open switch

(f) A closed switch

Page No 11:

A circuit diagram is a diagram that indicates how different components in a circuit have been connected, using electrical symbols for the components.
Here is the diagram of an electric circuit comprising a cell, a resistance, an ammeter, a voltmeter and a closed switch.

A voltmeter has a high resistance so that it takes only a negligible amount of current from the circuit.

Page No 11:

Given:
Current, I = 1 A
Time, t = 1 s
Using the formula:

Now, if the charge is $1.6×{10}^{-19}\mathrm{C}$, the number of electrons is 1.
So, if the charge is 1 C, the number of electrons is given by:

$\frac{1}{1.6×{10}^{-19}}×1=6.25×{10}^{18}$
Thus, electrons should pass through a conductor in 1 second to constitute 1 ampere current.

Page No 11:

Amount of energy = Work done
(a) Using the formula, VWQ, we have, W = V x Q
= 12 x 1 = 12 J

(b) W = V x Q = 12 x 5 = 60 J

(c) W = V x Q
= V x (I x t)
= 12 x 2 x 10 = 240 J

Page No 11:

The electric current is the physical quantity that is expressed as coulomb/second since current, IQ/t.

Page No 11:

The flow of electric charge in a conductor is called electric current. If a charge of Q coulombs flows through a conductor in time t seconds, the magnitude of electric current is given by, IQ/t.

Page No 11:

When we switch on a light, electrons travel through the wires. The flow of these electrons is called electric current.

Page No 11:

The flow of negatively charged particles (electrons) in a metallic conductor constitutes electric current.

Page No 11:

(a) The direction of conventional current is from the positive terminal of a cell or battery to the negative terminal through the outer circuit.
(b) The flow of electrons is from the negative terminal to the positive terminal of a cell.

Page No 11:

Because current, I = Q/t, the correct relationship is .

Page No 11:

The SI unit of electric current is ampere and is denoted by the letter, A.

Page No 11:

(a) 1 ampere = 1000 milliamperes =
(b) 1 ampere = 1,000,000 microamperes = ${10}^{6}\mathrm{microamperes}$

Page No 11:

Ammeter.
Ammeter is connected in series with the circuit, whereas voltmeter is connected in parallel across the points where the p.d. is to be measured.

Page No 11:

The ammeter is connected in series with the circuit in which the current is to be measured, whereas the voltmeter is connected across the points where the potential difference is to be measured.

Page No 11:

(i) Variable resistance or rheostat. A rheostat is used to change the resistance in a circuit.
(ii) A closed switch. When the switch is closed, the circuit is completed and current flows through it.

Page No 11:

Here:
Charge, Q = 20 C
Time, t = 1 s
We know that, IQ/t
So:
Current, I = 20/1 = 20 A
Thus, the electric current flowing through the circuit is 20 A.

Page No 11:

Here:
Current, I = 4 A
Time, t = 10 s
We know that IQ/t.
So:
4 = Q/10
or Q = 4 x 10 = 40 C
Thus, the amount of electric charge that flows through the circuit is 40 coulombs.

Page No 11:

Here:
Charge, Q = 20 C
Time, t = 40 s
We know that, IQ/t.
So:
I = 20/40 = 0.5 A
Therefore, the electric current flowing through the circuit is 0.5 A.

Page No 11:

(a) A current is a flow of electrons. For this to happen, there must be a closed circuit.
(b) Current is measured in amperes using an ammeter placed in series in a circuit.

Page No 12:

Given:
Time, t = 10 s
Charge, Q = 25 C
Energy delivered = Work done, W = 200 J
(a) Using the relation, VW/Q
V = 200/25 = 8 V
So the p.d. across the battery is 8 volts.
(b) Using the relation, IQ/t,
I = 25/10 = 2.5 A
So, a current of 2.5 A flows from the battery.

Page No 12:

(a) Electric current is the flow of electric charges (called electrons) in a conductor such as a metal wire and its magnitude is the amount of electric charge passing through a given point of the conductor in one second.
i.e. IQ/t
The SI unit of electric current is ampere.

(b) Here:
Charge, Q = 1 C
Time, t = 1 s
Then the current is given by:
IQ/t
= 1/1 = 1 A

(c) Current is measured by an instrument called ammeter. An ammeter should be connected in series with the circuit in which the current is to be measured.

(d) The conventional direction of the flow of electric current is from the positive terminal of a cell or a battery to its negative terminal through the outer circuit, whereas the direction of flow of electrons is from the negative terminal to the positive terminal of a cell.

(e) Given:
Charge, Q = 10 C
Time, t = 0.01 s
Then the current, IQt
So I = 10/0.01 = 1000 A
Also given that:
V = 10 mV = 1,00,00,000 V
So energy = Work done, W = V x Q
= 1,00,00,000 x 10 = 1,00,000,000 J = 100 MJ

Page No 12:

(c) voltage
The term voltage came from volt, which is the SI unit of potential difference.

Page No 12:

The correct statement is:
(d) 1
An ammeter is connected in series in a circuit and a voltmeter is connected in parallel. The other two statements are incorrect as an ammeter should have a very low resistance and a voltmeter should have a high resistance.

Page No 12:

(d) Ampere
It is the SI unit of electric current.

Page No 12:

(d) 50 C

Here, current, I = 5 A
Time, t = 10 s
So, using the relation, IQ/t:
Q = I x t
Charge, Q = 5 x 10 = 50 C
Thus, a charge of 50 C is passed in 10 s.
Hence, the correct option is (d) 50 C.

Page No 12:

(c) 0.5 A

Here, charge, Q = 300 C
Time, t = 10 minutes = (10 x 60) s = 600 s
Then, current, IQ / t
= 300 / 600 = 0.5 A
Hence, the correct option is (c) 0.5 A.

Page No 12:

(a) The lamps are connected in series.
(b) The student has connected the ammeter in parallel with the lamps. It should be connected in series.
(c) The correct diagram that shows how to connect the circuit:

Page No 12:

Here is the circuit diagram that shows how three bulbs can be lit from a battery so that two bulbs are controlled by the same switch. The third bulb has its own switch.

Page No 12:

Here, potential, V = 230 V
Current, I = 8 A
(a) The magnitude of the electric current is the amount of charge passing through a given point in one second. So, in this case, a charge of eight coulombs flows around the circuit each second.

(b) Energy transferred = Work done, W
Using the relation, VW/Q:
W = V x Q
= 230 x 8 = 1840 J
Thus, 1840 joules of energy is transferred to the heater each second.

Page No 12:

Current, I = 5 A
Time, t = 1 s
Using the relation, IQt:
Q = I x t
= 5 x 1 = 5 C
Now, if the charge is $1.6×{10}^{-19}C$, the number of electrons is one.
So, if the charge is 5 C, the number of electrons is given by:
$\frac{1}{1.6×{10}^{-19}}×5=31.25×10$18

Page No 18:

Ohm's law gives the relationship between the current in a conductor and the potential difference across its ends. According to this law, at constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends.

Page No 18:

The SI unit of electrical resistance is ohm and its symbol is $\mathrm{\Omega }$.

Page No 18:

Electrical resistance is the physical quantity whose SI unit is ohm.

Page No 18:

Objects that have infinitely high electrical resistance are called insulators. An insulator does not allow electricity to flow through it.

Page No 18:

From Ohm's law we have:
V / I = R
Or I = V/R
Now, according to the question, the resistance is constant and the potential difference is halved.
i.e. R' = R
and V' = V/2
Then the new current is given by:
I' = V'/R'
= V/2/R
= (1/2) (V/R)
= (1/2) I
It is clear from the above statement that the current becomes half.

Page No 18:

The strength of the electric current flowing in a given conductor depends on two factors:
1) The potential difference across the ends of the conductor
2) The resistance of the conductor

Page No 18:

A thick wire has a greater area of cross-section, whereas a thin wire has a smaller area of cross-section. Also, the resistance of a conductor is inversely proportional to its area of cross-section. This means that a thick wire has lesser resistance than a thin wire.

Page No 18:

From Ohm's law we have:
V/I = R
Or I = V/R
Now, according to the question, the resistance of the circuit is halved and the potential difference is constant.
i.e. R' = R/2
and V' = V
Then the new current is given by:
I' = V'/R'
= V/R/2
= 2V/R
= 2I
It is clear from the above statement that the current becomes double.

Page No 18:

Here:
Potential difference, V = 20 V
Resistance, R = 5 Ω
Substituting these values in Ohm's equation, V/I = R:
Current, I = V/R = 20/5 = 4 A
Thus, the current flowing in the circuit is 4 A.

Page No 18:

Here:
Resistance, R = 20 Ω
Current, I = 2 A
Substituting these values in the Ohm's equation, V/I = R:
Potential difference, V = I x R = 2 x 20 = 40 V

Page No 18:

Here:
Current, I = 5 A
Potential difference, V = 3 V
Substituting these values in the Ohm's equation, V/I = R:
Resistance of the wire, R = 3/5 = 0.6 Ω

Page No 18:

Ohm's law state a relation between potential difference and current.

Page No 18:

Substances that conduct electricity easily and offer very low resistance (opposition) to the flow of current are called good conductors. Examples: copper, silver

A resistor is an electrical component that limits or regulates the flow of electrical current in an electronic circuit. Resistors have comparatively high electrical resistance. Examples: constanan, nichrome

The substances that have infinitely high electrical resistance are called insulators. Examples: wood, rubber

Page No 18:

Conductors: mercury, aluminium, iron and metal coin

Resistors: nichrome and manganin

Insulators: rubber, polythene, wood, paper, thermocol and bakelite

Page No 19:

(a) Electricians wear rubber hand gloves while working with electricity because rubber is an insulator and it will prevent them from getting electric shock, in case of short circuit or overloading.

(b) We know that V = IR

Here, I = 6 A

R = 40 Ω

So, V = 6 x 40

V = 240 V

So 240 V is the potential difference needed to send a current of 6 A through an electrical appliance that has a resistance of 40 Ω.

Page No 19:

(i)

(ii) When potential difference = 0.8 V, current = 0.32 A.
$\frac{\mathrm{V}}{\mathrm{I}}=\frac{0.8}{0.32}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{V}}{\mathrm{I}}=2.5$
When potential difference = 1.2 V, current = 4.8 A.
$\phantom{\rule{0ex}{0ex}}\frac{\mathrm{V}}{\mathrm{I}}=\frac{1.2}{4.8}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{V}}{\mathrm{I}}=2.5$
When potential difference = 1.6 V, current = 6.4 A.
$\frac{\mathrm{V}}{\mathrm{I}}=\frac{1.6}{6.4}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{V}}{\mathrm{I}}=2.5$
The V-I graph is a straight line with a constant slope of 2.5.

(iii)  The resistance of the wire is 2.5 Ω.

Page No 19:

(a) The ratio of potential difference and current is known as resistance R.
(b)

From the graph, resistance, R =

(c) Ohm's law is illustrated by the above V−I graph.

(d)Relation between potential difference and current is
V = IR

(e) The potential difference between the terminals of an electric iron is 240 V and the current is 5.0 A.
So resistance
Thus, the resistance of the electric iron will be 48 Ω.

Page No 19:

(c) 2 A

If the p.d. across a 3 Ω resistor is 6 V, the current flowing in the resistor will be 2 A as current (I) is given by the equation, I = V (Voltage) / R (Resistance).

or  I = V/R

I = 6 V/3 Ω

I = 2 A

Page No 19:

(d) 24 Ω

Voltage, V = 12 V
Current, I = 0.5 A
Resistance, R = Voltage/Current
R = V/I
R = 12/0.5
R = 24 Ω

Thus, if a car's headlight bulb working on a 12 V car battery draws a current of 0.5 A, the resistance of the light bulb will be 24 Ω.

Page No 19:

(c) 9.2 A

Resistance = 25 Ω

Voltage = 230 V

Current = Voltage/Resistance

I = V/R

I = 230/25

I = 9.2 A

Thus, if an electrical appliance has a resistance of 25 Ω and when that electrical appliance is connected to a 230 V supply line, the current passing through it will be 9.2 A.

Page No 19:

(b) 3
The number of coulombs passing through the resistor is the current passing through it.
Current = Voltage/Resistance
I = V/R
I = 12/4
I = 3 A
Thus, when a 4 Ω resistor is connected across the terminals of a 12 V battery, the number of coulombs passing through the resistor per second will be 3.

Page No 19:

According to Ohm's law, the current (I) flowing through a wire is directly proportional to the potential difference (V) across it, provided its temperature remains the same.
Thus I α V.
V / I = Constant = R
V = IR
Here R is the resistance.
The unit of resistance is ohm (â„¦).

One ohm (â„¦) is the resistance of a conductor when a potential difference of one volt is applied to its ends and a current of one ampere flows through it.

Page No 19:

(a) Resistance is the property of a conductor to resist the flow of charge through it. The relation between resistance, current and potential difference is V = IR.
(b) Voltage, V = 12 V
Resistance, R = ?
Current, I = 2.5  A
VIR
R = V/I
R = 12/2.5
R = 4800 Î

Page No 19:

(a) One ohm (Ω) is the resistance of a conductor when a potential difference of one volt is applied to its ends and a current of one ampere flows through it.

(b) The resistance of a wire can be represented by the following equation:

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

So, when the conductor is made thinner, its area of cross-section will decrease, since resistance is inversely proportional to the area of the conductor. Therefore, its resistance will increase.

(c) We know that VIR

If V is constant, resistance is doubled  R' = 2R.

Then V' = I'R'

V' = V since V is constant

$V=I\text{'}×2×R$

$I\text{'}=\frac{V}{2R}=\frac{I}{2}$

Thus, the current will get reduced to half.

Page No 20:

(d) current and potential difference
Ohm's law gives the relationship between current and potential difference.

Page No 20:

(d) ohm
The unit of electrical resistance is ohm.

Page No 20:

(d) insulator
A substance having infinitely high electrical resistance is called an insulator.

Page No 20:

(b) half

As we know from Ohm’s law:
Voltage = Current x Resistance
V = IR
If the voltage is constant, the resistance is doubled and the current becomes half.

Page No 20:

(d) double

As we know from Ohm’s law:
Voltage = Current x Resistance
V = IR
If the voltage is constant, the resistance of the circuit is halved. That is, it becomes R / 2.
Current, I = V/R
I = V/R
I = V/R/2 = 2 I
Thus by keeping the p.d. constant, the resistance of a circuit is halved and the current is doubled.

Page No 20:

An electric room heater draws a current of 2.4 A from the 120 V supply line.

So current, I = 2.4 A

Voltage, V = 120 V

Resistance, R = V/I

R = 120/2.4 = 50 Ω

The resistance will remain constant unless the room heater is changed.

Now when the room heater is connected to the 240 V supply line, the current drawn is:

I = V/R

I = 240/50 = 4.8 A

So the room heater will draw a 4.8 A current when connected to the 240 V supply line.

Page No 20:

Resistance is represented by the symbol Omega, â„¦. Resistance has the following electrical properties:

1. The resistance of a conductor depends on its length. It is directly proportional to the length, R α L.
2. The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A.
3. Resistance depends on the nature of the material of the conductor.
4. Resistance also depends on the temperature of the conductor.

Page No 20:

(a) Ohm's law states that the graph between V and I for a conductor is a straight line passing through the origin.

(b) Ohm's law is true only if the temperature of the conductor is constant.

Page No 20:

A potential difference of 10 V is needed to make a current of 0.02 A flow through a wire.
So potential difference, V = 10 V
Current, I = 0.02 A
Resistance of the wire, R = V/I
R = 10/0.02 = 500 â„¦
The resistance will remain the same unless the wire is changed.
Now to make a current of 250 mA to flow through the wire, the potential difference required is:
V = IR
V = 250 x 10-3 x 500
V = 125 V
So a potential difference of 125 V is needed to make a current of 250 mA flow through the wire.

Page No 20:

A current of 200 mA flows through a 4 kΩ resistor.
So current, I = 200 mA = 200 x 10-3A
Resistance, R = 4 kΩ = 4000 Ω
Potential difference, V = IR
V = 200 x 10-3 x 4000
V = 800 V
So the potential difference across the resistor will be 800 V.

Page No 26:

The resistance of a conductor is inversely proportional to its area of cross-section, i.e.R α 1 / A. So when the conductor is made thicker, its resistance decreases.

Page No 26:

We know that the resistance of the wire, R = ρl / A where the resistivity of the wire is ρ.
The length of the wire = l
The area of cross-section of the wire, A = πr2
So when the length of the wire is doubled (l' = 2l) by taking more of the same wire, (it will keep the resistivity and the area of cross-section the same), the length is doubled.
New resistance = R'
R' = ρl' / A
R' = 2ρl / A
R' = 2R

Page No 26:

The resistance of a conductor depends on the following factors:

1. The resistance of a conductor depends on its length. It is directly proportional to the length, R α l.
2. The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A.
3. Resistance depends on the nature of the material of the conductor.
4. Resistance also depends on the temperature of the conductor.

Page No 26:

Silver is the best conductor of electricity.

Page No 26:

Iron is a better conductor of electricity than mercury because the resistivity of iron is less than that of mercury.

Page No 26:

Copper and aluminium wires are used for electrical transmission because they have a low resistance, which makes them good conductors of electricity.

Page No 26:

The coils of the heating element of an electric iron is made of alloys like nichrome because alloys have a high resistance and a high melting point.

Page No 26:

Nichrome is an alloy of nickel, chromium, manganese and iron. It is used to make the heating elements of electrical appliances because it has a high resistance and a high melting point.

Page No 26:

Nichrome is used to make the heating elements of electrical appliances because it has a high resistance and a high melting point.

Page No 26:

The coils of the heating elements of electric irons and electric toasters are made of alloys like nichrome rather than a pure metal because alloys have a high resistance and a high melting point.

Page No 26:

(a) A long piece of nichrome wire has more resistance because the resistance of a conductor is directly proportional to its length, R α l.
(b) A thin piece of nichrome wire has more resistance because the resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A
. The lesser the area of cross-section, the more the resistance.

Page No 26:

(a) The resistance of all pure metals increases on raising the temperature and decreases on lowering the temperature. Therefore, the resistance of a pure metal will decrease on decreasing the temperature.
(b) The presence of impurities in a metal increases its resistivity (also resistance) and it does not undergo oxidation easily even at a high temperature.

Page No 26:

Resistance is measured in ohm (Ω). The resistance of a wire increases as the length increases; as the temperature increases; and as the cross-sectional area decreases.

Page No 26:

â€‹

(a) We know that resistance,  $R=\frac{\mathrm{\rho }l}{A}$

So resistivity,  $\mathrm{\rho }=\frac{RA}{l}$

When the area of cross-section, A is 1 m2 and the length of the wire, l is 1 m, resistance is equal to resistivity.

Resistivity may be defined as the resistance offered by a conductor of length 1 m and area of cross-section 1 m2.

(b) Given length of the wire, l = 1 m

Diameter of the wire, d = 0.2 mm

So the radius of the wire, r = 0.1 mm = 0.0001 m

Area of cross-section of the wire =

Resistance of the wire, R = 10 Ω

So resistivity = $\mathrm{\rho }=\frac{RA}{l}$

Thus, the resistivity of the material, â€‹.

Page No 26:

(a) Length of the cable wire =

Area of cross-section = A
Resistivity of the aluminium cable wire = $\mathrm{\rho }$
Resistance of the wire = R

Then,  $R=\frac{\mathrm{\rho }l}{A}$

(b) Length of the cable wire, l = 2 m
Area of cross-section, A = 1.55 × 10−6 m2
Resistivity of the aluminium cable wire, $\mathrm{\rho }$ = 2.8 × 10−8 Ωm

We know that:

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

$R=\frac{2.8×{10}^{-8}×2}{1.55×{10}^{-6}}$

R = 0.036 Ω
Thus, the resistance of the metal wire, R = 0.036 Ω

Page No 26:

(a) Copper and aluminium are good conductors of electricity because their resistivity is very low.

(b) Length of the cable wire, l = 1 km = 1000 m

Diameter of the cable wire, d = 0.5 mm
Radius of the cable wire, r = 0.25 mm = 0.00025 m = 2.5 x 10-4 m
Area of cross-section,
Resistivity of the aluminium cable wire = 1.7 × 10−8 Ω m.

To find the resistance of the aluminium cable, R:

We know that $R=\frac{\mathrm{\rho }l}{A}$

$R=\frac{1.7×{10}^{-8}×1000}{19.6×{10}^{-8}}$

So, R = 86.5 Ω

Thus, the resistance of the copper wire, R = 86.5 Ω

Page No 26:

The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A. So as the conductor is a thick wire, its resistance decreases and current will flow easily through it, in comparison with a thin wire.

Page No 26:

1. The resistance of a conductor depends on its length. It is directly proportional to the length, R α l.
2. The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A.
3. The resistance also depends on the temperature of the conductor. It increases on raising the temperature and decreases on lowering the temperature.

Page No 26:

(a) If we take two similar wires with equal length and diameter, one made of copper (metal) and the other made of nichrome (alloy), we will find that the resistance of the nichrome wire is about 60 times more than that of the copper wire. This shows that the resistance of a conductor depends on the nature of the material of the conductor.

(b) Length of the cable wire, l = 10 km = 10000 m
Diameter of the cable wire, d = 2.0 mm
Radius of the cable wire, r = 0.1 mm = 0.0001 m = 10-4 m
Area of cross-section, $A=\mathrm{\pi }{r}^{2}$ = 3.14 × 10-8 m2
Resistivity of the aluminium cable wire = 27 × 10−8 Ω m.

Let the resistance of the aluminium cable be R.

We know that $R=\frac{\mathrm{\rho }l}{A}$

$R=\frac{27×{10}^{-8}×10000}{3.14×{10}^{-8}}$

R = 86 Ω

Page No 26:

(a) The resistance of a conductor depends on its length. It is directly proportional to the length, R α l. So on increasing the length of the wire, its resistance will increase.
(b) The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A. So on increasing its diameter, its resistance will decrease.
(c) Resistance also depends on the temperature of the conductor. So on increasing its temperature, its resistance will also increase.

Page No 26:

(a)The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A. So on increasing the area of cross-section, its resistance will decrease.
(b) We know that the area of cross-section, Aπr2 = π (d / 2)2, where r is the radius and d is the diameter of the wire. So on increasing the diameter of the wire, its resistance decreases.

Page No 27:

(a) We know that resistance,
$R=\frac{\mathrm{\rho }l}{A}$

So resistivity,

$\mathrm{\rho }=\frac{RA}{l}$
When the area of cross-section, A = 1 m2 and the length of the wire, l = 1 m, $\mathrm{\rho }=R$.

Resistivity is the resistance of the conductor whose area of cross-section, A is 1 m2 and length, l is 1 m.

(b) The SI unit of resistivity is Ω m or ohm-metre.
1 ohm-metre is the resistance of a conductor that has a resistance of 1 ohm and whose area of cross-section, A is 1 m2 and length, l is 1 m.

(c) Difference between resistance and resistivity:

 Resistance Resistivity 1. It depends on the dimensions of the conductor like radius and length. 2. Resistance of a given material may change at constant temperature. 1. It does not depend on the dimensions of the conductor. 2. Resistivity of a given material does not change at constant temperature.

(d) Resistivity of a substance depends on the material and the temperature of the substance. It does not depend on the dimensions of the substance like length, radius etc.

(e) Length of the wire, l = 1 m

Diameter of the wire, d = 0.3 mm

So the radius of the wire, r = 0.15 mm = 0.00015 m = 1.5 x 10-4 m

Area of cross-section of the wire,

Resistance of the wire, R = 26 Ω

So resistivity,
$\rho =\frac{RA}{l}$

Thus, the resistivity of the material is .

Page No 27:

(d) 30 Ω

Resistance, $R=\frac{\mathrm{\rho }l}{A}$

Length, l = 300 m

Cross-section area, A = 1.0 mm2 = 10-6 m2

Resistivity, $\mathrm{\rho }$ = 1.0 × 10−7 Ωm

Resistance, $R=\frac{{10}^{-7}×300}{{10}^{-6}}$

R = 30 Ω

Page No 27:

(d) one-fourth
R = ρ $\frac{\mathit{l}}{\mathit{A}}$

When the diameter is doubled, d' = 2d

Radius, r' = 2r

Area of cross-section, $A\text{'}=\mathrm{\pi }r{\text{'}}^{2}=\mathrm{\pi }\left(2r{\right)}^{2}=4\mathrm{\pi }{r}^{2}=4A$

The area of cross-section will increase by four times.

Then the new resistance,$R\text{'}=\frac{\mathrm{\rho }l}{A\text{'}}$

$R\text{'}=\frac{\mathrm{\rho }l}{4A}$

$R\text{'}=\frac{R}{4}$

Thus, the resistance will get reduced by four times.

Page No 27:

(c) 60 Ω

If the resistance of a certain copper wire is 1 Ω, the resistance of a similar nichrome wire will be about 60 Ω because the resistivity of nichrome is 60 times the resistivity of copper.

Page No 27:

(a) four times
Resistance of the wire is given by:
R = ρ $\frac{\mathit{l}}{\mathit{A}}$
When the diameter is halved:
$d\text{'}=\frac{d}{2}$

$r\text{'}=\frac{r}{2}$

Area of cross-section:
$A\text{'}=\mathrm{\pi }r{\text{'}}^{2}=\mathrm{\pi }{\left(\frac{r}{2}\right)}^{2}=\frac{\mathrm{\pi }{r}^{2}}{4}=\frac{A}{4}$

Area of cross-section will get reduced by four times.

Then the new resistance:
$R\text{'}=\frac{\mathrm{\rho }l}{A\text{'}}$

$R\text{'}=\frac{4\mathrm{\rho }l}{A}$

$R\text{'}=4R$

Thus, the resistance will increase by four times.

Page No 27:

(d) a semiconductor
The resistivity of a certain material is 0.6 Ωm. The material is most likely to be a semiconductor because it has moderate resistivity.

Page No 27:

(b) half
We know that the resistance of a conductor is given by:
$\mathrm{R}=\mathrm{\rho }\frac{\mathrm{l}}{\mathrm{A}}$
where ρ = resistivity
l = length of the conductor
A = area of the cross-section of the conductor
Let the new resistance be R' when the area of cross-section of the conductor is doubled
$\phantom{\rule{0ex}{0ex}}{\mathrm{R}}^{,}=\mathrm{\rho }\frac{\mathrm{l}}{2\mathrm{A}}$

Thus, the new resistance becomes half of the previous one.

Page No 27:

(c) temperature
The resistivity of copper depends only on temperature.

Page No 27:

(b) 2 times
The resistance of a conductor is inversely proportional to its area of cross-section, i.e. R α 1 / A. So, when the area of cross-section of a resistance wire is halved, its resistance will increase by two times.

Page No 27:

We know that the resistance of a conductor is given by:

where ρ = resistivity
l = length of the conductor
A = area of cross-section of the conductor
Now the length is increased to twice the original length. Then let the new resistance be denoted by R'.

Thus, the new resistance will become four times.

Page No 27:

(a) electric wires: They are made of conductors. Conductors have a low resistivity and so, we will use material Q.

 Q 2.63 × 10−8 Ω m:  low resistivity

(b) handle for soldering iron: It is made of an insulator. An insulator has a high resistivity and so, we will use material R.

 R 1.0 × 1015 Ω: high resistivity

(c) solar cells: They are made of semiconductors. A semiconductor has moderate resistivity and so, we will use material P.

 P 2.3 × 103 Ω m: moderate resistivity

Page No 27:

The electrical resistivities of the four materials A, B, C and D are given. We can classify them as follows:

 A Semiconductor 110 × 10−8 Ωm B Insulator 1.0 × 1010 Ωm C Good conductor 10.0 × 10−8 Ωm D Resistor 2.3 × 103 Ωm

Page No 27:

(a) E is the best conductor of electricity because its resistivity is the least among the given values.
(b
) C
is a better conductor than A because its resistivity is less than that of A.
(c) B should be used for making heating elements of electric iron because its resistivity is the highest among the given values.
(d) E and C should be used for making electric wires because their resistivity is less.

Page No 27:

Resistance,
$R=\frac{\mathrm{\rho }l}{A}$
where l = length of  wire
A= area of cross-section of the wire

(i) When the length is tripled, l' = 3l

New resistance,
$R\text{'}=\frac{\mathrm{\rho }l\text{'}}{A}$

$R\text{'}=\frac{\mathrm{\rho }\left(3l\right)}{A}=\frac{3\mathrm{\rho }l}{A}$

$R\text{'}=3R$

Thus, the resistance will become three times the original resistance.

(ii) When the diameter is tripled, d' = 3d

Radius, r' = 3r

Area of cross-section, $A\text{'}=\mathrm{\pi }r{\text{'}}^{2}=\mathrm{\pi }\left(3r{\right)}^{2}=9\mathrm{\pi }{r}^{2}=9A$

Thus, the area of cross-section will become nine times.

Then the new resistance,
$R\text{'}=\frac{\mathrm{\rho }l}{A}$

$R\text{'}=\frac{\mathrm{\rho }l}{9A}$

$R\text{'}=\frac{R}{9}$

Thus, the resistance will be reduced by nine times.

(iii) When the material is changed to one whose resistivity is three times:

New resistivity, $\mathrm{\rho }\text{'}=3\mathrm{\rho }$
The new resistance,
$R\text{'}=\frac{\mathrm{\rho }\text{'}l}{A}$
â€‹
$R\text{'}=\frac{3\mathrm{\rho }l}{A}=3R$
Thus, the resistance will become three times the original resistance.

Page No 27:

Length of the wire, l =1.0 m
Resistance of the wire = 23 Ω
Resistivity of the material of the wire = 1.84 × 10−6 Ωm.
We know that,
$R=\frac{\mathrm{\rho }l}{A}$
Area of cross-section,
$A=\frac{1.84×{10}^{-6}×1}{23}$
A = 8 x 10-8 m2
Thus, the area the of cross-section of the wire, A = 8 x 10-8 m2

Page No 37:

In a series combination, the same current flows through each resistor. Let the supplied voltage, V be divided as V1, V2 and V3 in resistors R1, R2 and R3, respectively.
V = V1 + V2 + V3
IR = IR1 + IR2 + IR3
IR = I(R1 + R2 + R3)
R = R1 + R2 + R3
Thus, in the case of a series combination, the total resistance, R is the sum of the individual resistance R1, R2 and R3.

Page No 37:

If five resistors, each of value 0.2 Ω, are connected in series, the resultant resistance will be the sum of the individual resistances.
R = R1 + R2 + R3 + R4 + R5
R = 0.2 + 0.2 +0.2 + 0.2 + 0.2
R = 5 x 0.2
R = 1 Ω

Page No 37:

In a parallel combination, the voltage remains the same across each resistor but the current gets divided. Let I1, I2 and I3 be the current across resistors R1, R2 and R3, respectively.

I = I1 + I2 + I3

V = IR, V = I1R1, V = I2R2, V = I3R3

So, the total resistance is the sum of the reciprocal of the individual resistances.

Page No 37:

If three resistors, each of 3 Ω, are connected in parallel combination, the total resistance of the parallel combination will be:

The total resistance of the parallel combination is 1 Ω.

Page No 37:

The two resistors of 2 Ω each should be connected in parallel combination to produce equivalent resistance of 1 Ω.

R = 1

Total resistance, R = 1 Ω

Page No 37:

When two resistors, X and Y are connected in parallel combination, the total resistance will be less than the individual resistance. The total resistance in a parallel arrangement is given by:
$\frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}$
where as in a series combination, the total resistance is more than the individual resistances. The total resistance in a series combination is given by:
R = R1+ R2

Page No 37:

Case 1
Connecting a 2 Ω and 6 Ω resistor in series combination:
R = R1 + R2
R =2 $\mathrm{\Omega }$ + 6 $\mathrm{\Omega }$ =8 $\mathrm{\Omega }$
Thus, we will get an 8 Ω resistor.

Case 2
Connecting a 2 Ω and 6 Ω resistor in parallel combination:

Thus, we will get a 1.5 Ω resistor.

Page No 38:

(a) Case 1
Connecting two 4 Ω resistors in parallel combination:

(b) Case 2
Connecting two 4 Ω resistors in series combination:
R = R1 + R2
R = 4 $\mathrm{\Omega }$ + 4 $\mathrm{\Omega }$ = 8 $\mathrm{\Omega }$
Thus, we will get an 8 Ω resistor.
We should connect two 4 $\mathrm{\Omega }$ resistors in parallel to get a resistance of 8 $\mathrm{\Omega }$.

Page No 38:

In figure A, the resistance, RA = 10 Ω

In figure B, the resistors of 10 Ω and 1000 Ω are connected in parallel combination.
Therefore:

RB = 9.9 Ω
So, the arrangement B has lower resistance.

Page No 38:

A wire that has a resistance, R is cut into two equal pieces.
The resistance of the each piece will be  $\frac{\mathrm{R}}{2}$.

The two parts, each of resistance, $\frac{\mathrm{R}}{2}$ are joined in parallel.

The resistance of the parallel combination:

Thus, the resistance of the combination will be R/4.

Page No 38:

Case 1
Let the total resistance of the combination shown in the diagram below be R.
Here 500 Ω and 1000 Ω resistors are connected in series combination. Their net resistance is given by R = R1+ R2
Here:
R1=500 Ω
R2=1000 Ω
So:
R = 500 + 1000 = 1500 Ω
Thus, the total resistance of the combination is 1500 Ω.

Case 2
In the circuit diagram shown above, two 2 â„¦ resistors are connected in parallel combination.
Therefore:

Thus the total resistance of the combination is 1 Ω.

Case 3
In the circuit diagram shown above, two 4 â„¦ resistors are connected in parallel combination.

In the above circuit diagram, this 2 â„¦ and 3 â„¦ resistors are connected in series combination. So, the total resistance of the circuit can be calculated as:
R = R1 + R2
here
R1= 2 â„¦
R2= 3 â„¦
So:
R = 2 â„¦ + 3 â„¦ = 5 â„¦
Thus, the total resistance of the combination is 5 â„¦.

Page No 38:

In the given circuit, two 6 Ω and 4 Ω resistors are connected in parallel combination across a cell of 24 V. We know that in a parallel combination, the voltage across the resistor remains the same and only the current gets divided. So the current (I1) through the 6 â„¦ resistor is:

The current (I2) through the 4 â„¦ resistor is:

Page No 38:

The series resistance is given by R = R1 + R2 + R3 + R4. Hence, it is more than the individual resistance.
The parallel resistance is given by 1 / R = 1 / R1 + 1 / R2 + 1 / R3 + 1 / R4. Hence, it is less than the individual resistance.

Page No 38:

When five resistors are connected in series, the total resistance is given by:
R = R1+ R2+ R3+ R4+ R5
Here,
R1= 0.2 $\mathrm{\Omega }$,
R2= 0.3 $\mathrm{\Omega }$,
R3= 0.4 $\mathrm{\Omega }$,
R4= 0.5 $\mathrm{\Omega }$,
R5= 0.12 $\mathrm{\Omega }$

Therefore:
R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
R = 13.4 Ω
Total resistance of the circuit = 13.4 Ω
The current flowing through this series combination is given by I = V / R.
or I = 9 / 13.4 = 0.67 A
Now since the resistors are connected in series, the current flowing through each resistance is the same. Hence, the current through the 12 Ω resistor is equal to 0.67 A.

Page No 38:

(a) Total resistance, R = 20 Ω + 4 Ω = 24 Ω
(b) Current through the circuit,
I = V/R
I = 6/24
I = 0.25 A
(c) Potential difference across the electric bulb,
Vbulb = IRbulb
=0.25$×$20
= 5 V
(d) Potential difference across the resistance wire,
Vwire = IRwire
= 0.25$×$4
=  1 V

Page No 38:

(i) The resistance between B and C are in parallel. Hence the total resistance, R between B and C is given by:
RBC= [15 x 10 ] / [ 15 + 10 ] = 6 Ω ... (a)
Potential drop across B and C = IR = 1 x 6 = 6 V
Current through the 10 Ω resistor = V / R = 6 / 10 = 0.6 A
Current through the 15 Ω resistor = V / R = 6 / 15 = 0.4 A
(ii) p.d. across AB:
VAB = 5 x 1 = 5 V
p.d. across AC:
VAC = I x RAC
RAC = R + 5 = 11 Ω (see equation (a) )
VAC = 1 x 11 = 11 V
(iii) Total resistance, RAC = RBC + RAC
= 6 + 5
= 11 Î

Page No 38:

(1) The resistors of 6 Ω and 3 Ω are connected in series. Therefore, their net resistance can be calculated as:
R = R1 + R2
Here, R1 = 6 Ω
R2 = 3 Ω
So:
R = 6 Ω + 3 Ω = 9 Ω
The current through this branch, I = V/R
I = 4/9 = 0.44 A
In a series combination, the current remains the same. So the current through the 6 Ω resistor is 0.44 A.

(2) The current through the branch with resistors of 12 Ω and 3 Ω:
I = V/R
I = 4/(12 + 3) = 4 / 15 A
The potential difference across the 12 Ω resistance can be obtained by using the equation,
V = IR.
V = (4 / 15) x 12 = 3.2 V

Page No 39:

(a) For the minimum current flowing in the circuit, the resistors should be connected in series and for the maximum current in the circuit, the resistors should be connected in parallel with the battery.
(b) When the resistors are connected in parallel:

Total resistance = 3.33 Ω
Therefore, strength of the total current, I = V / R
I = 6 / 3.33
I = 1.8 A
When the resistors are connected in series, the resultant resistance is given by R = R1 + R2
Here, R1 = 5 $\mathrm{\Omega }$
R2= 10 $\mathrm{\Omega }$
So, R = 5 $\mathrm{\Omega }$ + 10 $\mathrm{\Omega }$ = 15 $\mathrm{\Omega }$
Total resistance = 15 Ω
Therefore, strength of the total current, I = 6 / 15 = 0.4 A

Page No 39:

(i) As shown in the figure, the resistors R2 and R3 are connected in parallel. Their total resistance is given by:

1/R = 1/R2+ 1/R3

Here, R2
= 3 Ω
R3 = 6 Ω
So,

1/R = 1/3 + 1/6
Or
1/R  = (2+1)/6
1/R
= 3/6
R = 2 Ω
This resistance is in series with the resistor,
R1
T
otal resistance = 2 Ω + R1
R1 = 4 Ω
Therefore, total resistance = 2 Ω + 4 Ω = 6 Ω

(ii) The current through the circuit can be calculated as follows:
Current, I = V / R
I = (12 / 6) A
I = 2 A

(iii) The potential difference across R1 = 2 A x 4 Ω = 8 V

Page No 39:

Given I = 1 A (Across 5 Ω)
R = 5 Ω
The potential drop across AB, V = I $×$ R
or  V = 5 Ω x 1 A = 5 V
In a parallel circuit, the potential difference across the ends of all resistors remains the same. Therefore, the potential difference across 4 Ω and 10 Ω will be 5 V:
The current flowing through the 4 Ω resistor, I = V/R.
Here, V = 5 V
R = 4 Ω
So,  I = 5/4 = 1.25 A
The current flowing through the 10 Ω resistor, I = V/R.
Here, V = 5 V
R = 10 Ω
So.   I = 5/10 = 0.5 A

Page No 39:

Suppose x resistors should be connected in parallel to draw a current of 5 A.
For a parallel combination, 1 / R = 1 / R1 + 1 / R2 + 1 / R3 + ... (x times)

Here, R1 = R2 = R3 = ... = 176 Ω
Then the resultant resistance, R = 176 / x Ω

Given that the current, I = 5 A and voltage, V = 220 V

Then by Ohm's law, V = IR

By substituting the values, we get:

x = 4

Hence, four resistors of 176 Ω each should be connected in parallel to draw a current of 5 A from a 220 V supply line.

Page No 39:

(a) When only one coil, A is used:
V = IR
220 = 24 I
I = 9.2 A

(b) When coils, A and B are used in series:
Total resistance R = RA+ RB = 48 Ω
V = IR
220 = 48 I
I = 4.58 A

(c) When coils, A and B are used in parallel:
1/R = 1/RA+ 1/RB
Here, RA = 24 Ω and RB = 24 Ω
1/R = 1/24 + 1/24
or 1/R = 2/24
R = 12 Ω
Total resistance of parallel combination, R = 12 Ω
Now, V = IR
220 = 12 I
I = 18.33 A

Page No 39:

(a) When three resistors are connected in parallel, the net resistance can be obtained as followed:
1/R = 1/R1 + 1/R2 + 1/R3
The resistors of 30 Ω, 20 Ω and 60 Ω are connected in parallel. Therefore, the net resistance R will be:
1/R = 1/30 + 1/20 +1/60
1/R = (2 + 3 + 1)/60
1/R = 6/60
R =10 Ω
Two more resistors of 10 Ω and 40 Ω are connected in parallel to each other. Therefore, their net resistance, R' will be:
1/R' = (1/10) + (1/40)
1/R' = (4 + 1)/40
1/R' = 5/40
R' = 8 Ω
The resistors of 8 Ω and 10 Ω are connected in series. Therefore, the net resistance of the circuit =R+R,=8 Ω + 10 Ω = 18 Ω

(b) The total current flowing through the circuit can be calculated as:
I = V/R
I = 12/18
I = 0.67 A

Page No 40:

I(a) Let I1, I2 and I3 be the current flowing through the resistors of 5 Ω, 100 Ω and 30 Ω, respectively.
According to Ohm's law, V = IR
Here, V = 12 V and R = 5 Ω
I = V/R
I1 = 12/5 = 2.4 A
I2 = 12/10 = 1.2 A
I3 = 12/30 = 0.4 A

(b) Total current through the circuit, I = I1 + I2 + I3
=  2.4 A + 1.2 A + 0.4 A
= 4 A

(c) The resistors of 5 Ω, 100 Ω and 30 Ω are connected in parallel. Therefore, the net resistance will be:
1/ R = 1/R1 + 1/R2 + 1/R3
1/R = (1/5) + (1/10) + (1/30)
1/R = (6 + 3 +1)/30
I/R = 10/30
R = 3 Ω

Page No 40:

(a) When two resistors are connected in series, their resultant resistance is given by
R = R1+ R2
Here, R1=6 $\mathrm{\Omega }$,
R2=2 $\mathrm{\Omega }$
Therefore
The resulatant resistance, R = 6 + 2 = 8 Ω

(b) The current flowing through circuit is:
V = IR
I = V/R = 4/8
I = 0.5 A

(c) The p.d. across the 6 Ω resistor is:
V = 0.5 A x 6 Ω = 3 V

Page No 40:

(a) The resistors of 3 Ω and 6 Ω are connected in parallel. Therefore, their net resistance can be calculated as:
1/R = 1/R1 + 1/R2
Here, R1 = 3 Î
R2 = 6 Î
So:
1/R = (1/3) + (1/6)
1/R = (2 + 1)/6
1/R = 3/6
R = 2 Ω

(b) The current flowing through the main circuit,
I = V/R
I = 6/2 A
I = 3 A

(c) The current flowing in the 3 Ω resistor,
I = V/R
I = 6/3 = 2 A

Page No 40:

(a) The resistors of 2 Ω and 3 Ω are connected in series.
Therefore their combined resistance, R = R1 + R2
Here, R1=2 Ω
R2=3 Ω
So, the combined resistance, R = 2 Ω + 3 Ω
R = 5 Ω

(b) The current, I in the circuit can be calculated as:
V = IR
or  I = 10 / 5 = 2 At)=R2R1+R2v(t)
(c) The p.d. across the 2 Ω resistor is:
V = IR
or V = 2$×$2= 4 V

(d) The p.d. across the 3 Ω resistor:
V = IR
or  V = 2 x 3
or V = 6 V

Page No 40:

(a) The resistors of 6 Ω and 3 Ω are connected in parallel. Therefore, their combined resistance can be calculated as:
1/R = 1/R1 + 1/R2
Here, R1 = 6 Ω ,
R2 = 3 Ω
1/R = (1/6) + (1/3 )
1/R = (1 + 2)/6
R = 6/3
R = 2 Ω

(b) Potential difference, V = IR
The potential difference across the combined resistance,
V = (6 $×$ 2) V
V = 12 V

(c) The potential difference across the 3 Ω resistor can be calculated as:
V = IR
V = 6$×$3 = 18 V

(d) The current across the 3 Ω resistor is:
I = V/R
I = 12/3
I = 4 A

(e) The current across the 6 Ω resistor is:
I = V/R
I = 12/6
I = 2 A

Page No 41:

(a)

(a) Let the current in the circuit be I amperes and the battery be of strength V volts.
Let the combined resistance of the three resistors be R ohms.
Therefore, according to Ohm's law, we have:
V = IR ... (1)
We know that when resistors are connected in series, the current is the same in all the resistors but the voltage is different across each resistor.
Therefore:
V = V1 + V2
V = IR1 + IR2
V = I (R1 + R2) ... (2)
From the equations (1) and (2) we have:
R = R1 + R2

(b)
(i) The current through the 5 Ω resistance can be obtained using the equation,
I = V/R
or I = 10/5 = 2 A
(ii) In a series arrangement, the current remains the same across each resistance. Therefore, the current through R also will be 2 A.

(iii) Value of R = V/I
R = 6/2 = 3 Ω
(iv) The value of V = IR
V = 2 x (3 + 5)  = 16 V

Page No 41:

(a)

Let the current in the circuit be I amperes and the battery be of strength V volts.
Let the combined resistance of the three resistors be R ohms.
Therefore, according to Ohm's law, we have:
V = IR .... (1)
We know that when the resistors are connected in series, the current is the same in all the resistors. Therefore:
V = V1 + V2 + V3
V = IR1 + IR2 + IR3
V = I (R1 + R2 + R3) ... (2)
From the equations (1) and (2) we have:
R = R1 + R2 + R3

(b) The current through the 5 Ω resistor can be calculated as:
V = IR
I = V/R
I = 6/5 = 1.2 A
The current through the 10 Ω resistor can be calculated as:
I = V/R
I = 6/10 = 0.6 A
The current through the 30 Ω resistor can be calculated as:
I = V/R
I = 6/30 = 0.2 A
(b)The total current in the circuit, I = 1.2 A + 0.6 A + 0.2 A = 2 A

(c) The three resistors are connected in parallel. Therefore:
1/R = 1/R1 + 1/R2 + 1/R3
Here, R1=5 $\mathrm{\Omega }$,
R2 = 10 $\mathrm{\Omega }$,
R3= 30 $\mathrm{\Omega }$
1/R = (1/5) + (1/10) + (1/30)
1/R = (6 + 3 + 1)/30
1/R = 10/30
R = 3 Ω

Page No 41:

(a) Circuit diagram

(b) Since two resistors of 20 Ω  are connected in series, the effective resistance will be
R = R1+ R2
Here, R1=20 Ω,
R2 =20 $\mathrm{\Omega }$
Therefore,
R = 20 Ω + 20 Ω = 40 $\mathrm{\Omega }$
The effective resistance of the two resistors, R= 40 Ω
(c) The current that flows from the battery,
I = V / R
I = 5 / 40
I = 0.125 A

(d) The p.d. across the 20 Ω resistor, V= I $×$ R
V = 20 x 0.125 V
V = 2.5 V
Similarly, the potential difference across the second 20 Ω resistor is equal to 2.5 V.

Page No 41:

(a) The resistors are connected in parallel.
(b) In a parallel arrangement, the voltage remains the same across each resistor. The battery is also connected in parallel with the resistors. Therefore, the p.d across each resistor is 6 V.
(c) The 2 Ω resistor will have the bigger share of the current since resistance is inversely proportional to the current.
(d) The effective resistance of the parallel combination is:

(e) The current flowing through the battery, I = 5 A

Page No 41:

The coils of the resistors of 4 Ω and 2 Ω are connected in parallel. Therefore:
$\frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\phantom{\rule{0ex}{0ex}}\frac{1}{R}=\frac{1}{4}+\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{1}{R}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}R=\frac{4}{3}\mathrm{\Omega }$
Combined resistance of the 4 Ω coil and the 2 Ω coil = 1.33 Ω
Total current = 3 A
The voltage across the whole cicuit,as in case of parallel circuit volatge is same scross all resistances.
V = IR.
V = 3 x 4/3 = 4 V
Therefore, the current through the 2 Ω coil,
I = V/R1.
I = 4/2 A
I = 2 A

Page No 42:

(a)

(a) Let the individual resistance of the two resistors be R1 and R2 and their combined resistance be R. Let the total current flowing in the circuit be I and strength of the battery be V volts. Then, from Ohm's law, we have:
V = IR ... (1)
We know that when resistors are connected in parallel, the potential drop across each resistance is the same.
Therefore:
I  = I1 + I2
I = V/R1 + V/R2
I = V/(1/R1 + 1/R2) ... (2)
From the equations (1) and (2) we have:
1/R = 1/R1 + 1/R2

(b)
(1)

(ii) The current shown by the ammeter A, i.e. the current in the circuit can be calculated as:
I = $\frac{\mathrm{V}}{\mathrm{R}}$

I =$\frac{4}{2.5}$= 1.6 A

Page No 42:

(a)

Let the resistance of the three resistors be R1R2 and R3, respectively. Let their combined resistance be R. Let the total current flowing in the circuit be I and the strength of the battery be V.  Then from Ohm's law, we have:
V = IR       .....(1)
We know that when the resistors are connected in parallel, the potential drop across each resistance is the same.
Therefore:
I = I1 + I2 + I3
I = V/R1 + V/R2 + V/R3
I = V/(1/R1 + 1/R2 + 1/R3) ...... (2)
From equations (1) and (2) we have:
1/R = 1/R1 + 1/R2 + 1/R3

(b) Let be the voltage of the cell and R is the resistance of each resistor. When switch K is not closed then

Page No 42:

(d) 7$\frac{1}{2}$$\mathrm{\Omega }$
The resistors of 6 Î and 2 Î are connected in parallel.

This arrangement is further connected in series with the 6 Î resistor.
∴ Net resistance =$\frac{6}{4}+6$ = Ω

Page No 42:

(c) 20 Ω
When 25 Ω and 15 Ω are connected in series, then:
Total resistance, R = 25 + 15 = 40 Î
This 40 Î is connected in parallel with the 40 Î resistor.
Therefore, the net resistance = 40 / 2 = 20 Î

Page No 42:

(a) 9 Î
The two resistors of 6 Î are connected in parallel with each other. So, their net resistance 3 Î is connected in series with a resistance of 6 Î. So, the net resistance of the complete arrangement is 9 Ω.

Page No 43:

(c) 1V
The resistors of 1 $\mathrm{\Omega }$, 2 $\mathrm{\Omega }$ and 3 $\mathrm{\Omega }$  are connected in series. Therefore, the net resistance,
R =R1 +R2 +R3
R=1 $\mathrm{\Omega }$ + 2 $\mathrm{\Omega }$ + 3 $\mathrm{\Omega }$ = 6 $\mathrm{\Omega }$
Current in the circuit will be,
I = V/R
or I = 2/6 = 1/3 A
Current = $\frac{1}{3}$A
Therefore, the voltage across the 3 Î resistor,
V = IR
or V= $\frac{1}{3}$ x 3 = 1 V

Page No 43:

(b) The current at Y is greater than that at Z.
This is so because Y alone will offer less resistance.

Page No 43:

(b) V1 = 5, V2 = 10 and V3 = 15
Because V = IR, the net voltage can be obtained by multiplying current with resistance.

Page No 43:

(d) 25
If the resistance wire is cut into five pieces, the resistance of each wire is R/5. If we connect the pieces in parallel, we will get the net resistance as R/25. Therefore, the ratio will be 25.

Page No 43:

(i) To obtain 9 Ω and 4 Ω , we should connect the resistors as follows:

Page No 43:

Let the resistances of the two resistors be x and y.
When they are connected in parallel, we have:
1 / x + 1 / y = 1 / 2       ... (1)
When they are connected in series, we have:
x + y = 9                      ... (2)
On solving equations (1) and (2), we will obtain the values of the two resistors as 3 Ω and 6 Ω.

Page No 43:

When the two resistors are connected in parallel, we have:

x = 12 Ω
Therefore, the value of x = 12 Î

Page No 43:

(i) To get a value of 6 Ω, all the resistors are to be connected in series as shown below:

1 Î    2 Î   3 Î

Net resistance, R = R1 + R2 + R3
R = 1 Î + 2 Î + 3 Î = 6 Î

(ii) To get a value of 6 / 11 Ω, all the resistors are to be connected in parallel as shown below:

(iii) To get a value of 1.5 Ω, the 1 Ω and 2 Ω resistors should be connected in series and this arrangement should be connected in parallel with the 3 Ω resistor as shown below:

The resistors of resistance 1 Î and 2 Î are in series. Therefore, their net resistance is:
R = R1 + R2
R = 1 Î + 2 Î
R = 3 Î
This 3 Î is connected in parallel with another 3 Î resistance.
Therefore, net resistance will be:

Page No 44:

To get the resistance of 2.5 Î, we will have to connect the resistors of values 2 Î and 3 Î in series and this arrangement should be connected in parallel with the resistor of value 5 Î.

Page No 44:

(a) To obtain a 4 Ω resistance, connect the 2 Ω resistor in series with the parallel combination of the resistors of values 6 Ω and 3 Ω.
(b) To obtain a 1 Ω resistance, connect all the resistors in parallel.

Page No 44:

(a) To get the highest resistance, all the resistors must be connected in series.
Resistance in a series arrangement is given by R = R1+ R2 + R3 + R4

Therefore, the highest resistance is 48 Ω.
(b) To get the lowest resistance, all the resistors must be connected in parallel.
Resistance in a parallel arrangement is given by:
$\frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}+\frac{1}{{R}_{4}}$

Here R1= 4 $\mathrm{\Omega }$,
R2 = 8 $\mathrm{\Omega }$,
R3= 12 $\mathrm{\Omega }$,
R4 = 24 $\mathrm{\Omega }$

Therefore, the lowest resistance of the arrangement is 2 Ω.

Page No 44:

The resistors of values 20 Ω, 10 Ω and 20 Ω are connected in series. Therefore, their net resistance is:
R = 20 Ω + 10 Ω + 20 Ω
R = 50 Ω

This combination of resistors are connected in parallel with that of value 30 Ω. Therefore, the resistance R' is:
1/R' = (1/50) + (1/30)
1/R' = (3 + 5)/150
R' = 150/8
R' = 18.75 Ω

The resistance, R' is connected in series with the two resistors of values 10 Ω each. Hence, the total resistance between A and B is:
18.75 + 10 + 10 = 38.75 Ω

Page No 44:

Given: One hundred 1 Ω resistors
The largest value of resistance is obtained when the resistors are connected in series. Therefore, the largest resistance is 100 Ω.
The lowest resistance is obtained when all the resistors are connected in parallel. Therefore, the lowest resistance of the combination is 0.01 Ω.

Page No 44:

When two 100 Ω resistors are connected in parallel, the value of the combined resistance becomes 50 Ω.
Now to get a resistance of 250 Ω, we must connect two more 100 Ω resistors in series with the above combination.
Therefore, the configuration will be two 100 Ω resistors in series with a parallel combination of another two 100 Ω resistors.

Page No 44:

The p.d across all the resistors R1, R2, R3 and R4 is 12 V.
Since all the resistors are of the same value, the p.d at exactly half of the combination, i.e. the voltage shown by the voltmeter C will be 6 V. Similarly, the p.d. shown by the voltmeters A and B is one-fourth the p.d. across all the resistors. Therefore, A and B will have a reading of 3 V.

Page No 44:

When four resistors of 16 Ω each are connected in parallel, the effective resistance of the combination can be obtained as:

The effective resistance of combination is 4 Î.
When four such combinations are connected in series, the effective resistance becomes:
R = R1 + R2+ R3 + R4
R = 4 Î + 4 Î + 4 Î + 4 Î = 16 Ω

Page No 44:

Since the lamps are the same as given in the question, the resistance of both the lamps will be the same. Therefore, the current is distributed equally. Since the combination shows a 0.50 A current in the circuit by the ammeter A1, the current shown by A2, A3, A4 and A5 will be the same and equal to 0.25 A.

Page No 47:

No, the lights in the house are wired in parallel.

Page No 47:

When a bulb in a series circuit blows off, the power supply to all other bulbs is cut off. Hence, all the other bulbs are switched off.

Page No 47:

When one of the bulbs in a parallel circuit blows off, it does not affect the other bulbs.

Page No 47:

(a) Series circuit: This is so because in a series circuit, a large number of low-power decorating bulbs can be used effectively.
(b) Parallel circuit: This is so because in a parallel circuit, if one bulb does not work properly, it does not affect the working of the other bulbs.

Page No 47:

A series arrangement is not used for connecting domestic electrical appliances because different appliances have different resistances; this leads to the problem of unequal voltage distribution. Another reason why this arrangement is not preferred is that if one of the appliances gets damaged, the power supply to all other appliances is cut off.

Page No 47:

Three reasons why a parallel circuit is used in domestic wiring:
(a) A parallel circuit has a low resistance.
(b) It prevents unequal power supply as opposed to a series circuit.
(c) If one device in the series is damaged, it does not affect the other devices.

Page No 47:

(a) The parallel circuit will have the highest voltage across each bulb.
(b) Parallel circuit
(c) Series circuit
(d) Series circuit

Page No 47:

(a) The lamps will be the dimmest in the series circuit (ii) because the voltage will be divided between the two lamps.
(b) In circuit (iii), the lamps will be as bright as in circuit (i) because circuit (iii) is a parallel arrangement. In a parallel arrangement, the voltage remains the same.
(c) Circuit (iii) will give the maximum light as circuit (iii) has a parallel arrangement. A parallel arrangement offers less resistance and thus circuit (iii) will give more light.

Page No 48:

(a) A parallel arrangement will be a better way to connect lights and other electrical appliances in domestic wiring because a parallel arrangement has a low resistance. Also in case of any short circuiting in one appliance, the other electrical appliances remain unaffected.
(b) In a series arrangement, if a lamp blows off, the power supply to the other lamps is cut off. Hence, all the lamps will get switched off.
(c) The electrician has wired the house in a series circuit.
(d)

Page No 48:

(b) If one lamp fails, the others remain lit.
This is so because in a parallel circuit, if one device fails, it does not affect the working of the other devices.

Page No 48:

(c) With S2 open and S1 closed, A and B are lit.
This is so because if the switch, S2 is open, the current will flow through lamp A because it is a parallel circuit.

Page No 48:

(a)

(b) The brightness of the lamp can be changed by connecting the lamps in parallel.

Page No 48:

(a) The brightness of A and B is the same. C is brighter since in A and B, the same amount of current flows but the voltage gets divided into two parts. Therefore, bulbs A and B glow dimmer.
(b) The brightness of A will remain the same because voltage distribution will have no effect if bulb C burns out.
(c) Bulb A will get switched off because the current will not flow to it if bulb B burns out since they are in a series circuit. The brightness of C will remain the same because the same current will continuously flow through bulb C.

Page No 48:

The brightness of two lamps arranged in parallel will be more compared with the brightness of two lamps arranged in series because a parallel circuit offers less resistance.

Page No 48:

In a parallel circuit, if the filament of one lamp breaks, it will have no effect on the other lamp. If the same happens in a series circuit, the other lamp will stop glowing. This is because in a series circuit, if one of the bulbs blows off, the circuit is broken and the power supply to the other bulb is cut off.

Page No 48:

A parallel arrangement will be preferred in this case because a parallel arrangement has less resistance than a series arrangement. Also, in case of any short circuiting, the other bulb remains unaffected. A series arrangement will draw more current from the battery.

Page No 49:

The switch is turned to the right for brighter light and is turned to the left for dimmer light. This happens because when the switch is turned to the right, it offers less resistance and when turned to the left, it offers more resistance.

Page No 58:

Two factors on which electrical energy consumed by an electrical appliance depends on are:
(1) power
(2) time

Page No 58:

A 60-watt bulb has a higher electrical resistance than a 100-watt bulb. Because power is inversely proportional to resistance, when the power is less, the resistance is high.

Page No 58:

The commercial unit of electric energy is kilowatt hour or KWh in short.

Page No 58:

Given V = 220 V and P = 100 W

We know that power, P = IV = V2 / R
R = V2 / P

R = $\frac{220×220}{100}$
R = 484 Î

Page No 58:

(i) The SI unit of electric energy is joule.
(ii) The SI unit of electric power is watt.

Page No 58:

(i) Kilowatt is the unit of power.
(ii) Kilowatt-hour is the unit of energy.

Page No 58:

Watt is the unit of power.

Page No 58:

The symbol KWh stands for kilowatt-hour. It is the commercial unit of electrical energy.

Page No 58:

We know that power and potential difference are related to each other as given in the equation P = V2/R.

Now, keeping R the same, if V is doubled, power gets increased by four times.

Page No 58:

The label further tells us that the total power consumed by the electric lamp will be 36 W and the current derived by it is 3 A (using the equation, P = VI).

Page No 58:

Given:
P = 920 W
V = 230 V
We know that power is given by P = VI, where I is the current consumed by the device.
Using the given values of P and V, we get:
920 = 230 I
I = $\frac{920}{230}=4$ A
Thus, the current consumed by the appliance is 4 A.

Page No 58:

Watt is the unit of power. One watt can be defined as the power consumed by an electrical device if it is operated at a potential difference of 1 V and it carries a current of 1 A. That is:
1 watt = 1 volt x 1 ampere

Page No 58:

One watt-hour is the amount of electrical energy consumed when an electrical appliance of power equal to 1 watt is used for 1 hour.
1 watt-hour = 3600 joules

Page No 58:

Given:
I = 5 A
R = 100 Î
Power consumed, P= I2R
= (5)2 100 watt
= 2500 watts=2.5 kW

Energy consumed, E = P x time
= 2.5 x 2
= 5 kWh

Page No 58:

Here, V = 220 V and I = 0.5 A                  (given)
Power drawn by the device is given by:
P = VI

Therefore:
P = 220 x 0.5
= 110 W

Page No 58:

(i) I = 1 A
R = 300 Î
t = 1 h
Electrical energy consumed = I2 $×$ R $×$ t
= (1)2 (300) (1)
= 300 Wh
(ii) I = 2 A
R = 100 Î
Electrical energy consumed = I2 $×$ R $×$ t = (2)2 (100) (1) = 400 Wh

Thus, more electrical energy is consumed in case (ii).

Page No 58:

Given:
V = 220 V
P = 2.2 kW = 2.2 x 103 w
t = 3 h

Electrical energy, E = P x t
= 2.2 kW x 3 h
= 6.6 kWh

Current drawn, I = P/V
=
= 10 A

Page No 59:

(a)  Electrical power is the electric work done per unit time.
In terms of potential difference and current, electric power is given as:
Electric power, P = potential difference, V x current, I
P = VI

(b) Given:
Potential difference = reading of the voltmeter, V = 3 V
Current in the circuit = reading of the ammeter, I= 0.5 A
(i)  Resistance of the lamp, R =
(ii) Power, P = VI = 3 V x 0.5 A = 1.5 W

(c) One kilowatt-hour is the amount of electrical energy consumed when an electrical appliance having a power rating of one kilowatt is used for one hour.

1 kilowatt-hour = 36,00,000 joules

(d) Power rating of the heater, P = 500 W
Duration of the operation, t = 20 hours
Electrical energy consumed = P x t
= 500 W x 20 hours
= 10000 W = 10 kW
Rate for 1 kW = Rs. 3.90
Total cost of operating = units consumed x rate = 10 x Rs. 3.90 = Rs. 39

Page No 59:

(b) 6 W
Power, P = VI = 12 V x 0.5 A = 6 W

Page No 59:

(d) watt
Watt is the unit of electric power.

Page No 59:

(c) 850 W
An electric iron may use a power of 850 watts.

Page No 59:

(b) Rs. 24
Electrical energy consumed in three hours = P x t = 2 kW x 3 hour = 6 kWh
Unit cost = Rs. 4 per kWh
Therefore, the cost of energy used for three hours = 4 x 6 = Rs. 24

Page No 59:

Power of bulb 1, P1 = 60 W
Duration that it was lit, t1= 4 h
Electric energy consumed by two bulbs, E1 = 2 x P1 x t1
= 2 x 60 x 4
= 480 Wh

Power of bulb 2, P2 = 100 W
Duration that it was lit, t2= 5 h
Electric energy consumed by three bulbs, E2= 3 x P2x t2 = 3 x 100 x 5 = 1500 Wh

Total electric energy consumed in 30 days, E = 30 x (E1+ E2)
E = 30 x (480 + 1500)
E = 30 x 1980
E = 59400 Wh
E = 59.4 kWh

Page No 59:

Given:
V = 250 V
I = 0.4 A

Power, P = VI
= 250 x 0.4
= 100 W

Resistance, R = $\frac{P}{{I}^{2}}$
= $\frac{100}{\left(0.4{\right)}^{2}}$
= 625 Ω

Page No 59:

Given:
V = 220 V
P = 4 kW = 4 $×$ 103 W

(a)  I = A

(b) R = Î

(c) Energy consumed in two hours, E = P $×$ t
= 4 kW $×$ 2 h
= 8 kWh

(d) Cost of 1 kWh = Rs 4.6

Total cost = Energy consumed $×$ Price
= 8 $×$ 4.6
= Rs 36.80

Page No 59:

Given:
I = 5 A
V = 220 V

Power of the motor,
P = VI
= 5 $×$ 220
= 1100 W = 1.1 kW
Energy consumed in two hours,
E = P $×$ t
= 1.1 kW $×$ 2 h
= 2.2 kWh

Page No 59:

Energy consumed by the TV set,
E = P $×$ t
= 250 $×$ 1 = 250 Wh
Energy consumed by the toaster,
E = P $×$ t
= 1200 W $×$ $\frac{10}{60}$ h   (10 min = $\frac{10}{60}$ h )
= 200 Wh
Thus, it is clear that the TV set consumes more electrical energy.

Page No 59:

(i) As the 6 V battery is connected in series, the potential difference will be different across the two given resistors. So we need the current to calculate the power.
Therefore, the net resistance, R of the circuit = (1 + 2) Î = 3 Î
Current, I  = $\frac{\mathrm{V}}{\mathrm{R}}$ = $\frac{6}{3}$ = 2 A
Power consumed by the 2-ohm resistor,
P = I2 R
= 22 x 2
= 8 W

(ii) As the battery is connected in parallel to the 12 Î and 2 Î resistors, the potential difference across them will be the same.
Therefore, the power,

Page No 59:

(a)    220 V

(b) Current drawn by the 40 W bulb, I1 =

Current drawn by the 60 W bulb, I2=

Total current drawn, I = I1 + I2= 0.18 A + 0.27 A = 0.45 A

(c) Total energy consumed, E = Σ (P x t) = (40 x 1) + ( 60 x 1) = 100 Wh = 0.1 kWh

Page No 59:

(a) Power, P = VI = 230 $×$ 10 = 2300 W
(b) Energy transferred in one minute = $×$ t
= 2300 W $×$ 60 s         (1 minute = 60 s)
= 138000 J
= 138 kJ

Page No 59:

Energy consumed by the TV in 1 hour, E1 = P x t = 200 W x 1 hour = 200 Wh = 0.2 kWh
Energy consumed by the three bulbs in 1 hour, E2 = 3 x P x t = 3 x 100 W x 1 hour = 300 Wh = 0.3 kWh
Energy consumed by the heater in 1 hour, E3 = 2 kWh
Total energy consumed, E = E1+ E2 + E3
= 0.2 kWh + 0.3 kWh + 2 kWh
= 2.5 kWh

Total energy, consumed from 6 p.m. to 10 p.m., E = 4 x 2.5 kW = 10 kWh

Total cost at base price, Rs. 5.5 per kWh = E x 5.5 = 10 x 5.5 = Rs. 55

Page No 59:

Maximum power, P = VI
= 13 A 230 V
= 2990 W = 2.99 kW

Page No 59:

Rate of electrical energy transfer = Power consumed
∴ Rate of electrical energy transfer, P = VI
= 230 V $×$ 0.4 A
= 92 W = 92 J/s

Page No 60:

(a) joule
Joule is the SI unit of energy.

Page No 60:

(c) kilowatt-hour
Kilowatt-hour is the commercial unit of electrical energy.

Page No 60:

(d) 6000 J
Energy transferred, E = power x time = 100 W x 60 s = 6000 J

Page No 60:

(d) 15 A
Current drawn by the kettle, I =
So the cable should be able to carry more current than this and hence the answer is 15 A.

Page No 60:

(d) 3 J / s
Electrical energy transferred per second, E = V x I x t = =  3 J/s

Page No 60:

(c) 293
Total power provided in the house, P = VI = 100 A $×$ 220 V = 22000 W
This is the maximum power that can be drawn in the house.
Total power drawn by x bulbs = 75x W
∴ Maximum power x bulbs can draw = 22000 W

$⇒$ 75x = 22000

$⇒$x = $\frac{22000}{75}\approx 293$

Page No 60:

(d) one-fourth
Since P = $\frac{{\mathrm{V}}^{2}}{\mathrm{R}}$, when the potential difference is halved, the power becomes one-fourth.

Page No 60:

The electric heater will consume more electrical energy when the length of its heating element is reduced because resistance is directly proportional to length (R = ρ$\frac{l}{A}$). Hence, resistance will decrease. Power is inversely proportional to resistance (P = $\frac{{V}^{2}}{R}$). Hence, power will increase.
Thus, a decrease in resistance means an increase in the power consumed.

Page No 60:

(a) The lamp has the greatest electrical resistance because the current drawn by it is the least. R = $\frac{\mathrm{V}}{\mathrm{I}}$ and hence resistance varies inversely with the current.

(b) The reasons for this are as follows:

1. The kettle draws a large amount of current. So the wire may melt down due to heating.
2. For the device to be safe, it needs to be have an earthing system. This is missing here.
(c) The power, P will be:
P = VI = 240 V x 8.5 A = 2040 W

(d) As current is directly proportional to potential difference, halving the potential will halve the current too. Therefore, the new current value will be 4.25 A.

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(a) The meter reading on Sunday was 42919.
(b) The meter reading on Monday was 42935.
(c) Total units consumed = Meter reading on Monday − Meter reading on Sunday
= 42935 − 42919
= 16
(d) These units have been consumed in 24 hours.
(e) Price of one unit = Rs. 5
Total units consumed = Meter reading on Sunday − Meter reading on Monday
= 42935 − 42919
= 16
Total cost of units consumed = 16 $×$ 5 = Rs. 80

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Maximum power that can be drawn from the circuit = VI = 220 V x 5 A = 1100 W
Let x be the number of 10 W bulbs used.
Then the maximum power x bulbs can draw from the circuit = 1100 W
10x = 1100
x = $\frac{1100}{10}=110$ bulbs

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Let R be the resistance of the lamp and V be the potential difference applied.

In a series connection:
Net resistance of the circuit, R1 = R + R = 2R
Power drawn, P1 = V2/R1 = V2/2R
In a parallel connection:
Net resistance of the circuit, R2 will be:
$\frac{1}{{R}_{2}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}$
or   R2 = $\frac{R}{2}$
Power drawn, P2 = V2/R2 = 2V2/R
Therefore, P2:P1 = 4:1

Page No 66:

(a) Since a conductor offers resistance to the flow of current, some work must be done by the current continuously to keep itself flowing. When an electric charge, Q moves against a potential difference, V, the amount of work done is given by:
W = Q x V                                            (1)

From the definition of current, we know that:
Current, I = $\frac{Q}{t}$
So, Q = I x t                                     (2)

And from Ohm's law, we have:
$\frac{\mathrm{V}}{\mathrm{I}}=\mathrm{R}$
or                    V = I x R                                 (3)

Now, substituting Q = I x t and V = I x R in equation (1), we get:
W = I x t x I x R
So, the work done, W = I2 x R x t

Assuming that all the electrical work done is converted into heat energy:
Heat produced = Work done in the above equation
Thus heat produced, H = I2 x R x t joules
This is known as Joule's law of heating.

(b) Given: Power rating (P) = 12 W
Time interval (t) = 1 minute = 60 s                              (1 minute = 60 seconds)
Potential difference (V) = 12 V
Heat = I2Rt
But, we know that P = I2R
Therefore, H = P x t = 12 W x 60 s = 720 joules

(c) The heat produced by the room heater will become one-fourth the previous quantity when the current passing through it is halved because the heat produced is directly proportional to the square of the current.

(d) When an electric current is passed through a high resistance wire, like nichrome wire, the resistance wire becomes very hot and produces heat. This is called the heating effect of current.
The heating effect of current is used in electric irons, electric heaters, electric ovens, filament bulbs etc. and it is also utilised in electric fuses.

(e) Tungsten is used for making the filament of an electric bulb.

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(c) magnitude of current

We know that:
H = I2Rt

It shows that the heat produced is proportional to the square of the current.

Page No 66:

Heat produced by a current passing through a wire with a fixed resistance depends on the square of current, I.
Heat, H = I2 x R x t

Page No 66:

If the current passing through a conductor is doubled, the heat produced will increase by four times because the heat produced is directly proportional to the square of the current.

Page No 66:

The two effects produced by an electric current are:
(i) Heating effect
(ii) Chemical effect

Page No 66:

The heating effect of current is utilised in an electric bulb.

Page No 66:

The heating effect of current is utilised in the working of an electric fuse.

Page No 66:

The two devices that work on the heating effect of electric current are a bulb and an electric fuse.

Page No 66:

Two gases that are filled in the filament type of electric bulbs are:
(i) argon
(ii) nitrogen

Page No 66:

Filament type of electric bulbs are not power efficient because a lot of power gets wasted as heat.

Page No 66:

It is so because the connecting cord has a very low resistance. So, it does not get enough heat to glow.

Page No 66:

(a) H = I2Rt where H is the heat energy produced, I is the current, R is the resistance and t is the time.
(b) I = 5 A, R = 20 Î and t = 30 s
H = I2Rt = 52 x 20 x 30 = 25 x 20 x 30 = 15000 J

Page No 66:

The heat produced by an electric current depends on the square of the current, the resistance of the device and the time for which the current is passed through the device, i.e.:
H = I2Rt

Page No 66:

(a) Joule's law of heating states that heat, H produced in a wire is directly proportional to:
(i) the square of the current, I2
(ii) the resistance of the wire, R
(iii) the time, t for which the current is passed
That is, H = I2Rt
(b) The net resistance of the circuit, R = 40 + 60 = 100 Ω
Time for which current is passed = $\frac{1}{2}$ minute = 30 s
Current in the circuit =
Heat produced = I2Rt = (2.2)2 (100) (30) = 4.84 x 100 x 30 = 14520 J

Page No 66:

If air is filled in an electric bulb, the extremely hot tungsten filament would burn up quickly because of the oxygen present in the air. So, a bulb is filled with a chemically unreactive gas like argon or nitrogen.

Page No 66:

Tungsten is used for making the filaments of electric bulbs because it has a very high melting point (3380°C). It does not melt down and thus remains white hot.

Page No 66:

The current that makes the heater element very hot, only slightly warms the connecting wires leading to the heater because the connecting wires have a low resistance. So they do not get hot. But the heater element has a very high resistance and so it gets very hot.

Page No 66:

I = 4.0 A
H = 2.88 x 104 J
t = 10 minutes = 10 x 60 = 600 s

(a) Power =

(b) V =

Page No 66:

I = 2.5 A
R = 200 Ω
Rate of heat produced = Power = I2R = (2.5)2(200) = 1250 J/s

Page No 66:

I = 15 A, R = 8 Ω
Rate of heat produced = Power = I2R = (15)2(8) = 1800 J/s

Page No 66:

V =12 V
R = 25 Ω
t = 1 minute = 60 s

Current, I =

Heat generated = I2Rt = (0.48)2(25)(60) = 345.6 J

Page No 66:

Given R = 4 Ω
H = 100 J
t = 1 s
We know that, H = I2Rt

I2 =

I = 5 A
Now:
V = IR
V = 5 x 4 = 20 V

Page No 67:

(c) four times
Heat produced is directly proportional to the square of the current.

Page No 67:

(d) heating effect of current
An electric fuse works on the heating effect of current.

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(c) nichrome
Nichrome has a high resistance.

Page No 67:

(b) x $×$ z $×{y}^{2}$
This is so because heat, H = I2Rt

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(b) thick and short
This is so because in this case, the resistance of the wire will be low.

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(b) infra-red rays
This is so because of the heat produced when a filament type light bulb glows.

Page No 67:

(c) 2500°C

It is the temperature of an electric bulb filament when it glows.

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(c) one-fourth
This is so because heat produced is directly proportional to the square of the current passing through the resistor.

Page No 67:

(a) Material S will be used for making the heating element of an electric iron as it has a high resistivity .
(b) Material Q will be used for making the connecting wires of an electric iron as it has a low resistivity.
(c) Material R will be used for making the covering of the connecting wires as it has an extremely high resistivity.

Page No 67:

(a)

 Filament of a light bulb Connecting wire The wire in the filament of a light bulb has a high resistance and a high melting point. So, it heats up without melting and glows when the current flows through it. The connecting wire in a circuit has a low resistance and a low melting point. So, it does not heat up much when the current flows through it and thus does not glow.

(b) It is the high melting point of the filament wire that is accountable for this difference.

Page No 67:

Let the resistance be R.
In a series combination:

Net resistance = R + R = 2R

In a parallel combination:

Net resistance = $\frac{R}{2}$

It is clear that the net resistance is more in the series combination. Therefore, more heat will be obtained in a series connection, as the heat produced is directly proportional to the resistance in the circuit.

Page No 67:

In the case of minimum heating:

P = 360 W
V = 220 V
where the symbols used have their usual meanings.
We know that  P = VI
I = $\frac{\mathrm{P}}{\mathrm{V}}$

Resistance, R =$\frac{V}{I}$
R = Ω

In the case of maximum heating:

P = 840 W
V = 220 V
where the symbols used have their usual meanings.
We know that P = VI
I = $\frac{\mathrm{P}}{\mathrm{V}}$

Further, R =$\frac{\mathrm{V}}{\mathrm{I}}$

R = $\frac{220}{3.82}=57.60$ Ω

Page No 67:

The following electric heating devices have resistors to control the flow of electricity:

1. Electric iron
2. Electric geyser
3. Electric oven
4. Electric heater

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