CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
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.
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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.
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.
Download the chapter-wise formula of NCERT class 10 science by clicking on the link given in the box.
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.
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.
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.
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.
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.
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?
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.
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 :
Answer:
The diagram is as shown :
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.
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}
$
$\begin{aligned} & \Rightarrow \frac{1}{R}=\frac{1}{1}+\frac{1}{10^3}+\frac{1}{10^6} \\ & \Rightarrow \frac{1}{R}=\frac{10^6+10^3+1}{10^6} \\ & \Rightarrow R=\frac{10^6}{10^6+10^3+1}=0.999 \Omega \approx 1 \Omega\end{aligned}$
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.
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.
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}
$
(b) 1 Ohm
R=Equivalent resistance
Connect all three resistors in parallel
$\begin{aligned} \frac{1}{R}= & \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \\ & \Rightarrow \frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6} \\ & \Rightarrow \frac{1}{R}=\frac{6+3+1}{6} \\ & \Rightarrow R=\frac{16}{6}=1 \Omega\end{aligned}$
Answer:
The highest total resistance is obtained when all the resistors are connected in a series
$4+8+12+24=48 \Omega$
Answer:
The lowest total resistance is R is obtained when all the resistors are connected in parallel.
$\begin{aligned} & \frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24} \\ & \frac{1}{R}=\frac{6+3+2+1}{24}=\frac{12}{24} \\ & R=\frac{24}{12}=2\end{aligned}$
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.
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}
$
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.
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 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.
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.
(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.
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.
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
Answer:
The plot between voltage and current is as shown :
The slope of the line gives resistance (R)
$R=\frac{6.8}{2}=3.4 \Omega$
Answer:
Given : V=12V , I = 2.5 mA = 0.0025 A
$R=\frac{12}{0.0025}=4800 \Omega=4.8 k \Omega$
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}
$
(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}
$
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.
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
$
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.
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.
Answer:
Copper and aluminum wires are usually employed for electricity transmission because they have low resistivity and they are good conductors of electricity.
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 – 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,
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.
$
V=I R
$
Where:
$V=$ potential difference (volts)
$I=$ current (amperes)
$R=$ resistance (ohms, $\Omega$)
$
R=\rho \frac{l}{A}
$
Where:
$R=$ resistance
$\rho=$ resistivity of the material
$l=$ length of the conductor
$A=$ cross-sectional area
$R_{\text {total }}=R_1+R_2+R_3+\ldots$
$\frac{1}{R_{\text {total }}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots$
$
P=V I=I^2 R=\frac{V^2}{R}
$
Where:
$P=$ power (watts)
$
E=P t=V I t=I^2 R t=\frac{V^2 t}{R}
$
Where:
$E=$ electrical energy (in joules)
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.
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)
Class 10 Science Chapter 12 Electricity carries 6 to 8 marks.
Along with questions and concepts of class 10 chapter, practicing CBSE previous year question papers are mandatory for board exams.
Yes, it helps in higher studies if students chooses the Science stream because class 12 has current electricity chapter in syllabus.
On Question asked by student community
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Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any
