CBSE Class 12th Exam Date:17 Feb' 26 - 17 Feb' 26
Have you ever wondered how your phone or bulbs charge or why they have different bright lights? Chapter 3 Current Electricity in Class 12 Physics discusses these electrical phenomena in your day to day life in detail. NCERT Solutions of Chapter 3 help the students in understanding the practical aspects of electric current, circuit behavior, resistance and internal resistances in a clear and effortless manner.
The JEE Main is a national-level, computer-based engineering entrance examination conducted by the National Testing Agency (NTA). It is held for admission to undergraduate programmes, including Bachelor of Engineering (BE), Bachelor of Technology (BTech), Bachelor of Planning (BPlanning), and Bachelor of Architecture (BArch), at premier institutions such as National Institutes of Technology (NITs), Indian Institutes of Information Technology (IIITs), Centrally Funded Technical Institutions (CFTIs), and other state-participating universities across India.
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These NCERT solutions are written by subject experts and make difficult topic understandable and are of great assistance in both CBSE board exams and various competitive exams such as JEE, NEET, etc. When studying current electricity as part of revision work, or when attempting entrance tests, working through every solved example presented, and every other practice problem will strengthen your knowledge base. The NCERT Solutions of Class 12 Physics chapter 3 comprises of: Exercise questions that are solved so as to get a clear picture and practice on the concepts of the textbook, Additional and HOTS questions to test your reasoning and application skills, Key Topics like Ohm Law, resistivity, resistor combination, Kirchoffs laws and Wheatstone bridge, and approach to Solve Questions that assist students to confidently work on both numericals and theory
Download the free PDF of NCERT solution for Class 12 Physics Chapter 3 Exercise Solutions and get exam-ready for your CBSE board. These detailed solutions cover all the important questions, helping you strengthen your understanding of Current Electricity and score high in exams.
Answer :
Given, the emf of battery, E = 12 V
Internal resistance of battery, r = 0.4 Ohm
Let I be the maximum current drawn from the battery.
We know, according to Ohm's law
E = Ir
I = E/r = 12/0.4 =30 A
Hence the maximum current drawn from the battery is 30 A.
Answer:
Given, The emf of the battery, E = 10 V
The internal resistance of the battery, r = 3 Ohm
Current in the circuit, I = 0.5 A
Let R be the resistance of the resistor.
Therefore, according to Ohm's law:
E = IR' = I(R + r)
10 = 0.5(R + 3)
R = 1 Ohm
Also,
V = IR (Across the resistor)
= 0.5 x 17 = 8.5 V
Hence, the terminal voltage across the resistor = 8.5 V
3.3 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.7 x 10-4 oC-1?
Answer:
Given,
temperature coefficient of filament,
α = 1.7 x 10-4 oC-1
$T_1= 27^\circ C$ ; $R_1 = 100 \Omega$
Let T2 be the temperature of element, R2=117 Ω
(Positive alpha means that the resistance increases with temperature. Hence we can deduce that T2 will be greater than T1 )
We know,
$R_2 = R_1[1 + \alpha \Delta T]$
$\implies$ $117 = 100[1 +(1.70\times 10^{-4})(T_2 - 27) ]$
$\\ \Rightarrow T_2-27=\frac{117-100}{1.7\times10^{-4}}\\$
$\Rightarrow T_2-27=1000\\\Rightarrow T_2=1027^\circ C$
Hence, the temperature of the element is 1027 °C.
Answer:
$R = \rho l/ A$ , where $\rho$ is the resistivity of the material
$\implies \rho = RA/l = 5 \times6\times10^{-7}/15$
$\implies \rho = 2\times10^{-7}$
Hence, the resistivity of the material of wire is $\rho = 2\times10^{-7}\ ohm-m$
Answer:
Given,
$T_1= 27.5 ^\circ C$ ; $R_1 = 2.1\Omega$
$T_2= 100^\circ C$ ; $R_2= 2.7\Omega$
We know,
$R_2 = R_1[1 + \alpha \Delta T]$
2.7 = 2.1[1 + $\alpha$ (100 - 27.5) ]
$\alpha$ = (2.7 - 2.1) / 2.1(100 -27.5)
$\alpha$ = 0.0039 $^\circ C^{-1}$
Hence, the temperature coefficient of silver wire is 0.0039 $^\circ C^{-1}$
Answer:
For the given voltage, the two values of current will correspond to two different values of resistance which will correspond to two different temperature.
V = 230 V
$I_1= 3.2 A$ and $I_2 = 2.8 A$
Using Ohm's law:
$R_1 = 230/3.2 = 71.87\Omega$
and
$R_2= 230/2.8 = 82.14 \Omega$
$T_1= 27^\circ C$
Let $T_2$ be the steady temperature of the heating element.
We know,
$R_2 = R_1[1 + \alpha \Delta T]$
That is
230/2.8 = 230/3.2[1 + ( $1.70\times 10^{-4}$ ) ( T2 - 27) ]
T2 = (840.5 + 27) °C
Hence, steady temperature of the element is 867.5 °C.
3.7 Determine the current in each branch of the network shown in Fig. 3.30:
Answer:
The current in the circuit is distributed like
where I1, I2, and I3 are the different currents through shown branches.
Now, applying KVL in Loop
$10-I10-I_25-(I_2+I_3)10=0$
Also, we have $I=I_1+I_2$
So putting it in KVL equation
$10-(I_1+I_2)10-I_25-(I_2+I_3)10=0$
$10-10I_1-10I_2-5I_2-10I_2-10I_3=0$
$10-10I_1-25I_2-10I_3=0$ .................................(1)
Now let's apply KVL in the loop involving I1, I2 and I3
$5I_2-10I_1-5I_3=0$ .................................(2)
now, the third equation of KVL
$5I_3-5(I_1-I_3)+10(I_2+I_3)=0$
$-5I_1+10I_2+20I_3=0$ ..............................(3)
Now we have 3 equation and 3 variable, on solving we get
$I_1= \frac{4}{17}A$
$I_2= \frac{6}{17}A$
$I_3= \frac{-2}{17}A$
Now the total current
$I=I_1+I_2=\frac{4}{17}+\frac{6}{17}=\frac{10}{17}$
Answer:
Emf of battery, E = 8 V
Internal resistance of battery, r = 0.5 Ω
Supply Voltage, V = 120 V
The resistance of the resistor, R = 15.5 Ω
Let V' be the effective voltage in the circuit.
Now, V' = V - E
V' = 120 - 8 = 112 V
Now, current flowing in the circuit is:
I = V' / (R + r)
$\implies I = \frac{112}{15.5 + 0.5} = 7 A$
Now, using Ohm’s Law:
Voltage across resistor R is v = IR
V = 7 x 15.5 = 108.5 V
Now, the voltage supplied, V = Terminal voltage of battery + V
$\therefore$ Terminal voltage of battery = 120 -108.5 = 11.5 V
The purpose of having a series resistor is to limit the current drawn from the supply.
Answer:
We know,
$I = neAv_d$
vd :drift Velocity = length of wire(l) / time taken to cover
$I = neA \frac{l}{t}$
by substituting the given values
$\implies$ t = 2.7 x 104 s
Therefore, the time required by an electron to drift from one end of a wire to its other end is $2.7\times 10^4$ s.
Answer:
Given,
The surface charge density of Earth
$\rho$ = $10^{-9}C m^{-2}$
Current over the entire globe = 1800 A
Radius of earth, r = 6.37 x 106 m
$\therefore$ The surface area of earth A = $4\pi r^2$
A= $4\pi (6.37\times10^6 )^2$ = $5.09 \times 10^{14} m^2$
Now, charge on the earth surface,
$q=\rho\times A$
Therefore,
$q=1005.09 \times 10^5 \mathrm{C}$
Let the time taken to neutralize earth surface be t
$\therefore$ Current I = q / t
t = 282.78 s.
Therefore, time take to neutralize the Earth's surface is 282.78 s
Answer:
Given,
There are 6 secondary cells.
Emf of each cell, E = 2 V (In series)
The internal resistance of each cell, r = 0.015 Ω (In series)
And the resistance of the resistor, R = 8.5 Ω
Let I be the current drawn in the circuit.
$I = \frac{nE}{R + nr}$
$\implies I = \frac{6(2)}{8.5 + 6(0.015)} = \frac{12}{8.59}$
$\implies$ I = 1.4 A
Hence current drawn from the supply is 1.4 A
Therefore, terminal voltage, V = IR = 1.4 x 8.5 = 11.9 V
Answer:
Given,
Emf , E = 1.9 V
Internal resistance, r =380 Ω
The maximum current that can drawn is I = E/r = 1.9/380 = 0.005 A
The motor requires a large value of current to start and hence this cell cannot be used for a motor of a car.
Answer:
We know,
$R = \rho l / A$
The wires have the same resistance and also are of the same length.
Hence,
$\frac{\rho_{Al}}{A_{Al}} = \frac{\rho_{Cu}}{A_{Cu}}$
$\implies \frac{A_{Al}}{A_{Cu}} = \frac{\rho_{Al}}{\rho_{Cu}} = \frac{2.63}{1.73}$
Now, mass = Density x Volume = Density x Area x length
Taking the ratio of their masses for the same length
$\implies \frac{m_{Al}}{m_{Cu}} = \frac{d_{AL}A_{Al}}{d_{Cu}A_{Cu}} = \frac{2.7\times2.63}{8.9\times1.73} < 1$
Hence, $m_{Al} < m_{Cu}$
Therefore, for the same resistance and length, the aluminium wire is lighter.
Since aluminium wire is lighter, it is used as power cables.
5. What conclusion can you draw from the following observations on a resistor made of alloy manganin?
Current A |
Voltage V |
Current A |
Voltage V |
0.2 |
3.94 |
3.0 |
59.2 |
0.4 |
7.87 |
4.0 |
78.8 |
0.6 |
11.8 |
5.0 |
98.6 |
0.8 |
15.7 |
6.0 |
118.5 |
1.0 |
19.7 |
7.0 |
138.2 |
2.0 |
39.4 |
8.0 |
158.0 |
Answer:
The ratio of Voltage to current for the various values comes out to be nearly constant which is around 19.7.
Hence the resistor made of alloy manganin follows Ohm's law.
Answer:
The current flowing through the conductor is constant for a steady current flow.
Also, current density, electric field, and drift speed are inversely proportional to the area of cross-section. Hence, not constant.
Answer the following questions:
Answer:
No. Ohm’s law is not universally applicable for all conducting elements.
A semiconductor diode is such an example.
Answer the following questions
8. A low voltage supply from which one needs high currents must have very low internal resistance. Why?
Answer:
Ohm's law states that: V = I x R
Hence for a low voltage V, resistance R must be very low for a high value of current.
Answer the following questions:
9. A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
Answer:
A very high internal resistance is required for a high tension supply to limit the current drawn for safety purposes.
10. Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of ( $10^{22} / 10^{23}$ ).
Answer:
(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of 1022 .
Answer:
To get maximum effective resistance, combine them in series. The effective resistance will be nR.
Answer:
To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.
11.(c) What is the ratio of the maximum to minimum resistance?
Answer:
The ratio is nR/(R/n) = $n^2$
Answer:
We have, equivalent resistance = 11/3
Let's break this algebraically so that we can represent it in terms of 1, 2 and 3
$\frac{11}{3}=\frac{9+2}{3}=3+\frac{2}{3}=3+\frac{1*2}{1+2}$
this expression is expressed in terms of 1, 2 and 3. and hence we can make a circuit which consist only of 1 ohm, 2 ohms and 3 ohms and whose equivalent resistance is 11/3. that is :
Answer:
Connect 2 Ω and 3 Ω resistor in parallel and 1 Ω resistor in series to it
Equivalent Resistance R = {1/(1/2 + 1/3)} + 1 = 6/5 + 1
R = 11/5 Ω
Answer:
1 Ω+2 Ω+ 3 Ω= 6 Ω, so we will combine the resistance in series.
Answer:
Connect all three resistors in parallel.
Equivalent resistance is R = 1/(1/1 + 1/2 + 1/3) = (1x 2 x 3)/(6 + 3 + 2)
R = 6/11 Ω
13.
(a) Determine the equivalent resistance of networks shown in Fig..
Answer:
It can be seen that in every small loop resistor 1 ohm is in series with another 1 ohm resistor and two 2 ohms are also in series and we have 4 loops,
so equivalent resistance of one loop is equal to the parallel combination of 2 ohms and 4 ohm that is
$Equivalent\ R_{loop}=\frac{2*4}{2+4}=\frac{8}{6}=\frac{4}{3}$
now we have 4 such loops in series so,
$Total\ Equivalent\ R_{loop}=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=\frac{16}{3}$
Hence equivalent resistance of the circuit is 16/3 ohm.
13.
(b) Determine the equivalent resistance of networks shown in Fig
Answer:
It can be seen that all 5 resistors are in series, so
Equivalent Resistance = R + R + R + R + R = 5R
Hence equivalent resistance is 5R.
Answer:
First, let us find the equivalent of the infinite network,
let equivalent resistance = R'
Here from the figure, We can consider the box as a resistance of R'
Now, we can write,
equivalent resistance = R' =[( R')Parallel with (1)] + 1 + 1
$\frac{R'*1}{R'+1}+2=R'$
$R'+2R'+2=R'^2+R'$
$R'^2-2R'-2=0$
$R'=1+\sqrt{3},or1-\sqrt{3}$
Since resistance can never be negative we accept
$R'=1+\sqrt{3}$
, We have calculated the equivalent resistance of infinite network,
Now
Total equivalent resistance = internal resistance of battery+ equivalent resistance of the infinite network
= 0.5+1+1.73
=3.23 ohm
$V=IR$
$I=\frac{V}{R}=\frac{12}{3.23}=3.72A$
Hence current drawn from the 12V battery is 3.72 Ampere.
(a) What is the value of $\epsilon$?
Answer:
Given
Maintained constant emf of standard cell = 1.02V, balanced point of this cell = 67.3cm
Now when the standard cell is replaced by another cell with emf = $\varepsilon$, balanced point for this cell = 82.3cm
Now as we know the relation
$\frac{\varepsilon}{l} =\frac{E}{L}$
$\varepsilon =\frac{E}{L}*l=\frac{1.02}{67.3}*82.3=1.247V$
Hence, emf of another cell is 1.247V.
(b) What purpose does the high resistance of 600 k $\Omega$ have?
Answer:
If a sufficiently high current passes through galvanometer then it can get damaged. So we limit the current by adding a high resistance of 600 k $\Omega$.
(c) Is the balance point affected by this high resistance?
Answer:
No, the Balance point is not affected by high resistance. High resistance limits the current to galvanometer wire. The balance point is obtained by moving the joe key on the potentiometer wire and current through potentiometer wire is constant. The balance point is the point when the current through galvanometer becomes zero. The only duty of high resistance is to supply limited constant current to potentiometer wire.
Answer:
No, the method would not have worked if the driver cell of the potentiometer had an emf of 1.0V instead of 2, because when emf of the driving point is less than the other cell, there won't be any balance point in the wire.
Answer:
No, the circuit would not work properly for very low order of Voltage because the balance points would be near point A and there will be more percentage error in measuring it. If we add a series resistance with wire AB. It will increase the potential difference of wire AB which will lead to a decrease in percentage error.
20.
Answer:
Given,
the balance point of cell in open circuit = $l_1=76.3cm$
value of external resistance added = $R=9.5\Omega$
new balance point = $l_2=64.8cm$
let the internal resistance of the cell be $r$.
Now as we know, in a potentiometer,
$r=\frac{l_1-l_2}{l_2}*R$
$r=\frac{76.3-64.8}{64.8}*9.5=1.68\Omega$
Hence, the internal resistance of the cell will be 1.68 $\Omega$.
Q.1 The total current supplied to the circuit by the battery is
Answer:
The equivalent circuit is
The total resistance of the circuit is
$1.5+\frac{2 \times 6}{2+6}$ in parallel with $3 \Omega$
$\therefore$ Total resistance is $\frac{3}{2}=1.5 \Omega$
Current in the circuit $=\frac{6}{1.5}=4 \mathrm{~A}$
Q.2 What should be the value of E for which the galvanometer shows no deflection?
Answer:
If there will be no current in G . As the potential difference across $10 \Omega$ resistance is equal to 10 V then
$
\begin{aligned}
& \frac{\frac{10}{15}+\frac{E}{5}}{\frac{1}{15}+\frac{1}{15}}=10 \\
& \Rightarrow E=10 \mathrm{~V}
\end{aligned}
$
Q.3 A circuit consists of five identical conductors as shown in the figure. The two similar conductors are added as indicated by the dotted lines. The ratio of resistances before and after addition will be:
Answer:
Before addition,
Total resistance of the circuit is $\mathrm{R}_1=5 \Omega$
After the addition of the two conductors, the circuit will acquire the form shown in the figure
It is a balanced Wheatstone bridge. Resistance of the central conductor is ineffective.
$\therefore$ Total resistance of the circuit is
$
\begin{aligned}
& \mathrm{R}_2=1 \Omega+\frac{(2 \Omega)(2 \Omega)}{(2 \Omega+2 \Omega)}+1 \Omega=1 \Omega+1 \Omega+1 \Omega=3 \Omega \\
& \therefore \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{5}{3}
\end{aligned}
$
Q.4 A current of $2A$ flows in the system of conductors as shown in the figure. The potential difference $V_P-V_R$ will be:
Answer:
Current through QRS = current through $\mathrm{}$ QPS=1$ \mathrm{}$ amp.
$
\begin{aligned}
& V_D-V_A=2 \times 1=2 \text { volt } \\
& V_D-V_B=3 \times 1=3 \text { volt } \\
& \therefore V_A-V_B=\left(V_D-V_B\right)-\left(V_D-V_A\right) \\
& =3-2 \\
& =1 \text { volt }
\end{aligned}
$
Q.5 In the network shown in the figure below the potential difference between A and B is
Answer:
The distribution of current is shown in Fig. Keeping in view that the inflow and outflow of current in a cell must be same. Applying the loop rule to left and right loops.
$\begin{aligned} & 2 \mathrm{i}_1+6 \mathrm{i}_1=4 \text { or } 2 \mathrm{i}_1=0.5 \mathrm{~A} \\ & 3 \mathrm{i}_1+1 \mathrm{i}_2=6 \text { or } \mathrm{i}_2=1.5 \mathrm{~A} \\ & \mathrm{~V}_{\mathrm{AB}}=\sum \mathrm{ir}-\sum e \\ & (=-2 \times 0.5+3 \times 1.5-4=-0.5 \mathrm{~V}) \\ & \mathrm{V}_{\mathrm{AB}}=-0.5 \mathrm{~V}\end{aligned}$
3.1 Introduction
3.2 Electric Current
3.3 Electric Currents In Conductors
3.4 Ohm’S Law
3.5 Drift Of Electrons And The Origin Of Resistivity
3.5.1 Mobility
3.6 Limitations Of Ohm’S Law
3.7 Resistivity Of Various Materials
3.8 Temperature Dependence Of Resistivity
3.9 Electrical Energy, Power
3.10 Cells, Emf, Internal Resistance
3.11 Cells In Series And In Parallel
3.12 Kirchhoff’S Rules
3.13 Wheatstone Bridge
1. $d i=J d A \cos \theta=\vec{J} \cdot d \vec{A}$
Where: is the angle between the normal to the area and direction of current
2. Relation between current density and electric field-
$\vec{J}=\sigma \vec{E}=\frac{\vec{E}}{\rho}$
where = conductivity and = resistivity or specific resistance of the substance
$\begin{aligned} & V \propto I \\ & V=I R\end{aligned}$
$\rho=\frac{m}{n e^2 \tau}$
Where:
m is the mass, n is the number of electrons per unit volume, e is the charge of the electron and is the relaxation time
1. Series Grouping of resistance
$R_{e q}=R_1+R_2+\ldots \ldots \ldots+R_n$
2. Parallel Grouping of Resistance
$\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\ldots \ldots \ldots . .+\frac{1}{R_n}$
The power developed $=\frac{\text { energy }}{\text { time }}=i^2 R=i R=\frac{V^2}{R}$
$
\sum i=0
$
$
i_1+i_3=i_2+i_4
$
$\begin{aligned} & \sum V=0 \\ & -i_1 R_1+i_2 R_2-E_1-i_3 R_3+E_2+E_3-i_4 R_4=0\end{aligned}$
$\begin{aligned} & \frac{P}{Q}=\frac{R}{S} \\ & V_B=V_D(\text { Balanced condition })\end{aligned}$
1. Know What Current Is:
Current is simply the flow of electric charge, such as water in a pipe. It flows due to a push named voltage.
2. Learn Ohm's Law(V=IR):
This states $V=I R$ (Voltage $=$ Current $\\times$ Resistance). It's the most important formula - used in the majority of questions.
3. Know About Resistance and Drift of Electrons
Electrons travel slowly in a wire - slow movement is referred to as drift. Resistance is determined by material, length, and thickness of the wire.
4. Energy and Power:
Understand how to calculate energy and power consumed by devices such as bulbs or fans. Utilize formulas such as $P=$ $V I$ and $E=P t$.
5. Cells and EMF (Voltage):
Real batteries lose some energy within. That's internal resistance. Learn how to calculate total voltage with cells in series or parallel.
6. Kirchhoff's Rules:
These rules assist you in solving circuit problems. One is the current at junctions rule, and the other is the voltage in loops rule.
7. Wheatstone Bridge:
If the bridge is balanced, you can directly use a simple ratio formula to easily determine unknown resistance.
8. Practice NCERT Questions:
All questions of the board exam are based on NCERT book mostly. Attempt all examples and exercises. Practice few previous years papers also.
NCERT covers the basics of Current Electricity well, but for JEE, students should go beyond and master complex circuits, Kirchhoff’s laws, and internal resistance. Please check the below JEE topics for complete preparation.
When it comes to CBSE board exams and competitive exams like NEET and JEE Main, the solutions for NCERT Class 12 Physics Chapter 3 are super important. 7-10% of the questions in NEET and JEE Main are based on this chapter. In the 2019 CBSE board exam, there was even a 6-mark question on Current Electricity. So, using these NCERT solutions for Chapter 3 will help you score better in both your board exams and competitive exams.
Also, check the NCERT Books and NCERT Syllabus here:
Subject wise solutions
Also, check
Frequently Asked Questions (FAQs)
Focus on understanding the key concepts, practising numerical problems, and solving previous year questions. Refer to NCERT solutions for detailed explanations and practice regularly.
Kirchhoff’s laws — Current Law (KCL) and Voltage Law (KVL) — help in analyzing complex electrical circuits. These laws are crucial for solving circuit problems, especially in JEE and NEET exams.
The concepts in the current electricity chapter of NCERT Class 12 Physics are integral parts of NEET exam. Along with NCERT, practicing with previous year papers and mock tests would be enough.
In CBSE board exam, around 8 to 10% questions can be expected from the chapter Current Electricity. Certain papers of CBSE ask around 15% questions from NCERT chapter 3 Current Electricity. To score well in the exam follow NCERT syllabus and the exercise given in the NCERT Book. To practice more problems can refer to NCERT exemplar.
In JEE main, 2 to 3 questions can be expected from the chapter on Current Electricity. The topics like Potentiometer, meter bridge, KVL and KCL are important.
On Question asked by student community
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