NCERT Solutions For Class 10 Science Chapter 11 Electricity

NCERT Solutions For Class 10 Science Chapter 11 Electricity

Vishal kumarUpdated on 19 Aug 2025, 07:17 PM IST

Electricity is an important part of our lives every day-lights and fans, mobile phones, laptops and other important appliances all run due to electricity. Its applications greatly touch on people living at home, school, hospital, factory, business and even in agriculture. We have become heavily reliant on the power that it can erupt our daily schedule even by a momentary power outage.

This Story also Contains

  1. NCERT Solution for Class 10 Science chapter 11 solutions: Download Solution PDF and Formulae Handbook
  2. NCERT Class 10 Science Chapter 11 :Electricity - Intext Questions Solution
  3. NCERT Class 10 Science Electricity - Exercise Solution
  4. Class 10 Physics NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions
  5. NCERT Solutions for Class 10 Science Chapter 11: Important Topics
  6. NCERT Solutions for Class 10 Science Chapter 11: Important Formulae
  7. Approach to Solve Questions of Current Electricity
  8. Benefits of NCERT Solutions Class 10 Chapter 11: Electricity
  9. NCERT Solutions for Class 10 Science: Chapter Wise
NCERT Solutions For Class 10 Science Chapter 11 Electricity
NCERT Solutions for Class 10 Science Chapter Electricity(Image: Shutterstock)


In Class 10 Science Chapter 11- Electricity, students learn how electric circuits work and the fundamental concepts students can use including the electric current, voltage, resistance, Ohm Law, and power. Through NCERT solutions, students are able to tackle numerically some of the issues that relate to the present calculation, resistance, and power consumption in devices with a lot of confidence. The grasping of these topics is not only important in exams, but also in gaining a clear understanding of how to use the electrical systems in real life. The NCERT solutions for class 10 Science Chapter 11 contains: the PDF in form that can be downloaded, answers to intext questions, solutions to end of chapter exercises, Higher Order Thinking Skills (HOTS) questions, covers all the significant topics of this chapter.


NCERT Solution for Class 10 Science chapter 11 solutions: Download Solution PDF and Formulae Handbook

The NCERT Solutions of Class 10 Science Chapter 11- Electricity provides step wise as well as clear explanations to all the textbook questions and thus making the concepts of electricity fundamental to the students easy to grasp. All these solutions are provided in PDF format so that one can download all solutions and read conveniently offline.

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Download the chapter-wise formula of NCERT class 10 science by clicking on the link given in the box.

Download Formulae Handbook

NCERT Class 10 Science Chapter 11 :Electricity - Intext Questions Solution

NCERT Solutions of Class 10 Science Chapter 11 Electricity give a clear as well as detailed explanation of the answers to all intext questions present in the NCERT text book. Through these solutions the students will be able to grasp some important concepts without prior knowledge of electric currents, potential differences, resistance and Ohm s law in a sequential manner guaranteeing efficient board preparation.

Class 10 science Chapter 11 ncert Solutions - Topic:11.1

Q.1 What does an electric circuit mean?

Answer:

A closed and continuous path of electric current is known as the electric circuit. It consists of electric circuit elements like batteries, resistors, etc, and electric devices like a switch and measuring devices like ammeters, etc.

Q.2 Define the unit of currcurrent ent.

Answer:

Ampere(A) is unit of current.1A is flow of 1C of charge through a wire in 1s of time.

current $I=\frac{q}{t}$

$
1 A=\frac{1 C}{1 \sec }
$

Q.3 Calculate the number of electrons constituting one coulomb of charge.

Answer:

Given: Q=1C

We know that the charge of an electron $e=1.6 \times 10^{-19} \mathrm{C}$

$
\begin{aligned}
& Q=n e \\
& \Rightarrow n=\frac{Q}{e} \\
& \Rightarrow n=\frac{1}{1.6 \times 10^{-19}}=6.25 \times 10^{18}
\end{aligned}
$


Thus, the number of electrons constituting one coulomb of charge is $6 \times 10^{18}$ electrons.

Class 10 science Chapter 11 ncert Solutions - Topic 11.2

Q.1 Name a device that helps to maintain a potential difference across a conductor.

Answer:

Battery, cell or power supply source helps to maintain a potential difference across a conductor.

Q.2 What is meant by saying that the potential difference between two points is 1 V?

Answer:

The potential difference between two points is 1 V means 1 J of work is required to move a charge of amount 1C from one point to another.

Q.3 How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer:

Given: potential difference = 6V and charge =1C.

$
\begin{aligned}
& \text { Potential difference }=\frac{\text { work done }}{\text { charge }} \\
& \Rightarrow 6=\frac{\text { work done }}{1} \\
& \Rightarrow \text { work done }=6 \times 1=6 \mathrm{~J}
\end{aligned}
$


Thus, 6J energy is given to each coulomb of charge passing through a 6 V battery.

NCERT solutions for class 10 science chapter 11 Electricity: Topic 11.5

Q.1 On what factors does the resistance of a conductor depend?

Answer:

The resistance of a conductor depends on :

1. Cross-section area of the conductor.

2. Length of conductor

3. The temperature of the conductor.

4. Nature of material of the conductor.

Q.2 Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer:

We know that resistance is given as

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

$\mathrm{R}=$ resistance
$\rho=$ resistivity
$l$=length of wire
$A=$ area of cross-section
Resistance is inversely proportional to the area of the cross-section.
Thicker the wire more is cross-sectional area resulting in less resistance resulting in more current flow.

Q.3 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?

Answer:

By Ohm's law,

$
\begin{aligned}
& \mathrm{V}=\mathrm{IR} \\
& I=\frac{V}{R}
\end{aligned}
$

$\mathrm{V}=$ potential difference
I =current
$\mathrm{R}=$ resistance
Now, the potential difference is reduced to half i.e.

$
V^{\prime}=\frac{V}{2}
$

$\mathrm{R}^{\prime}=\mathrm{R}=$ resistance
I' =current

$
I^{\prime}=\frac{\frac{V}{2}}{R}=\frac{V}{2 R}=\frac{I}{2}
$

Q.4 Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer:

The resistivity of an alloy is higher than pure metal so alloy does not melt at high temperature. Thus, coils of electric toasters and electric irons are made of an alloy rather than a pure metal.

Q.5 Use the data in Table to answer the following –

(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

1644209855219

Answer:

Lower the value of resistivity lower will be the resistance for a material of given area and length. If resistance is law current flow will be high when a potential difference is applied across the conductor.

(a) resistivity of iron $=10.0 \times 10^{-8} \Omega \mathrm{~m}$
the resistivity of mercury $=94 \times 10^{-8} \Omega \mathrm{~m}$
the resistivity of mercury is more than iron so iron is a better conductor than mercury.

(b) From the table, we can observe silver has the lowest resistivity so it is the best conductor.

NCERT solutions for class 10 science chapter 11 Electricity: Topic 11.6

Q.1 Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a $5 \Omega$ resistor, an $8 \Omega$ resistor, a $12 \Omega$ resistor, and a plug key, all connected in series.

Answer:

The schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a $5 \Omega$ resistor, an $8 \Omega$ resistor, and a $12 \Omega$ resistor, and a plug key, all connected in series is as shown below :

1644209887262

Q.2 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 $12 \Omega$ resistor. What would be the readings in the ammeter and the voltmeter?

Answer:

The diagram is as shown :

1644210295069

Resistance of circuit, R=5+8+12=25

Potential = 6V

$
\begin{aligned}
& V=I R \\
& \Rightarrow I=\frac{V}{R}=\frac{6}{25}=0.24 \mathrm{~A}
\end{aligned}
$


Now, for 12 ohm resistor, current $=0.24 \mathrm{~A}$.
By Ohm's law,

$
V=I R=0.24 \times 12=2.88 \mathrm{~V}
$


The reading of the ammeter is 0.24 A and the voltmeter is 2.88 V.

NCERT solutions for class 10 science chapter 11 Electricity Topic 11.6

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

Answer:
(a) $1 \Omega$ and $10^6 \Omega$
$R=$ Equivalent resistance

$
\begin{aligned}
& \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow \frac{1}{R}=\frac{1}{1}+\frac{1}{10^6} \\
& \Rightarrow \frac{1}{R}=\frac{10^6+1}{10^6} \\
& \Rightarrow R=\frac{10^6}{10^6+1} \approx 1 \Omega
\end{aligned}
$

(b) $1 \Omega$ and $10^3 \Omega$, and $10^6 \Omega$,
$\mathrm{R}=$ Equivalent resistance

$
\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}
$

$\begin{aligned} & \Rightarrow \frac{1}{R}=\frac{1}{1}+\frac{1}{10^3}+\frac{1}{10^6} \\ & \Rightarrow \frac{1}{R}=\frac{10^6+10^3+1}{10^6} \\ & \Rightarrow R=\frac{10^6}{10^6+10^3+1}=0.999 \Omega \approx 1 \Omega\end{aligned}$

Q.2 An electric lamp of $100 \Omega$ , a toaster of resistance $50 \Omega$ and a water filter of resistance $500 \Omega$ are connected in parallel to a 220 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?

Answer:

Given : $R_1=100 \Omega, R_2=50 \Omega, R_3=500 \Omega$
$\mathrm{R}=$ Equivalent resistance

$
\begin{aligned}
& \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \\
& \Rightarrow \frac{1}{R}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500} \\
& \Rightarrow \frac{1}{R}=\frac{5+10+1}{500}=\frac{16}{500} \\
& \Rightarrow R=\frac{500}{16}=31.25 \Omega
\end{aligned}
$


By Ohm's law,

$
I=\frac{V}{R}=\frac{220}{31.25}=7.04 A
$


Hence, the resistance of electric iron is 31.25, and the current through it is 7.04 A.

Q.3 What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer:

In parallel, there is no division of voltage among the appliances so the potential difference across all appliances is equal and the total effective resistance of a circuit can be reduced. Hence, connecting electrical devices in parallel with the battery is beneficial instead of connecting them in series.

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

Answer:
(a) $R_1=3, R_2=6, R_3=2$
$\mathrm{R}=$ Equivalent resistance

$
\begin{aligned}
& \frac{1}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow \frac{1}{R_{12}}=\frac{1}{3}+\frac{1}{6} \\
& \Rightarrow \frac{1}{R_{12}}=\frac{2+1}{6} \\
& \Rightarrow R_{12}=\frac{6}{3}=2 \\
& R=R_{12}+R_3=2+2=4
\end{aligned}
$

1644210349354

(b) 1 Ohm

R=Equivalent resistance

Connect all three resistors in parallel

$\begin{aligned} \frac{1}{R}= & \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \\ & \Rightarrow \frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6} \\ & \Rightarrow \frac{1}{R}=\frac{6+3+1}{6} \\ & \Rightarrow R=\frac{16}{6}=1 \Omega\end{aligned}$

Q.5 (a) What is the highest total resistance that can be secured by combinations of four coils of resistance $4 \Omega, 8 \Omega, 12 \Omega, 24 \Omega ?$

Answer:

The highest total resistance is obtained when all the resistors are connected in a series

$4+8+12+24=48 \Omega$

Q.5 (b) What is the lowest total resistance that can be secured by combinations of four coils of resistance $4 \Omega, 8 \Omega, 12 \Omega, 24 \Omega ?$

Answer:

The lowest total resistance is R is obtained when all the resistors are connected in parallel.

$\begin{aligned} & \frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24} \\ & \frac{1}{R}=\frac{6+3+2+1}{24}=\frac{12}{24} \\ & R=\frac{24}{12}=2\end{aligned}$

NCERT solutions for class 10 science chapter 11 Electricity topic 11.6

Q. 1. Why does the cord of an electric heater not glow while the heating element does?

Answer:

The heating element of an electric heater is a resistor.

The amount of heat production is given as, $H=I^2 R t$.

The resistance of elements (alloys) of an electric heater is high. As current flows through this element, it becomes hot and glows red.

The resistance of cord (metal like Cu or Al)of an electric heater is low. As current flow through this element, it does not becomes hot and does not glow red.

Q. 2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer:

Given: potential difference = 50 V.

Charge = 9600 C

time = 1 hr=3600 s

$\begin{aligned} & \quad I=\frac{\text { charge }}{\text { time }(\text { seconds })}=\frac{9600}{3600}=\frac{80}{3} \mathrm{~A} \\ & \text { So, } H=\text { VIt } \\ & \Rightarrow H=50 \times \frac{80}{3} \times 3600 \\ & \Rightarrow H=4800000 \mathrm{~J}=4.8 \times 10^6 \mathrm{~J}\end{aligned}$

Q. 3 An electric iron of resistance $20 \Omega$ takes a current of 5 A. Calculate the heat developed in 30 s.

Answer:

Given : resistance $=\mathrm{R}=20 \Omega$, Time $=30$ s, current $=\mathrm{I}=5 \mathrm{~A}$

$
\begin{aligned}
& V=I R \\
& \Rightarrow V=5 \times 20=100 \mathrm{~V} \\
& \\
& \quad H=V I t \\
& \quad \Rightarrow H=100 \times 5 \times 30 \\
& \quad \Rightarrow H=15000 \mathrm{~J}=1.5 \times 10^4 \mathrm{~J}
\end{aligned}
$

NCERT solutions for class 10 science chapter 11 Electricity topic 11.8

Q.1 What determines the rate at which energy is delivered by a current?

Answer:

The rate at which energy is delivered by a current is power.

Power P=I2 R, where I is the current and R is the resistance of the circuit/ appliance. Power depends on the current drawn by the appliance and the resistance of the appliance.

Q.2 An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer:

Given: I= 5 A and V= 220 V.

Power=P=VI

$P=220 \times 5=1100 \mathrm{~W}$
Time $=2 h r=2 \times 60 \times 60=7200 s$
The energy consumed $=$ power $\times$ time$=1100 \times 7200 \mathrm{~J}=7920000 \mathrm{~J}$
Power of motor $=1100 \mathrm{~W}$
Energy consumed by motor $=7920000 \mathrm{~J}$

NCERT Class 10 Science Electricity - Exercise Solution

NCERT Class 10 Science Chapter 11 Electricity Exercise Solutions comprise the well-organized answer to all the exercises presented in the textbook, including key topics such as resistance, resistivity, electrical power, heating effects of current etc. The solutions facilitate the solving of problems and enhance better exam preparations with proper explanations.

Q. 1. 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 $R^{\prime}$ , then the ratio $R / R^{\prime}$ is –

(a) 1/25 (b) 1/5 (c) 5 (d) 25

Answer:

Given: A piece of wire of resistance R is cut into five equal parts.

Resistance of each part is $\frac{R}{5}$.

$\frac{1}{R^{\prime}} =\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R} \\$


$\frac{1}{R^{\prime}} =\frac{25}{R} \\$

$R^{\prime} =\frac{R}{25} \\$

$\text { Hence, } \frac{R}{R^{\prime}} =\frac{R}{\frac{R}{25}}=25$

Thus, option d is correct.

Q. 2. Which of the following terms does not represent electrical power in a circuit?

(a) $I^2 R(b) I R^2(c) V I(d) V^2 / R$

Answer:
We know that power $=\mathrm{P}=\mathrm{VI}$-------------- (1)

Put, $\mathrm{V}=\mathrm{IR}$ in equation 1

$
P=I^2 R
$


Pur, $I=\frac{V}{R}$ in equation 1 ,

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

$\mathrm{P}=$ power, $\mathrm{V}=$ potencial difference, $\mathrm{I}=$ current, $\mathrm{R}=$ resistance
P cannot be $I R^2$.

Thus, option b is correct.

Q. 3. 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

Answer:

Given : $\mathrm{V}=220 \mathrm{~V}, \mathrm{P}=100 \mathrm{~W}$

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


The resistance of the bulb

$
\Rightarrow R=\frac{V^2}{P}=\frac{(220)^2}{100}=484 \Omega
$


If bulb is operated on 110 Vand resistance is the same, the power consumed will be $\mathrm{P}^{\prime}$

$
P^{\prime}=\frac{V^2}{R}=\frac{(110)^2}{484}=25 W
$

Hence, option d is correct.

Q. 4. 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

Answer:

If resistors are connected in parallel, the net resistance is given as

$
\begin{aligned}
& \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow \frac{1}{R_p}=\frac{1}{R}+\frac{1}{R} \\
& \Rightarrow \frac{1}{R_p}=\frac{2}{R} \\
& \Rightarrow R_p=\frac{R}{2}
\end{aligned}
$


If resistors are connected in series, the net resistance is given as

$
\Rightarrow R_s=R_1+R_2=R+R=2 R
$


Heat produced $=\mathrm{H}=\frac{V^2}{R} t$

$
\begin{aligned}
& \frac{H_s}{H_p}=\frac{\frac{V^2 t}{2 R}}{\frac{V^{2 t}}{\frac{R}{2}}} \\
& \Rightarrow \frac{H_s}{H_p}=\frac{1}{4} \\
& \Rightarrow H_s: H_p=1: 4
\end{aligned}
$

Thus, option c is correct.

Q. 5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer:

A voltmeter should be connected in parallel, to measure the potential difference between two points.

Q. 6. A copper wire has a diameter of 0.5 mm and resistivity of $1.6 \times 10^{-8} \Omega$ m. What will be the length of this wire to make its resistance $10 \Omega$ ? How much does the resistance change if the diameter is doubled?

Answer:

Given : diameter $=\mathrm{d}=0.5 \mathrm{~mm}$ and resistivity $=\rho=1.6 \times 10^{-8} \Omega \mathrm{~m}$, resistance $=\mathrm{R}=10 \Omega$
Area $=A$

$
\begin{aligned}
& A=\frac{\pi d^2}{4}=\frac{3.14 \times 0.5 \times 0.5}{4} \\
& \Rightarrow A=0.000000019625 \mathrm{~m}^2
\end{aligned}
$


We know

$
\begin{aligned}
& R=\frac{\rho l}{A} \\
& \Rightarrow l=\frac{R A}{\rho}=\frac{10 \times 0.000000019625}{1.6 \times 10^{-8}} \\
& =122.72 \mathrm{~m}
\end{aligned}
$


If the diameter is doubled.

$
\mathrm{d}=1 \mathrm{~mm}
$


Area $=A^{\prime}$

$
\begin{aligned}
& A^{\prime}=\frac{\pi d^2}{4}=\frac{3.14 \times 1 \times 1}{4} \\
& \Rightarrow A=0.000000785 \mathrm{~m}^2
\end{aligned}
$


We know

$
\begin{aligned}
& R^{\prime}=\frac{\rho l}{A^{\prime}} \\
& \Rightarrow R^{\prime}=\frac{1.6 \times 10^{-8} \times 122.72}{0.000000785} \\
& \Rightarrow R^{\prime}=2.5 \Omega \\
& \frac{R^{\prime}}{R}=\frac{2.5}{10}=\frac{1}{4}
\end{aligned}
$


Hence, new resistance is $\frac{1}{4}$ of original resistance.

Q.9. A battery of 9 V is connected in series with resistors of $02 \Omega$ , $0.3 \Omega$ , $0.4 \Omega$ , $0.5 \Omega$ and $12 \Omega$ respectively. How much current would flow through the $12 \Omega$ resistor?

Answer: Total resistance $=\mathrm{R}$

$
R=0.2+0.3+0.4+0.5+12=13.4 \Omega
$


$
\begin{aligned}
& \mathrm{V}=9 \mathrm{~V} \\
& I=\frac{V}{R}=\frac{9}{13.4}=0.67 \mathrm{~A}
\end{aligned}
$


Hence, All resistors are in series so 0.67 A current would flow through the $12 \Omega$ resistor.

Q.10. How many $176 \Omega$ resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer:

Given: V=220V and I =5A

$
R=\frac{V}{I}=\frac{220}{5}=44 \Omega
$


Let $x$ number of resistors be connected in parallel to obtain 44 Ohm equivalent resistance.
Thus,

$
\begin{aligned}
& \frac{1}{R}=\frac{1}{176}+\frac{1}{176}+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \text {............ } x \text { times } \\
& \Rightarrow \frac{1}{44}=\frac{x}{176} \\
& \Rightarrow x=\frac{176}{44}=4
\end{aligned}
$


Hence, 4 resistors of 176 Ohm are connected in parallel to obtain 44 Ohm.

Q.11. Show how you would connect three resistors, each of resistance $6 \Omega$ , so that the combination has a resistance of (i) $92 \Omega$ , (ii) $4 \Omega$ .

Answer:

(i) $R_1=R_2=R_3=6$
$\mathrm{R}=$ Equivalent resistance

$
\begin{aligned}
\frac{1}{R_{12}} & =\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow \frac{1}{R_{12}}=\frac{1}{6}+\frac{1}{6} \\
& \Rightarrow \frac{1}{R_{12}}=\frac{1+1}{6} \\
& \Rightarrow R_{12}=\frac{6}{2}=3 \\
& R=R_{12}+R_3=3+6=9 \Omega
\end{aligned}
$

1644210424046

(ii)

$
R_1=R_2=R_3=6
$

$\mathrm{R}=$ Equivalent resistance

$
\begin{aligned}
& R_{12}=R_1+R_2=6+6=12 \Omega \\
& \frac{1}{R}=\frac{1}{R_{12}}+\frac{1}{R_3} \\
& \Rightarrow \frac{1}{R}=\frac{1}{12}+\frac{1}{6} \\
& \Rightarrow \frac{1}{R}=\frac{1+2}{12} \\
& \Rightarrow R=\frac{12}{3}=4
\end{aligned}
$

Q.12. 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?

Answer:

Given: $\mathrm{V}=220 \mathrm{~V}$ and $\mathrm{P}=10 \mathrm{~W}$

$
R=\frac{V^2}{P}=\frac{220^2}{10}=4840 \Omega
$


Let x be the number of bulbs.

$
\begin{aligned}
& \mathrm{I}=5 \mathrm{~A} \text { and } \mathrm{V}=220 \mathrm{~V} \\
& \qquad R=\frac{V}{I}=\frac{220}{5}=44 \Omega
\end{aligned}
$


For $x$ bulbs of resistance 4680 Ohm, are connected in parallel to obtain 44 Ohm equivalent resistance.

Thus,

$
\begin{aligned}
& \frac{1}{R}=\frac{1}{4840}+\frac{1}{4840}+\ldots \ldots \ldots \ldots \ldots \ldots . . . \text { to } x \text { times } \\
& \Rightarrow \frac{1}{44}=\frac{x}{4840} \\
& \Rightarrow x=\frac{4840}{44}=110
\end{aligned}
$

Hence, 110 bulbs of 4840 Ohm are connected in parallel to obtain 44 Ohm.

Q.13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of $24 \Omega$ resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer:

Given : V=220V and Resistance of each coil=R =24A

When coil is used separately,current in coil is

$
I=\frac{V}{R}=\frac{220}{24}=9.16 A
$


When two coils are connected in series, net resistance is

$
R=R_1+R_2=24+24=48 \Omega
$

current in coil is I'

$
I^{\prime}=\frac{V}{R}=\frac{220}{48}=4.58 \mathrm{~A}
$


When two coils are connected in parallel, net resistance is

$
\begin{aligned}
& \frac{1}{R}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24} \\
& \Rightarrow R=12 \Omega
\end{aligned}
$

current in the coil is I"

$
I^{\prime \prime}=\frac{V}{R}=\frac{220}{12}=18.33 \mathrm{~A}
$

Q.14. Compare the power used in the $2 \Omega$ resistor in each of the following circuits:

(i) a 6 V battery in series with $1 \Omega$ and $2 \Omega$ resistors

(ii) a 4 V battery in parallel with $12 \Omega$ and $2 \Omega$ resistors.

Answer:

i) Given: V=6V

$\mathrm{R}=1+2=30 \mathrm{hm}$

$
I=\frac{V}{R}=\frac{6}{3}=2 A
$


In, a series current is constant.
So, power $=P$

$
P=I^2 R=2^2 \times 2=8 W
$

ii) Given: $V=6 \mathrm{~V}, \mathrm{R}=2 \mathrm{Ohm}$

In, parallel combination voltage in the circuit is constant.
So, power $=P$

$
P=\frac{V^2}{R}=\frac{4^2}{2}=8 W
$

Q.15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to an electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer:

Given :

For lamp one: Power $=\mathrm{P_1} =100 \mathrm{~W}$ and $\mathrm{V}=220 \mathrm{~V}$

$
I_1=\frac{P_1}{V}=\frac{100}{220}=0.455 \mathrm{~A}
$


For lamp two: Power $=\mathrm{P_2} =60 \mathrm{~W}$ and $\mathrm{V}=220 \mathrm{~V}$

$
I_2=\frac{P_2}{V}=\frac{60}{220}=0.273 \mathrm{~A}
$


Thus, the net current drawn from the supply is $0.455+0.273=0.728 \mathrm{~A}$

Q.16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer:

For TV set :

Given : Power $=250 \mathrm{~W}$ and time $=1 \mathrm{hr}=3600$ seconds
Energy consumed $=\mathrm{H}=\mathrm{PI}$

$
H=250 \times 3600=900000 J
$


For toaster :
Given : Power $=1200 \mathrm{~W}$ and time $=10$ minutes $=600$ seconds
Energy consumed $=\mathrm{H}=\mathrm{PI}$

$
H=1200 \times 600=720000 J
$

Thus, the TV set uses more energy than a toaster.

Q.17. An electric heater of resistance $8 \Omega$ draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Answer:

Given: R=8 Ohm ,I=15A and t= 2hr

The heat developed in the heater is H.

The heat developed in the heater is H .

$
H=I^2 R t
$


The rate at which heat is developed is given as

$
\frac{I^2 R t}{t}=I^2 R=15^2 \times 8=1800 \mathrm{~J} / \mathrm{s}
$

Q.18(a). Explain the following: Why is the tungsten used almost exclusively for filament of electric lamps?

Answer:

The tungsten used almost exclusively for filament of electric lamps because the melting point of tungsten is very high. The bulb glows at a very high temperature. So tungsten filament does not melt when bulb glows.

Q.18.(b) Explain the following: Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

Answer:

The conductors of electric heating devices, such as bread toasters and electric irons, are made of an alloy rather than a pure metal because the resistivity of the alloy is more than a pure metal. So the resistance will be hight and the heating effect will be high.

Q.18.(c) Explain the following: Why is the series arrangement not used for domestic circuits?

Answer:

If any of the elements in the circuit get damaged the entire circuit will be affected. If an element brake there will not be any current flow through the circuit. So the series circuit is not preferred in the domestic circuit.

Q.18.(d) Explain the following: How does the resistance of a wire vary with its area of cross-section?

Answer:

We know that

$\begin{aligned} & R=\frac{\rho l}{A} \\ & R \alpha \frac{1}{A}\end{aligned}$

Thus, the resistance of a wire is inversely proportional to its area of cross-section.

Q.18.(e) Explain the following: Why are copper and aluminum wires usually employed for electricity transmission?

Answer:

Copper and aluminum wires are usually employed for electricity transmission because they have low resistivity and they are good conductors of electricity.

Class 10 Physics NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions

The NCERT Class 10 Physics Chapter 11 Electricity HOTS (Higher Order Thinking Skills) questions aim at challenging high-order conceptual insight in topics such as ohm law, series and parallel circuits, and power consumption. These complicated questions lead to increased analytical thinking and problem solving abilities, which also helps the students in all types of competitive exams.

Q1:

If the electronic charge is $1.6 \times 10^{-19} \mathrm{C}$, then the number of electrons passing through a section of wire per second when the wire carries a current of 2 A, is

Answer:

Given : $\mathrm{e}=1.6 \times 1 \mathrm{o}^{-19} \mathrm{C}, I=2 \mathrm{A}, t=1 \mathrm{~s}$
Using $I=\frac{\text { ne }}{t}$
$\therefore$ Number of electrons $\mathrm{n}=\frac{\mathrm{It}}{\mathrm{e}}=\frac{2 \times 1}{1.6 \times 10^{-19}}=1.25 \times 10^{19}$ electrons


Q2:

When two bulbs of power 60 W and 40 W are connected in series, then the power of their combination will be:

Answer:

Given:- Power of two bulbs $P_1=60$ and $P_2=40 \mathrm{~W}$

To find the Power of their combination,
We know, Power $P=\frac{V^2}{R}$
where $V$ denotes voltage across bulb and $R$ denotes resistance of bulb.
So, Resistance of first bulb $\quad R_1=\frac{V^2}{p_1}=\frac{V^2}{60}....(1)$

Similarly,
Resistance of and bulb, $R_2=\frac{V^2}{P_2}=\frac{V^2}{40}......(2)$
Now, $R_1$ and $R_2$ are Joined in Series
then,
equivalent resistance $R_{e q}=R_1+R_2$

$
\begin{aligned}
& \frac{V^2}{P_{e q}}=\frac{V^2}{60}+\frac{V^2}{40} \\
& \frac{1}{P_{e q}}=\frac{1}{60}+\frac{1}{40} \\
& P_{e q}=\frac{60 \times 40}{60+40}=24 \mathrm{~W} \\
& P_{e q}=24 \mathrm{~W}
\end{aligned}
$


Q3:

At a constant potential difference, the resistance of any electric circuit is halved. The value of heat produced will be:

Answer:

We know,
The heat produced $=i^2 R t....(1)$
where $i$ is current, $R$ is resistance and is tee.

$
\begin{aligned}
\because \quad V & =i R \\
i & =V / R
\end{aligned}
$

putt in (1)

$
H=\left(\frac{V}{R}\right)^2 R t=\frac{V^2}{R} \cdot t
$

So, $V$ is constant
and $R$ is now $R / 2$ given.
and $t$ remains same

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

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

Hence, the heating produced will be doubled


Q4:

A wire of resistance $9 \Omega$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be ____ ohm.

Answer:

$9 \Omega$ is the resistance of whole wire
$\therefore$ resistance of each wire $=3 \Omega$.
$\therefore$ Equivalent resistance $=2 \Omega$


Q5:

A wire of resistance $R$ is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two endpoints of an edge of the triangle to that of the square is

Answer:

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

So, $R \propto \ell$
Side length of triangle is $1 / 3$ of total length.

$\left(\mathrm{R}_{\mathrm{eq}}\right)_1=\frac{2 \mathrm{r} / 3 \times \mathrm{r} / 3}{2 \mathrm{r} / 3+\mathrm{r} / 3} = 2\mathrm{r}/9$

$\left(\mathrm{R}_{\mathrm{eq}}\right)_2=\frac{3 \mathrm{r} / 4 \times \mathrm{r} / 4}{3 \mathrm{r} / 4+\mathrm{r} / 4} = 3\mathrm{r}/16$

$\frac{\left(\mathrm{R}_{\mathrm{eq}}\right)_1}{\left(\mathrm{R}_{\mathrm{eq}}\right)_2}=\frac{2 \mathrm{r} / 9}{3 \mathrm{r}/16}=\frac{32}{27}$


NCERT Solutions for Class 10 Science Chapter 11: Important Topics

NCERT Solutions for Class 10 Science Chapter 11 – Electricity cover all the important topics needed to understand how electric circuits work and how electrical quantities like current, voltage, resistance, and power are calculated. These topics are essential for both board exam preparation and real-life applications.

11.1 Electric Current And Circuit
11.2 Electric Potential And Potential Difference
11.3 Circuit Diagram
11.4 Ohm’s Law
11.5 Factors On Which The Resistance of A Conductor Depends
11.6 Resistance of A System Of Resistors
11.6.1 Resistors In Series
11.6.2 Resistors In Parallel
11.7 Heating Effect Of Electric Current
11.7.1 Practical Applications Of Heating Effect Of Electric Current
11.8 Electric Power

Also, students can refer,

NCERT Solutions for Class 10 Science Chapter 11: Important Formulae

The NCERT Solutions for Class 10 Science Chapter 11 Electricity provide all the important formulae related to current, voltage, resistance, power, and energy. These formulas make problem-solving easier, help in quick revision, and are essential for scoring well in board as well as competitive exams.

1. Ohm’s Law

$
V=I R
$
Where:
$V=$ potential difference (volts)
$I=$ current (amperes)
$R=$ resistance (ohms, $\Omega$)

2. Resistance of a Conductor

$
R=\rho \frac{l}{A}
$
Where:
$R=$ resistance
$\rho=$ resistivity of the material
$l=$ length of the conductor
$A=$ cross-sectional area

3. Resistance in Series

$R_{\text {total }}=R_1+R_2+R_3+\ldots$

4. Resistance in Parallel

$\frac{1}{R_{\text {total }}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots$

5. Electric Power

$
P=V I=I^2 R=\frac{V^2}{R}
$
Where:
$P=$ power (watts)

6. Electrical Energy

$
E=P t=V I t=I^2 R t=\frac{V^2 t}{R}
$
Where:
$E=$ electrical energy (in joules)

Approach to Solve Questions of Current Electricity

  • Carefully read the question to determine what you are given and what to find.

  • Write down the known variables which can be current, voltage, resistance or power or time.

  • Get familiar with the nature of the circuit either series or parallel.

  • Use the proper idea relative to the type of circuits and provided values.

  • Use proper units and convert as appropriate to standard SI units.

  • Carry out calculations in steps with the concept needed.

  • Check your last answer against being correct and related to the question.

Benefits of NCERT Solutions Class 10 Chapter 11: Electricity

  1. Concept Clarity: Uses simple yet step-by-step explanations of Ohm laws, resistance, power, and electric energy, thereby making these concepts easier to understand.
  2. Exam-Oriented: Solutions are designed according to CBSE syllabus and pattern and this aid in students preparing to appear in board examinations.
  3. Quick Revision: The entire chapter can be revised within a short period of time using important formulae and solved examples.
  4. Enhances Problem-Resolving Ability: Practicing questions enhance problem solving and aid in resolving both numerical and theoretical problems without fear.
  5. Helps in Competitive Exams and tests: Creates excellent background towards entrance exams such as NTSE, Olympiads, NEET, and IIT-JEE.

NCERT Solutions for Class 10 Science: Chapter Wise

Chapter-wise NCERT solutions can make it extremely easy to prepare notes for the Class 10 Science exams. These solutions include explanations to all intext and exercise questions, and they assist students to develop firm concepts and get along with the problem solving abilities. These links serve as a comprehensive guide to studying all board exams and competitive tests with step-by-step explanations, relevant formulas.

Also, check the NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

Q: What is the theme of Chapter 11 in Class 10 science?
A:

The chapter describes the functioning of the human eye and the occurrence of natural phenomenon such as rainbow and the color of the sky because of light behaviour.

Q: Why is this chapter significant in the board exams?
A:

It discusses important topics such as refraction, dispersion, scattering, and vision defects that come regularly in board exam questions.

Q: Do they ask numerical questions on this chapter?
A:

Yes, basic questions involving lens power and focal length, based on numbers, can be asked.

Q: Is there need to practice ray diagram in this chapter?
A:

Yes, particularly to understand myopia, hypermetropia, and how the eye lens works.

Q: What real-life phenomenon is described in the chapter?
A:

Brightness of stars, rainbows, blue sky, red sunset and the formation of images by the eye.

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Questions related to CBSE Class 10th

On Question asked by student community

Have a question related to CBSE Class 10th ?

Hello,

Yes, you can give the CBSE board exam in 2027.

If your date of birth is 25.05.2013, then in 2027 you will be around 14 years old, which is the right age for Class 10 as per CBSE rules. So, there is no problem.

Hope it helps !

Hello! If you selected “None” while creating your APAAR ID and forgot to mention CBSE as your institution, it may cause issues later when linking your academic records or applying for exams and scholarships that require school details. It’s important that your APAAR ID correctly reflects your institution to avoid verification problems. You should log in to the portal and update your profile to select CBSE as your school. If the system doesn’t allow editing, contact your school’s administration or the APAAR support team immediately so they can correct it for you.

Hello Aspirant,

Here's how you can find it:

  • School ID Card: Your registration number is often printed on your school ID card.

  • Admit Card (Hall Ticket): If you've received your board exam admit card, the registration number will be prominently displayed on it. This is the most reliable place to find it for board exams.

  • School Records/Office: The easiest and most reliable way is to contact your school office or your class teacher. They have access to all your official records and can provide you with your registration number.

  • Previous Mark Sheets/Certificates: If you have any previous official documents from your school or board (like a Class 9 report card that might have a student ID or registration number that carries over), you can check those.

Your school is the best place to get this information.

Hello,

It appears you are asking if you can fill out a form after passing your 10th grade examination in the 2024-2025 academic session.

The answer depends on what form you are referring to. Some forms might be for courses or examinations where passing 10th grade is a prerequisite or an eligibility criteria, such as applying for further education or specific entrance exams. Other forms might be related to other purposes, like applying for a job, which may also have age and educational requirements.

For example, if you are looking to apply for JEE Main 2025 (a competitive exam in India), having passed class 12 or appearing for it in 2025 are mentioned as eligibility criteria.

Let me know if you need imformation about any exam eligibility criteria.

good wishes for your future!!

Hello Aspirant,

"Real papers" for CBSE board exams are the previous year's question papers . You can find these, along with sample papers and their marking schemes , on the official CBSE Academic website (cbseacademic.nic.in).

For notes , refer to NCERT textbooks as they are the primary source for CBSE exams. Many educational websites also provide chapter-wise revision notes and study material that align with the NCERT syllabus. Focus on practicing previous papers and understanding concepts thoroughly.