This essential chapter from NCERT Solutions for Class 10 Science Chapter 11 , Electricity explores the fundamental principles of electric current and circuits that power our modern world. From understanding how electricity flows through conductors to calculating the energy consumed by household appliances, this chapter provides the foundation for comprehending the electrical devices we use every day. This chapter is part of the comprehensive NCERT Solutions Class 1o Science series, which covers all chapters in detail.
The chapter covers the basic concepts of electric current, potential difference, and resistance, along with Ohm's Law and its applications. Students learn about series and parallel circuits, the heating effect of electric current, and how to calculate electrical power and energy consumption. Every solution has been designed keeping CBSE board exam patterns in mind, with clear circuit diagrams, step-by-step numerical problem solutions, and practical applications that ensure students develop both conceptual clarity and problem-solving confidence for their examinations.
NCERT Solutions for Class 10 Science Chapter 11 - All Exercise Questions
Q.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio is
Q.
Show how would you connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω (ii) 4 Ω.
Q.
An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Q.
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Q.
Two lamps, one rated 100 W and 220 V, and the other 60 W at 220 W, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Q.
Compare the power used in the 2 Ω in each of the following circuits:
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Q.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 resistances which may be used separately, in series or in parallel. What are the currents in the three cases?
Q.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Q.
How many 176 resistors (in parallel) are required to carry 5 A on a 220 V line?
Q.
Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
Q.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 resistors?
Q.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Q.
The value of current (I) flowing in a given resistor for the corresponding values of potential difference (V) across the resistor are given below-
|
I (amperes)
|
0.5
|
1.0
|
2.0
|
3.0
|
4.0
|
|
V (volts)
|
1.6
|
3.4
|
6.7
|
10.2
|
13.2
|
Plot a graph between V and I and calculate the resistance of that resistor.
Q.
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8 Ωm. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Q.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Q.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be -
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Q.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Q.
Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Class 10 Chapter 11 Science Questions & Answers – Electricity
Q1. A piece of wire of resistance \(R\) is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is \(R'\), then the ratio \( \frac{R}{R'} \) is:
(a) 125
(b) 15
(c) 5
(d) 25
Solution: The correct option is (d).
The resistance of a piece of wire is \(R\), and it is proportional to the length of the wire. Since the wire is cut into five equal parts, the resistance of each part is:
\[
R_{\text{each}} = \frac{R}{5}
\]
Let the equivalent resistance of these pieces of wire in parallel be \(R'\). Then,
\[
\frac{1}{R'} = \frac{1}{R_{\text{each}}} + \frac{1}{R_{\text{each}}} + \frac{1}{R_{\text{each}}} + \frac{1}{R_{\text{each}}} + \frac{1}{R_{\text{each}}}
\]
\[
\frac{1}{R'} = 5 \times \frac{1}{\frac{R}{5}} = \frac{25}{R}
\]
\[
R' = \frac{R}{25}
\]
Therefore, the ratio \( \frac{R}{R'} = 25 \).
Q2. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of:
(i) 9 Ω
(ii) 4 Ω
Solution:
(i)When two resistors are connected in parallel and the third one is connected in series with these two, the equivalent resistance of the combination can be calculated as follows:
Two 6 Ω resistors are connected in parallel. Their equivalent resistance \( R' \) will be:
\[
\frac{1}{R'} = \frac{1}{6\,\Omega} + \frac{1}{6\,\Omega} = \frac{2}{6} = \frac{1}{3}\,\Omega
\]
So,
\[
R' = 3\,\Omega
\]
The third 6 Ω resistor is connected in series with the 3 Ω.
Hence, the total equivalent resistance \( R_1 \) is:
\[
R_1 = 6\,\Omega + 3\,\Omega = 9\,\Omega
\]
(ii)Two resistors of 6 Ω are connected in series. Their equivalent resistance is:
\[
R = 6\,\Omega + 6\,\Omega = 12\,\Omega
\]
The third 6 Ω resistor is connected in parallel with the 12 Ω.
Hence, the equivalent resistance \( R_2 \) is:
\[
\frac{1}{R_2} = \frac{1}{12\,\Omega} + \frac{1}{6\,\Omega} = \frac{3}{12} = \frac{1}{4}\,\Omega
\]
So,
\[
R_2 = 4\,\Omega
\]
Q3. An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Solution: Given:
Resistance, \( R = 8\,\Omega \)
Current, \( I = 15\,\text{A} \)
The rate of heat produced can be expressed as power:
\[
P = I^2 R
\]
Substituting the given values:
\[
P = (15\,\text{A})^2 \times 8 = 225 \times 8 = 1800\,\text{W}
\]
Thus, the rate at which heat is developed in the heater is 1800 W or 1800 J/s.
Q4. Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?
Solution:
The energy consumed by a TV set of power 250 W in 1 hour is given by the expression:
\[
H = P \times t = 250\,\text{W} \times 3600\,\text{s} = 9 \times 10^5\,\text{J}
\]
Similarly, the energy consumed by a toaster of power 1200 W in 10 minutes is:
\[
H = 1200\,\text{W} \times 600\,\text{s} = 7.2 \times 10^5\,\text{J}
\]
Hence, the energy consumed by a 250 W TV in 1 hour is more than the energy consumed by a 1200 W toaster in 10 minutes.
Q5.Two lamps, one rated 100 W and 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Solution:
The potential difference across each bulb will be 220 V because no division of voltage occurs in a parallel circuit.
The current drawn by the bulb with a rating of 100 W is given by:
\[
P = V \times I_1
\]
\[
I_1 = \frac{P}{V} = \frac{100\,\text{W}}{220\,\text{V}} = 0.4545\,\text{A}
\]
Similarly, the current drawn by the bulb with a rating of 60 W is:
\[
I_2 = \frac{P}{V} = \frac{60\,\text{W}}{220\,\text{V}} = 0.2727\,\text{A}
\]
Hence, the total current drawn from the line is:
\[
I = I_1 + I_2 = 0.4545\,\text{A} + 0.2727\,\text{A} = 0.727\,\text{A}
\]
Q6. Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Solution:
(i)
Given:
Potential difference, \( V = 6\,\text{V} \)
\( R_1 = 1\,\Omega \), \( R_2 = 2\,\Omega \)
The equivalent resistance in series is:
\[
R = R_1 + R_2 = (1 + 2)\,\Omega = 3\,\Omega
\]
Using the relation \( V = I R \) (where \( I \) is the current through the circuit), we get:
\[
I = \frac{V}{R} = \frac{6}{3} = 2\,\text{A}
\]
Now, power in the circuit:
\[
P = I^2 R = (2)^2 \times 2 = 8\,\text{W}
\]
(ii)
Given:
Potential difference, \( V = 4\,\text{V} \)
\( R_1 = 12\,\Omega \), \( R_2 = 2\,\Omega \)
In a parallel circuit, the voltage across each component remains the same. Hence, the voltage across the 2 Ω resistor is 4 V.
The power consumed by the 2 Ω resistor is:
\[
P = \frac{V^2}{R} = \frac{4^2}{2} = 8\,\text{W}
\]
Q7. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Solution:
Given:
Supply voltage, \( V = 220\,\text{V} \)
Resistance of each coil, \( R = 24\,\Omega \)
(i) When coils are used separately:
Using the relation \( V = I_1 R_1 \), where \( I_1 \) is the current flowing through the coil, we get:
\[
I_1 = \frac{V}{R_1} = \frac{220}{24} = 9.166\,\text{A}
\]
(ii) When coils are connected in series:
The total resistance of the two coils in series is:
\[
R_2 = R + R = 24\,\Omega + 24\,\Omega = 48\,\Omega
\]
Using the relation \( V = I_2 R_2 \), we get:
\[
I_2 = \frac{V}{R_2} = \frac{220}{48} = 4.58\,\text{A}
\]
(iii) When coils are connected in parallel:
The total resistance \( R_3 \) is given by:
\[
\frac{1}{R_3} = \frac{1}{R} + \frac{1}{R} = \frac{2}{24}
\]
\[
R_3 = 12\,\Omega
\]
Using the relation \( V = I_3 R_3 \), we get:
\[
I_3 = \frac{V}{R_3} = \frac{220}{12} = 18.33\,\text{A}
\]
Q8. Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of the 220 V line if the maximum allowable current is 5 A?
Solution:
Given:
Supply voltage, \( V = 220\,\text{V} \)
Maximum allowable current, \( i = 5\,\text{A} \)
Power of each bulb, \( P_1 = 10\,\text{W} \)
Using the relation \( P = \frac{V^2}{R_1} \), we can find the resistance of each bulb:
\[
R_1 = \frac{V^2}{P_1} = \frac{220^2}{10} = 4840\,\Omega
\]
As per Ohm's law, \( V = I R \), the total resistance of the circuit is:
\[
R = \frac{V}{i} = \frac{220}{5} = 44\,\Omega
\]
Now, for \( x \) electric bulbs connected in parallel, the total resistance is:
\[
\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_1} + \cdots \text{(x times)}
\]
This simplifies to:
\[
\frac{1}{R} = \frac{x}{R_1}
\]
Therefore:
\[
x = \frac{R_1}{R} = \frac{4840}{44} = 110
\]
Thus, 110 electric bulbs can be connected in parallel.
Q9. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Solution:
Given:
Supply voltage, \( V = 220\,\text{V} \)
Current, \( I = 5\,\text{A} \)
The equivalent resistance \( R \) of the combination is given by:
\[
\frac{1}{R} = x \times \frac{1}{176}
\]
\[
R = \frac{176}{x}
\]
From Ohm's law:
\[
V = I R
\]
\[
220 = 5 \times R
\]
\[
R = \frac{220}{5} = 44\,\Omega
\]
Now, substituting \( R = 44 \) into the equation:
\[
\frac{1}{44} = x \times \frac{1}{176}
\]
Solving for \( x \):
\[
x = \frac{176 \times 5}{220} = 4
\]
Thus, 4 resistors of 176 Ω are required to draw the given amount of current.
Q10. Which of the following terms does not represent electrical power in a circuit?
(a) \( I^2 R \)
(b) \( IR^2 \)
(c) \( VI \)
(d) \( \frac{V^2}{R} \)
Solution:
Electrical power is given by:
\[
P = VI \tag{1}
\]
According to Ohm's law:
\[
V = IR \tag{2}
\]
Where:
\( V \) = potential difference
\( I \) = current
\( R \) = resistance
Substituting from equation (2) into equation (1):
\[
P = VI = (IR) \times I = I^2 R
\]
From equation (2), we also have:
\[
I = \frac{V}{R}
\]
Substituting into the expression for power:
\[
P = \frac{V^2}{R}
\]
Thus, the correct expressions for electrical power are \( P = VI \), \( P = I^2 R \), and \( P = \frac{V^2}{R} \).
Therefore, \( IR^2 \) does not represent electrical power.
Q11. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω, and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Solution:
The current would be the same throughout the series circuit.
The equivalent resistance of the circuit is:
\[
R_{\text{eq}} = (0.2 + 0.3 + 0.4 + 0.5 + 12) \,\Omega = 13.4\,\Omega
\]
Given that \( V = 9\,\text{V} \), we use Ohm's law:
\[
V = IR
\]
Solving for \( I \):
\[
I = \frac{V}{R_{\text{eq}}} = \frac{9}{13.4} = 0.671\,\text{A}
\]
Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.
Q12. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Solution: According to Ohm's law:
\[
V = IR \quad \text{or} \quad R = \frac{V}{I}
\]
Given:
Potential difference, \( V = 12\,\text{V} \)
Current, \( I = 2.5\,\text{mA} = 2.5 \times 10^{-3}\,\text{A} \)
Substitute the values into the formula for \( R \):
\[
R = \frac{12}{2.5 \times 10^{-3}} = 4.8 \times 10^{3}\,\Omega
\]
Thus, the resistance of the resistor is 4.8 kΩ.
Q13. The value of current (I) flowing in a given resistor for the corresponding values of potential difference (V) across the resistor are given below-
| I (amperes) |
0.5 |
1.0 |
2.0 |
3.0 |
4.0 |
| V (volts) |
1.6 |
3.4 |
6.7 |
10.2 |
13.2 |
Plot a graph between V and I and calculate the resistance of that resistor.
Solution: When the graph is plotted between voltage and current then, it is called V-I characteristics. Here, the voltage is plotted on X axis and current is plotted on Y axis.
The slope of the line gives the value of resistance (\( R \)) as:
\[
\text{Slope} = \frac{1}{R} = \frac{BC}{AC} = \frac{2}{6.8}
\]
Therefore:
\[
R = \frac{6.8}{2} = 3.4\,\Omega
\]
Q14. A copper wire has a diameter of 0.5 mm and resistivity of \( 1.6 \times 10^{-8} \, \Omega\text{m} \). What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Solution:
Given:
Resistivity, \( \rho = 1.6 \times 10^{-8} \, \Omega\text{m} \)
Diameter, \( d = 0.5 \, \text{mm} = 0.0005 \, \text{m} \)
Resistance, \( R = 10 \, \Omega \)
The area of the wire's cross-section is given by:
\[
A = \pi r^2 = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{0.0005}{2} \right)^2 = 0.0005^2 \, \text{m}^2 = 2.5 \times 10^{-7} \, \text{m}^2
\]
Using the formula for resistance:
\[
R = \frac{\rho l}{A}
\]
Solving for the length \( l \):
\[
l = \frac{R A}{\rho} = \frac{10 \times 3.14 \times (0.0005)^2}{1.6 \times 10^{-8}} = 122.72 \, \text{m}
\]
Now, if the diameter of the wire is doubled, the new diameter is:
\[
d' = 2 \times 0.5 \, \text{mm} = 1 \, \text{mm} = 0.001 \, \text{m}
\]
The new area of the cross-section is:
\[
A' = \pi \left( \frac{d'}{2} \right)^2 = \pi \left( \frac{0.001}{2} \right)^2 = 2.5 \times 10^{-7} \, \text{m}^2
\]
Now, the new resistance \( R' \) is:
\[
R' = \frac{\rho l}{A'} = \frac{1.6 \times 10^{-8} \times 122.72 \, \text{m} \times (1 \times 10^{-3})^2}{2.5 \times 10^{-7}} = 2.5 \, \Omega
\]
Thus, when the diameter is doubled, the resistance decreases to 2.5 Ω.
Q15. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Solution:
A voltmeter need to be connected in parallel in a circuit to measure the potential difference between two points.
Q16. Two conducting wires of the same material, equal lengths, and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be:
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Solution: The correct option is (c).
Let \( R \) be the resistance of each wire.
### For Series Combination:
The equivalent resistance in series is:
\[
R_s = R + R = 2R
\]
If \( V \) is the applied potential difference, then the voltage across the equivalent resistance is:
\[
V = I_s \times 2R \quad \text{(where \( I_s \) is the current in the series combination)}
\]
Solving for \( I_s \):
\[
I_s = \frac{V}{2R}
\]
The heat dissipated in time \( t \) is given by:
\[
H_s = I_s^2 \times 2R \times t
\]
Substituting the value of \( I_s \):
\[
H_s = \left(\frac{V}{2R}\right)^2 \times 2R \times t = \frac{V^2 t}{2R} \tag{1}
\]
### For Parallel Combination:
The equivalent resistance of the parallel connection is:
\[
R_p = \frac{R}{2}
\]
The potential difference \( V \) is applied across \( R_p \). Thus, the current \( I_p \) is:
\[
V = I_p \times R_p \quad \Rightarrow \quad I_p = \frac{V}{R_p} = \frac{V}{R/2} = \frac{2V}{R}
\]
The heat dissipated in time \( t \) is:
\[
H_p = I_p^2 \times R_p \times t
\]
Substituting the value of \( I_p \):
\[
H_p = \left(\frac{2V}{R}\right)^2 \times \frac{R}{2} \times t = \frac{4V^2 t}{2R} = \frac{2V^2 t}{R} \tag{2}
\]
### Ratio of Heat Produced:
The ratio of heat produced in series to parallel is:
\[
\frac{H_s}{H_p} = \frac{\frac{V^2 t}{2R}}{\frac{2V^2 t}{R}} = \frac{1}{4}
\]
Thus, the ratio of heat produced in series and parallel combinations is 1:4.
Therefore, the correct answer is (c).
Q17 . An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Solution:
Given:
\( V_1 = 220\,\text{V} \), \( V_2 = 110\,\text{V} \)
\( P_1 = 100\,\text{W} \)
Using the relation:
\[
P_1 = \frac{V_1^2}{R} \quad \Rightarrow \quad R = \frac{V_1^2}{P_1} = \frac{(220\,\text{V})^2}{100\,\text{W}} = 484\,\Omega
\]
If the bulb is operated on 110 V, then the power consumed is:
\[
P_2 = \frac{V_2^2}{R} = \frac{(110\,\text{V})^2}{484\,\Omega} = 25\,\text{W}
\]
Therefore, the power consumed will be 25 W.
Q.18Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Solution:
(a) Due to high melting point and high resistivity of tungsten, it does not burn at a high temperature. As the electric lamps glow at very high temperatures hence, the tungsten is used as heating element of electric bulbs.
(b) Bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of pure metals. Due to high resistivity, they produce large amount of heat.
(c) In a series circuit, there is a voltage distribution. Each element of a series circuit receives a small voltage for a large supply voltage. Thus, the amount of current decreases and the device becomes hot. Therefore, series arrangement is not used in domestic circuits.
(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A).
(e) The wires made of copper or aluminium have low resistivity and they are good conductors of electricity. Therefore, they are frequently used for electricity transmission.