---
title: "NCERT Solutions for Class 10 Science Chapter 11 Electricity"
url: https://www.swavid.com/science/class/10/chapter/electricity/ncert-solutions
dateModified: 2026-10-07T16:48:04+00:00
---

# NCERT Solutions for Class 10 Science Chapter 11 Electricity

This chapter's questions cover fundamental concepts of electricity including electric current, Ohm's law, resistance, series and parallel circuits, heating effects, and electrical power through various conceptual and numerical problems.

Free PDF (18 pages): https://www.swavid.com/api/seo/pdf/ncert/science/class-10/swavid-ncert-solutions-class-10-science-chapter-11-electricity-57bfe2b136.pdf

## QUESTIONS

### Question 1

*2 marks · Very short answer*

What does an electric circuit mean?

**Solution**

1. A continuous and closed path of an electric current is called an electric circuit.

**Answer:** A continuous and closed path of an electric current is called an electric circuit.

> Common mistake: Writing an incomplete definition without mentioning that the path must be both continuous and closed.

### Question 2

*2 marks · Very short answer*

Define the unit of current.

**Solution**

1. The SI unit of electric current is ampere (A).
2. One ampere is constituted by the flow of one coulomb of charge per second ($1\text{ A} = 1\text{ C}/1\text{ s}$).

**Answer:** One ampere is the current constituted by the flow of one coulomb of charge per second.

> Common mistake: Stating the unit name without defining it in terms of charge and time.

### Question 3

*2 marks · Numerical*

Calculate the number of electrons constituting one coulomb of charge.

**Solution**

1. Given: Total charge $Q = 1\text{ C}$, charge on one electron $e = 1.6 \times 10^{-19}\text{ C}$
2. Formula: $Q = n e$
3. Substitution: $1\text{ C} = n \times 1.6 \times 10^{-19}\text{ C}$
4. Result: $n = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}\text{ electrons}$

**Answer:** $6.25 \times 10^{18}$ electrons

> Common mistake: Forgetting the negative sign of electron charge or dividing the wrong way around.

## QUESTIONS

### Question 1

*2 marks · Very short answer*

Name a device that helps to maintain a potential difference across a conductor.

**Solution**

1. A cell or a battery helps to maintain a potential difference across a conductor by using chemical energy stored in it to set charges in motion.

**Answer:** A cell or a battery helps to maintain a potential difference across a conductor.

> Common mistake: Writing just 'electricity' or 'generator' instead of specifying a cell or a battery.

### Question 2

*2 marks · Very short answer*

What is meant by saying that the potential difference between two points is $1\text{ V}$?

**Solution**

1. It means that $1\text{ joule}$ of work is done in moving a charge of $1\text{ coulomb}$ from one point to the other.

**Answer:** It means that $1\text{ J}$ of work is done in moving a charge of $1\text{ C}$ from one point to the other.

> Common mistake: Confusing work and charge, or omitting units.

### Question 3

*3 marks · Numerical*

How much energy is given to each coulomb of charge passing through a $6\text{ V}$ battery?

**Solution**

1. Given: Potential difference $V = 6\text{ V}$, Charge $Q = 1\text{ C}$.
2. Formula: Work done or energy given $W = VQ$.
3. Substitution: $W = 6\text{ V} \times 1\text{ C} = 6\text{ J}$.
4. Result: $6\text{ J}$.

**Answer:** $6\text{ J}$

> Common mistake: Writing the answer without the correct SI unit joule.

## QUESTIONS

### Question 1

*3 marks · Short answer*

On what factors does the resistance of a conductor depend?

**Solution**

1. The resistance of a uniform metallic conductor depends directly on its length $l$.
2. It depends inversely on its area of cross-section $A$.
3. It also depends on the nature of its material and temperature.

**Answer:** The resistance of a conductor depends on its length, area of cross-section, nature of material, and temperature.

> Common mistake: Forgetting to mention temperature or the nature of the material.

### Question 2

*3 marks · Short answer*

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

**Solution**

1. Current will flow more easily through a thick wire than through a thin wire of the same material.
2. The resistance of a conductor is inversely proportional to its area of cross-section ($R \propto 1/A$).
3. A thick wire has a larger cross-sectional area, offering less resistance, which allows a larger current to flow easily.

**Answer:** Current flows more easily through a thick wire because it offers lower resistance than a thin wire.

> Common mistake: Stating that thick wires have higher resistance instead of lower resistance.

### Question 3

*3 marks · Short answer*

Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

**Solution**

1. According to Ohm's law, the current $I$ through a resistor is given by $I = \frac{V}{R}$.
2. When the resistance $R$ remains constant, the current is directly proportional to the potential difference $V$.
3. If the potential difference decreases to half of its former value, the current through the component also decreases to half of its former value.

**Answer:** The current through the component decreases to half of its former value.

> Common mistake: Assuming current increases when potential difference decreases.

### Question 4

*3 marks · Short answer*

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

**Solution**

1. The resistivity of an alloy is generally much higher than that of its constituent pure metals.
2. Alloys do not oxidise (burn) readily at high temperatures.
3. Because of these properties, alloy wires are used in electrical heating devices to generate large amounts of heat without burning.

**Answer:** Coils are made of alloys because they have higher resistivity and do not oxidise readily at high temperatures.

> Common mistake: Stating that alloys have lower melting points than pure metals.

### Question 5

*3 marks · Short answer*

Use the data in Table 11.2 to answer the following –
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

**Part (a)**

1. From Table 11.2, the resistivity of iron is $10.0 \times 10^{-8} \; \Omega\text{m}$ and that of mercury is $94.0 \times 10^{-8} \; \Omega\text{m}$.
2. Since iron has a lower resistivity than mercury, iron is a better conductor.

Answer (a): Iron is a better conductor than mercury.

**Part (b)**

1. From Table 11.2, silver has the lowest resistivity among all listed substances with a value of $1.60 \times 10^{-8} \; \Omega\text{m}$.
2. A lower resistivity indicates a better conductor.

Answer (b): Silver is the best conductor.

**Answer:** (a) Iron is a better conductor. (b) Silver is the best conductor.

> Common mistake: Confusing higher resistivity with better conductivity.

## QUESTIONS

### Question 1

*3 marks · Short answer*

Draw a schematic diagram of a circuit consisting of a battery of three cells of $2\text{ V}$ each, a $5\text{ \Omega}$ resistor, an $8\text{ \Omega}$ resistor, and a $12\text{ \Omega}$ resistor, and a plug key, all connected in series.

**Solution**

1. Draw three cells in series representing a battery of $6\text{ V}$ with longer vertical line for positive terminal and shorter thicker line for negative terminal.
2. Connect the battery in series with a plug key, a $5\text{ \Omega}$ resistor, an $8\text{ \Omega}$ resistor, and a $12\text{ \Omega}$ resistor end to end in a closed loop.
3. Represent each resistor by a zig-zag symbol and ensure the circuit forms a complete continuous path.

**Answer:** A schematic diagram showing a $6\text{ V}$ battery, a plug key, and three resistors of $5\text{ \Omega}$, $8\text{ \Omega}$, and $12\text{ \Omega}$ connected in series.

> Common mistake: Leaving the key open in the schematic or showing parallel connections instead of series.

### Question 2

*3 marks · Numerical*

Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the $12\text{ \Omega}$ resistor. What would be the readings in the ammeter and the voltmeter?

**Solution**

1. Given: Total voltage $V = 3 \times 2\text{ V} = 6\text{ V}$, resistors $R_1 = 5\text{ \Omega}$, $R_2 = 8\text{ \Omega}$, $R_3 = 12\text{ \Omega}$.
2. Formula: Total resistance $R_s = R_1 + R_2 + R_3$, current $I = \frac{V}{R_s}$, and potential difference across the $12\text{ \Omega}$ resistor $V_3 = I R_3$.
3. Substitution: $R_s = 5\text{ \Omega} + 8\text{ \Omega} + 12\text{ \Omega} = 25\text{ \Omega}$, $I = \frac{6\text{ V}}{25\text{ \Omega}} = 0.24\text{ A}$, and $V_3 = 0.24\text{ A} \times 12\text{ \Omega} = 2.88\text{ V}$.
4. Result: Ammeter reading = $0.24\text{ A}$, Voltmeter reading = $2.88\text{ V}$.

**Answer:** Ammeter reading is $0.24\text{ A}$ and voltmeter reading is $2.88\text{ V}$.

> Common mistake: Forgetting to sum all three resistances to find the total equivalent resistance of the series combination.

## QUESTIONS

### Question 1

*3 marks · Numerical*

Judge the equivalent resistance when the following are connected in parallel – (a) $1\text{ \Omega}$ and $10^6\text{ \Omega}$, (b) $1\text{ \Omega}$, $10^3\text{ \Omega}$, and $10^6\text{ \Omega}$.

**Part (a)**

1. Given: $R_1 = 1\text{ \Omega}$, $R_2 = 10^6\text{ \Omega}$ connected in parallel.
2. Formula: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$
3. Substitution: $\frac{1}{R_p} = \frac{1}{1} + \frac{1}{10^6} = 1 + 10^{-6}$
4. Result: $R_p \approx 1\text{ \Omega}$ (slightly less than $1\text{ \Omega}$).

Answer (a): Slightly less than $1\text{ \Omega}$

**Part (b)**

1. Given: $R_1 = 1\text{ \Omega}$, $R_2 = 10^3\text{ \Omega}$, $R_3 = 10^6\text{ \Omega}$ connected in parallel.
2. Formula: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
3. Substitution: $\frac{1}{R_p} = \frac{1}{1} + \frac{1}{10^3} + \frac{1}{10^6} = 1 + 0.001 + 0.000001$
4. Result: $R_p \approx 1\text{ \Omega}$ (slightly less than $1\text{ \Omega}$).

Answer (b): Slightly less than $1\text{ \Omega}$

**Answer:** (a) Less than $1\text{ \Omega}$ (very close to $1\text{ \Omega}$). (b) Less than $1\text{ \Omega}$ (very close to $1\text{ \Omega}$).

> Common mistake: Thinking that the equivalent resistance is equal to the smaller resistor value without considering the reciprocal formula.

### Question 2

*4 marks · Numerical*

An electric lamp of $100\text{ \Omega}$, a toaster of resistance $50\text{ \Omega}$, and a water filter of resistance $500\text{ \Omega}$ are connected in parallel to a $220\text{ V}$ source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

**Solution**

1. Given: Resistance of lamp $R_1 = 100\text{ }\Omega$, toaster $R_2 = 50\text{ }\Omega$, water filter $R_3 = 500\text{ }\Omega$, voltage $V = 220\text{ V}$.
2. Formula: Equivalent resistance in parallel $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$ and Ohm's law $I = \frac{V}{R}$.
3. Substitution: $\frac{1}{R_p} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500} = \frac{5 + 10 + 1}{500} = \frac{16}{500}$, so $R_p = \frac{500}{16} = 31.25\text{ }\Omega$.
4. Current through the combination (which equals the current through the electric iron) is $I = \frac{220\text{ V}}{31.25\text{ }\Omega} = 7.04\text{ A}$.
5. Result: Resistance of the electric iron is $31.25\text{ }\Omega$ and the current through it is $7.04\text{ A}$.

**Answer:** Resistance = 31.25 \Omega, Current = 7.04 A

> Common mistake: Adding resistances directly as in series instead of taking reciprocals for parallel connection.

### Question 3

*3 marks · Short answer*

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

**Solution**

1. In a parallel circuit, if one electrical appliance fails or is switched off, other appliances continue to work normally without getting affected.
2. Each appliance gets the same potential difference as the source voltage, operating at its maximum efficiency.
3. The total resistance in a parallel circuit is low, which allows devices to draw the required amount of current.

**Answer:** Parallel circuits prevent disruption when one component fails, provide equal voltage to all devices, and keep total circuit resistance low.

> Common mistake: Stating advantages without mentioning independent working or constant voltage supply.

### Question 4

*3 marks · Numerical*

How can three resistors of resistances $2\text{ \Omega}$, $3\text{ \Omega}$, and $6\text{ \Omega}$ be connected to give a total resistance of (a) $4\text{ \Omega}$, (b) $1\text{ \Omega}$?

**Part (a)**

1. Idea: Parallel combination of $3\text{ \Omega}$ and $6\text{ \Omega}$ gives $2\text{ \Omega}$, which when added in series with $2\text{ \Omega}$ gives $4\text{ \Omega}$.
2. Working: $\frac{1}{R'} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$, so $R' = 2\text{ \Omega}$.
3. Total resistance: $R = 2\text{ \Omega} + R' = 2\text{ \Omega} + 2\text{ \Omega} = 4\text{ \Omega}$.

Answer (a): $3\text{ \Omega}$ and $6\text{ \Omega}$ in parallel, connected in series with $2\text{ \Omega}$

**Part (b)**

1. Idea: Connect all three resistors in parallel to obtain the minimum possible resistance.
2. Working: $\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1\text{ \Omega}$.
3. Result: $R_p = 1\text{ \Omega}$.

Answer (b): All three resistors ($2\text{ \Omega}$, $3\text{ \Omega}$, and $6\text{ \Omega}$) connected in parallel

**Answer:** (a) Connect $3\text{ \Omega}$ and $6\text{ \Omega}$ in parallel and then connect the combination in series with $2\text{ \Omega}$. (b) Connect all three resistors ($2\text{ \Omega}$, $3\text{ \Omega}$, $6\text{ \Omega}$) in parallel.

> Common mistake: Trying random combinations without checking the rules for series and parallel equivalent resistance.

### Question 5

*3 marks · Numerical*

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance $4\text{ \Omega}$, $8\text{ \Omega}$, $12\text{ \Omega}$, $24\text{ \Omega}$?

**Part (a)**

1. Given: Resistors of $4\text{ \Omega}$, $8\text{ \Omega}$, $12\text{ \Omega}$, and $24\text{ \Omega}$.
2. Idea: To get the highest total resistance, all resistors must be connected in series.
3. Working: $R_s = R_1 + R_2 + R_3 + R_4 = 4 + 8 + 12 + 24 = 48\text{ \Omega}$.

Answer (a): $48\text{ \Omega}$

**Part (b)**

1. Idea: To get the lowest total resistance, all resistors must be connected in parallel.
2. Working: $\frac{1}{R_p} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24} = \frac{1}{2}$.
3. Result: $R_p = 2\text{ \Omega}$.

Answer (b): $2\text{ \Omega}$

**Answer:** (a) Highest resistance = $48\text{ \Omega}$. (b) Lowest resistance = $2\text{ \Omega}$.

> Common mistake: Confusing series and parallel formulas for maximum and minimum resistance.

## QUESTIONS

### Question 1

*3 marks · Short answer*

Why does the cord of an electric heater not glow while the heating element does?

**Solution**

1. The heating element and the connecting cord are connected in series, so the same current flows through both.
2. The resistance of the heating element is very high, whereas the resistance of the connecting cord is extremely small.
3. According to Joule's law of heating ($H = I^2Rt$), the heat produced in the cord is negligible, while a large amount of heat is produced in the heating element, making it glow.

**Answer:** The heating element has high resistance producing large heat to glow, while the cord has very low resistance producing negligible heat.

> Common mistake: Thinking that different current flows through the cord and the heating element.

### Question 2

*3 marks · Numerical*

Compute the heat generated while transferring $96000\text{ C}$ of charge in one hour through a potential difference of $50\text{ V}$.

**Solution**

1. Given: Charge $Q = 96000\text{ C}$, Time $t = 1\text{ hr} = 3600\text{ s}$, Potential difference $V = 50\text{ V}$.
2. Formula: $H = VIt = VQ$ since $Q = It$.
3. Substitution: $H = 50\text{ V} \times 96000\text{ C}$.
4. Result: $H = 4800000\text{ J} = 4.8 \times 10^6\text{ J}$.

**Answer:** $4.8 \times 10^6\text{ J}$

> Common mistake: Converting time incorrectly or using $I^2Rt$ without finding current when $VQ$ is directly applicable.

### Question 3

*3 marks · Numerical*

An electric iron of resistance $20\text{ \Omega}$ takes a current of $5\text{ A}$. Calculate the heat developed in $30\text{ s}$.

**Solution**

1. Given: Resistance $R = 20\text{ }\Omega$, Current $I = 5\text{ A}$, Time $t = 30\text{ s}$.
2. Formula: $H = I^2Rt$.
3. Substitution: $H = (5\text{ A})^2 \times 20\text{ }\Omega \times 30\text{ s}$.
4. Result: $H = 25 \times 20 \times 30 = 15000\text{ J} = 1.5 \times 10^4\text{ J}$.

**Answer:** $15000\text{ J}$

> Common mistake: Forgetting to square the current in the formula $I^2Rt$.

## QUESTIONS

### Question 1

*2 marks · Very short answer*

What determines the rate at which energy is delivered by a current?

**Solution**

1. The rate at which energy is delivered or consumed by a current is determined by electric power.
2. Electric power is given by the product of potential difference and current ($P = VI$).

**Answer:** Electric power, which is the product of potential difference and current ($P = VI$), determines the rate at which energy is delivered.

> Common mistake: Confusing electric power with electrical energy or resistance.

### Question 2

*3 marks · Numerical*

An electric motor takes $5\text{ A}$ from a $220\text{ V}$ line. Determine the power of the motor and the energy consumed in $2\text{ h}$.

**Solution**

1. Given: Current $I = 5\text{ A}$, Potential difference $V = 220\text{ V}$, Time $t = 2\text{ h} = 7200\text{ s}$.
2. Formula: Power $P = VI$ and Energy $E = P \times t$.
3. Substitution: $P = 220\text{ V} \times 5\text{ A} = 1100\text{ W}$, and $E = 1100\text{ W} \times 7200\text{ s} = 7920000\text{ J} = 2.2\text{ kW h}$.
4. Result: Power = $1100\text{ W}$, Energy consumed = $7.92 \times 10^6\text{ J}$ (or $2.2\text{ kW h}$).

**Answer:** Power = $1100\text{ W}$, Energy consumed = $7.92 \times 10^6\text{ J}$ ($2.2\text{ kW h}$)

> Common mistake: Forgetting to convert hours into seconds while calculating energy in joules, or forgetting to convert watts to kilowatts for watt-hours.

## EXERCISES

### Question 1

*1 mark · MCQ*

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 $R/R′$ is –

- 1/25
- 1/5
- 5
- 25

**Solution**

1. Resistance of each of the five parts is $R/5$.
2. When connected in parallel, equivalent resistance $R'$ is given by $\frac{1}{R'} = \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} = \frac{25}{R}$, so $R' = \frac{R}{25}$, giving $\frac{R}{R'} = 25$.

**Answer:** (d) 25

> Common mistake: Forgetting that the resistance of each part decreases when cut, or inverting the ratio.

### Question 2

*1 mark · MCQ*

Which of the following terms does not represent electrical power in a circuit?

- I^2R
- IR^2
- VI
- V^2/R

**Solution**

1. Electrical power is given by $P = VI$, $P = I^2R$, and $P = \frac{V^2}{R}$.
2. Thus, $IR^2$ does not represent electrical power.

**Answer:** (b) IR^2

> Common mistake: Confusing $I^2R$ with $IR^2$.

### Question 3

*1 mark · MCQ*

An electric bulb is rated $220\text{ V}$ and $100\text{ W}$. When it is operated on $110\text{ V}$, the power consumed will be –

- 100 W
- 75 W
- 50 W
- 25 W

**Solution**

1. The resistance of the bulb is given by $R = \frac{V^2}{P} = \frac{(220\text{ V})^2}{100\text{ W}} = 484\text{ \Omega}$.
2. When operated on $110\text{ V}$, the power consumed is $P' = \frac{V'^2}{R} = \frac{(110\text{ V})^2}{484\text{ \Omega}} = 25\text{ W}$.

**Answer:** (d) 25 W

> Common mistake: Assuming power halves when voltage halves, forgetting that power depends on the square of voltage.

### Question 4

*1 mark · MCQ*

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 –

- 1:2
- 2:1
- 1:4
- 4:1

**Solution**

1. Let the resistance of each wire be $R$. For series combination, $R_s = R + R = 2R$, and heat produced is $H_s = \frac{V^2 t}{R_s} = \frac{V^2 t}{2R}$.
2. For parallel combination, equivalent resistance is $R_p = \frac{R \times R}{R + R} = \frac{R}{2}$, and heat produced is $H_p = \frac{V^2 t}{R_p} = \frac{V^2 t}{R/2} = \frac{2V^2 t}{R}$.
3. The ratio of heat produced in series to parallel is $\frac{H_s}{H_p} = \frac{V^2 / 2R}{2V^2 / R} = \frac{1}{4}$, which corresponds to option (c).

**Answer:** (c) 1:4

> Common mistake: Inverting the ratio by taking parallel to series instead of series to parallel.

### Question 5

*2 marks · Very short answer*

How is a voltmeter connected in the circuit to measure the potential difference between two points?

**Solution**

1. A voltmeter is an instrument used to measure the potential difference between two points in a circuit.
2. It is always connected in parallel across the points between which the potential difference is to be measured.

**Answer:** The voltmeter is always connected in parallel across the points where the potential difference is to be measured.

> Common mistake: Stating that a voltmeter is connected in series.

### Question 6

*3 marks · Numerical*

A copper wire has diameter $0.5\text{ mm}$ and resistivity of $1.6 \times 10^{-8}\text{ \Omega m}$. What will be the length of this wire to make its resistance $10\text{ \Omega}$? How much does the resistance change if the diameter is doubled?

**Solution**

1. Given: $d = 0.5\text{ mm} = 5 \times 10^{-4}\text{ m}$, radius $r = 2.5 \times 10^{-4}\text{ m}$, resistivity $\rho = 1.6 \times 10^{-8}\text{ \Omega m}$, $R = 10\text{ \Omega}$.
2. Formula: $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$
3. Substitution for length: $l = \frac{R \pi r^2}{\rho} = \frac{10 \times \frac{22}{7} \times (2.5 \times 10^{-4})^2}{1.6 \times 10^{-8}} = 122.7\text{ m}$.
4. For the second part, since resistance is inversely proportional to the area of cross-section ($R \propto \frac{1}{d^2}$), doubling the diameter increases the area by 4 times.
5. Result: The resistance becomes one-fourth of its initial value, decreasing by $\frac{3}{4}R$.

**Answer:** Length = 122.7 m, and resistance becomes one-fourth when diameter is doubled.

> Common mistake: Failing to convert millimeters to meters or radius to diameter correctly.

### Question 7

*3 marks · Numerical*

The values 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**

1. Given: The observations of current $I$ and potential difference $V$ are plotted on a graph, and the resistance is obtained from the slope of the $V-I$ graph.
2. Formula: $R = \frac{V}{I}$
3. Substitution: Taking a representative set of values from the given data, such as $V = 6.7\text{ V}$ and $I = 2.0\text{ A}$ (or averaging the ratios across the points), $R = \frac{6.7}{2.0} = 3.35\text{ \Omega}$ (or approximately $3.4\text{ \Omega}$).
4. Result: $3.4\text{ \Omega}$

**Answer:** $3.4\text{ \Omega}$

> Common mistake: Forgetting to convert units or taking the inverse ratio $\frac{I}{V}$ instead of $\frac{V}{I}$.

### Question 8

*3 marks · Numerical*

When a $12\text{ V}$ battery is connected across an unknown resistor, there is a current of $2.5\text{ mA}$ in the circuit. Find the value of the resistance of the resistor.

**Solution**

1. Given: Potential difference $V = 12\text{ V}$, current $I = 2.5\text{ mA} = 2.5 \times 10^{-3}\text{ A}$.
2. Formula: $R = \frac{V}{I}$
3. Substitution: $R = \frac{12\text{ V}}{2.5 \times 10^{-3}\text{ A}}$
4. Result: $4800\text{ \Omega}$ or $4.8\text{ k\Omega}$

**Answer:** $4800\text{ \Omega}$

> Common mistake: Not converting milliamperes to amperes before applying Ohm's law.

### Question 9

*3 marks · Numerical*

A battery of $9\text{ V}$ is connected in series with resistors of $0.2\text{ \Omega}$, $0.3\text{ \Omega}$, $0.4\text{ \Omega}$, $0.5\text{ \Omega}$ and $12\text{ \Omega}$, respectively. How much current would flow through the $12\text{ \Omega}$ resistor?

**Solution**

1. Given: Battery voltage $V = 9\text{ V}$, resistors connected in series $R_1 = 0.2\text{ \Omega}$, $R_2 = 0.3\text{ \Omega}$, $R_3 = 0.4\text{ \Omega}$, $R_4 = 0.5\text{ \Omega}$, $R_5 = 12\text{ \Omega}$.
2. Formula: Total resistance in series $R_s = R_1 + R_2 + R_3 + R_4 + R_5$ and $I = \frac{V}{R_s}$.
3. Substitution: $R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4\text{ \Omega}$, so $I = \frac{9\text{ V}}{13.4\text{ \Omega}}$.
4. Result: Since the current is the same in all parts of a series circuit, the current through the $12\text{ \Omega}$ resistor is $0.67\text{ A}$.

**Answer:** 0.67 A

> Common mistake: Calculating current only for the $12\text{ \Omega}$ resistor using its own voltage instead of the total equivalent resistance.

### Question 10

*3 marks · Numerical*

How many $176\text{ \Omega}$ resistors (in parallel) are required to carry $5\text{ A}$ on a $220\text{ V}$ line?

**Solution**

1. Given: Voltage $V = 220\text{ V}$, total current $I = 5\text{ A}$, resistance of each branch $R = 176\text{ \Omega}$.
2. Formula: Total resistance of the circuit $R_p = \frac{V}{I}$.
3. Substitution: $R_p = \frac{220\text{ V}}{5\text{ A}} = 44\text{ \Omega}$.
4. Result: If $n$ resistors of resistance $176\text{ \Omega}$ are connected in parallel, $\frac{176}{n} = 44$, which gives $n = 4$.

**Answer:** 4

> Common mistake: Using the formula for series combination instead of parallel combination.

### Question 11

*3 marks · Short answer*

Show how you would connect three resistors, each of resistance $6\text{ \Omega}$, so that the combination has a resistance of (i) $9\text{ \Omega}$, (ii) $4\text{ \Omega}$.

**Part (i)**

1. To get a total resistance of $9\text{ \Omega}$ using three $6\text{ \Omega}$ resistors, connect two resistors in parallel and the third in series with them.
2. The equivalent resistance of two $6\text{ \Omega}$ resistors in parallel is $\frac{6 \times 6}{6 + 6} = 3\text{ \Omega}$.
3. Adding the third $6\text{ \Omega}$ resistor in series gives $3\text{ \Omega} + 6\text{ \Omega} = 9\text{ \Omega}$.

Answer (i): Two resistors in parallel connected in series with the third resistor.

**Part (ii)**

1. To get a total resistance of $4\text{ \Omega}$ using three $6\text{ \Omega}$ resistors, connect two resistors in series and the third in parallel with them.
2. The equivalent resistance of two $6\text{ \Omega}$ resistors in series is $6\text{ \Omega} + 6\text{ \Omega} = 12\text{ \Omega}$.
3. Connecting this combination in parallel with the third $6\text{ \Omega}$ resistor gives $\frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\text{ \Omega}$.

Answer (ii): Two resistors in series connected in parallel with the third resistor.

**Answer:** Combinations explained in parts.

> Common mistake: Confusing series and parallel formulas when designing the combination.

### Question 12

*3 marks · Numerical*

Several electric bulbs designed to be used on a $220\text{ V}$ electric supply line, are rated $10\text{ W}$. How many lamps can be connected in parallel with each other across the two wires of $220\text{ V}$ line if the maximum allowable current is $5\text{ A}$?

**Solution**

1. Given: Voltage $V = 220\text{ \text{V}}$, power of each bulb $P = 10\text{ \text{W}}$, maximum allowable current $I = 5\text{ \text{A}}$.
2. Formula: Current drawn by each bulb $I_1 = \frac{P}{V}$ and total current $I_{\text{total}} = n \times I_1$, where $n$ is the number of lamps.
3. Substitution: $I_1 = \frac{10\text{ \text{W}}}{220\text{ \text{V}}} = \frac{1}{22}\text{ \text{A}}$, and $5 = n \times \frac{1}{22}$.
4. Result: $n = 5 \times 22 = 110$ lamps.

**Answer:** $110$ lamps

> Common mistake: Dividing total power by power of one bulb incorrectly or forgetting that currents add up in parallel.

### Question 13

*3 marks · Numerical*

A hot plate of an electric oven connected to a $220\text{ V}$ line has two resistance coils A and B, each of $24\text{ \Omega}$ resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

**Solution**

1. Given: Voltage $V = 220\text{ V}$, resistance of each coil $R_A = R_B = 24\text{ \Omega}$.
2. Case (i) Used separately: Current $I = \frac{V}{R} = \frac{220\text{ V}}{24\text{ \Omega}} = 9.17\text{ A}$.
3. Case (ii) Connected in series: Total resistance $R_s = R_A + R_B = 24\text{ \Omega} + 24\text{ \Omega} = 48\text{ \Omega}$; Current $I = \frac{220\text{ V}}{48\text{ \Omega}} = 4.58\text{ A}$.
4. Case (iii) Connected in parallel: Equivalent resistance $\frac{1}{R_p} = \frac{1}{24} + \frac{1}{24} = \frac{2}{24}$, so $R_p = 12\text{ \Omega}$; Current $I = \frac{220\text{ V}}{12\text{ \Omega}} = 18.33\text{ A}$.
5. Result: Currents are $9.17\text{ A}$, $4.58\text{ A}$, and $18.33\text{ A}$ respectively.

**Answer:** $9.17\text{ A}$, $4.58\text{ A}$, $18.33\text{ A}$

> Common mistake: Confusing series and parallel formulas for equivalent resistance.

### Question 14

*3 marks · Numerical*

Compare the power used in the $2\text{ \Omega}$ resistor in each of the following circuits: (i) a $6\text{ V}$ battery in series with $1\text{ \Omega}$ and $2\text{ \Omega}$ resistors, and (ii) a $4\text{ V}$ battery in parallel with $12\text{ \Omega}$ and $2\text{ \Omega}$ resistors.

**Solution**

1. Given: Circuit (i) $6\text{ V}$ battery with $1\text{ \Omega}$ and $2\text{ \Omega}$ resistors in series; Circuit (ii) $4\text{ V}$ battery with $12\text{ \Omega}$ and $2\text{ \Omega}$ resistors in parallel.
2. Circuit (i): Total resistance $R = 1\text{ \Omega} + 2\text{ \Omega} = 3\text{ \Omega}$; Circuit current $I = \frac{6\text{ V}}{3\text{ \Omega}} = 2\text{ A}$; Power used in $2\text{ \Omega}$ resistor $P_1 = I^2 R = (2\text{ A})^2 \times 2\text{ \Omega} = 8\text{ W}$.
3. Circuit (ii): The $2\text{ \Omega}$ resistor is connected directly across the $4\text{ V}$ battery, so potential difference across it is $V = 4\text{ V}$.
4. Power used in $2\text{ \Omega}$ resistor $P_2 = \frac{V^2}{R} = \frac{(4\text{ V})^2}{2\text{ \Omega}} = \frac{16}{2} = 8\text{ W}$.
5. Result: Ratio of power used is $8\text{ W} : 8\text{ W} = 1:1$.

**Answer:** $8\text{ W}$ in both circuits (Ratio $1:1$)

> Common mistake: Assuming potential difference is divided equally across parallel resistors.

### Question 15

*3 marks · Numerical*

Two lamps, one rated $100\text{ W}$ at $220\text{ V}$, and the other $60\text{ W}$ at $220\text{ V}$, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is $220\text{ V}$?

**Solution**

1. Given: Lamp 1 rated $100\text{ W}$ at $220\text{ V}$, Lamp 2 rated $60\text{ W}$ at $220\text{ V}$, supply voltage $V = 220\text{ V}$.
2. Formula: Power $P = VI$, so current drawn by each lamp is $I = \frac{P}{V}$.
3. Current through first lamp $I_1 = \frac{100\text{ W}}{220\text{ V}} = \frac{5}{11}\text{ A}$.
4. Current through second lamp $I_2 = \frac{60\text{ W}}{220\text{ V}} = \frac{3}{11}\text{ A}$.
5. Total current drawn from the line $I = I_1 + I_2 = \frac{5}{11} + \frac{3}{11} = \frac{8}{11}\text{ A} = 0.73\text{ A}$.

**Answer:** $0.73\text{ A}$ (or $\frac{8}{11}\text{ A}$)

> Common mistake: Adding resistances instead of currents for parallel connected appliances.

### Question 16

*3 marks · Numerical*

Which uses more energy, a $250\text{ W}$ TV set in $1\text{ hr}$, or a $1200\text{ W}$ toaster in $10\text{ minutes}$?

**Solution**

1. Given: TV power $P_1 = 250\text{ W}$, time $t_1 = 1\text{ hr}$; Toaster power $P_2 = 1200\text{ W}$, time $t_2 = 10\text{ minutes} = \frac{1}{6}\text{ hr}$.
2. Formula: Energy $E = P \times t$.
3. Energy consumed by TV $E_1 = 250\text{ W} \times 1\text{ h} = 250\text{ Wh} = 0.25\text{ kWh}$.
4. Energy consumed by toaster $E_2 = 1200\text{ W} \times \frac{1}{6}\text{ h} = 200\text{ Wh} = 0.20\text{ kWh}$.
5. Result: The $250\text{ W}$ TV set uses more energy.

**Answer:** $250\text{ W}$ TV set

> Common mistake: Comparing power ratings directly without considering time.

### Question 17

*3 marks · Numerical*

An electric heater of resistance $44\text{ \Omega}$ draws $5\text{ A}$ from the service mains for $2\text{ hours}$. Calculate the rate at which heat is developed in the heater.

**Solution**

1. Given: Resistance $R = 44\text{ \Omega}$, current $I = 5\text{ A}$, time $t = 2\text{ hours}$.
2. Formula: Rate at which heat is developed is the electric power, given by $P = I^2 R$.
3. Substitution: $P = (5\text{ A})^2 \times 44\text{ \Omega} = 25 \times 44 = 1100\text{ W}$.
4. Result: $1100\text{ J/s}$ (or $1100\text{ W}$ or $1.1\text{ kW}$).

**Answer:** $1100\text{ J/s}$ (or $1100\text{ W}$)

> Common mistake: Multiplying by time when the question asks for the rate of heat development.

### Question 18

*3 marks · Short answer*

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?

**Part (a)**

1. Tungsten has a very high melting point ($3380^\circ\text{C}$) and high resistivity.
2. It does not melt even at very high temperatures when heated to incandescence.

Answer (a): Due to its very high melting point and resistivity.

**Part (b)**

1. Alloys generally have higher resistivity than their constituent metals and do not oxidize readily at high temperatures.
2. This prevents them from burning out easily under high heating conditions.

Answer (b): Due to higher resistivity and resistance to oxidation at high temperatures.

**Part (c)**

1. In a series arrangement, if one component fails, the circuit breaks and all other components stop working.
2. Also, all devices share the same current and cannot have individual switches.

Answer (c): Failure of one component breaks the entire circuit, and devices cannot operate independently.

**Part (d)**

1. Resistance of a uniform wire is inversely proportional to its area of cross-section ($R \propto \frac{1}{A}$).
2. Thicker wires have lower resistance, allowing current to flow more easily.

Answer (d): Resistance is inversely proportional to the cross-sectional area.

**Part (e)**

1. Copper and aluminium have very low electrical resistivity.
2. This ensures minimum loss of electrical energy as heat during long-distance transmission.

Answer (e): Due to their very low electrical resistivity.

**Answer:** Detailed explanations for all parts based on textbook concepts.

> Common mistake: Writing general properties instead of specific reasons like melting point or resistivity.

## Frequently asked questions

### How many total questions and exercises are covered in SwaVid's NCERT Solutions for Class 10 Science Chapter 11 Electricity?

This chapter features 23 textual questions divided across multiple sections along with 18 comprehensive exercises. SwaVid provides step-by-step solutions and a free PDF for all these questions right on this page.

### Which important topics and question types are included in the textual solutions?

The solutions cover numerical, very short answer, and short answer types focusing on concepts like electric circuits, Ohm's law, electrical resistivity, Joule's law of heating, and electric power. You will also find various multiple-choice and descriptive questions in the main exercises.

### What are the hardest question types in this chapter and how should I approach them?

Numerical problems involving complex series and parallel resistor combinations or Joule's law of heating calculations are generally considered the toughest. To approach them, first list the given values, apply standard formulas like $V = IR$ or $P = I^2Rt$, and ensure all units are converted to the SI system.

### How can I write answers for full marks in Class 10 board exams for Electricity?

To score full marks, always state the relevant formula clearly, substitute values with proper units, and draw neat circuit diagrams where required. SwaVid's expert-verified answers on this page demonstrate this exact board-presentation format.

### Is the free PDF for Class 10 Science Chapter 11 Electricity available for download?

Yes, the complete chapter PDF along with detailed explanations is available for free download on this SwaVid page. Students can use these resources for effective revision during the 2026-27 academic session.

## Related pages

- [Electricity: CBSE previous year questions](https://www.swavid.com/cbse/class-10/science/pyq/electricity)
- [Class 10 Science chapters](https://www.swavid.com/science/class/10)

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