Chapter 3 Current Electricity

NCERT Solutions Class 12 Physics Chapter 3 - Current Electricity

Current Electricity is the third chapter of Class 12 Physics. It explains how electric current flows through a conductor and how current behaves in an electric circuit. The chapter covers important topics such as electric current, drift velocity, Ohm’s law, resistance, resistivity, EMF, and internal resistance. It also explains Kirchhoff’s rules, Wheatstone bridge, meter bridge, and potentiometer.

NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity provide simple, step-by-step answers to the questions in the NCERT textbook. They help students understand formulas, derivations, numerical problems, and circuit questions easily. Students can use these solutions to revise the chapter, practise important questions, and prepare for CBSE board exams, JEE, and NEET.

1.0NCERT Solutions Class 12 Physics Chapter 3 Current Electricity: Key Concepts

Class 12 Physics Chapter 3, Current Electricity, explains how electric charge flows through a conductor and the laws used to study electric circuits. The main concepts covered in this chapter include:

Electric Current and Drift Velocity: Understand electric current and how electrons move through a conductor. Learn the relation between electric current and the drift velocity of electrons.
Ohm’s Law and Resistivity: Learn Ohm’s law and understand the factors that affect resistance. Study resistivity and how it changes with temperature for different materials.
Combination of Resistors: Learn how resistors are connected in series and parallel and how to calculate their equivalent resistance.
Cells, EMF, and Internal Resistance: Understand the EMF and potential difference of a cell, along with internal resistance. Learn how cells behave when connected in series and parallel.
Kirchhoff’s Rules: Learn how to use Kirchhoff’s junction rule and loop rule to find current and voltage in electrical circuits.
Measuring Instruments: Study the Wheatstone bridge and potentiometer, including their basic principles and uses for finding unknown resistance and comparing the EMF of cells.

2.0NCERT Solutions Class 12 Physics Chapter 3 Current Electricity : Detailed Solutions

SOLVED EXAMPLES

(a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0 \times 1 0^{- 7} m^{2}$ carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is $9.0 \times 1 0^{3} kg / m^{3}$ , and its atomic mass is 63.5 u. (b) Compare the drift speed obtained above with, (i) thermal speeds of copper atoms at ordinary temperatures, (ii) speed of propagation of electric field along the conductor which causes the drift motion.

Sol.

(a) The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed $v_{d} = \frac{I}{neA}$ Now, $e = 1.6 \times 1 0^{- 19} C, A = 1.0 \times 1 0^{- 7} m^{2}$ , $I = 1.5 A$ . The density of conduction electrons, n is equal to the number of atoms per cubic metre. A cubic metre of copper has a mass of $9.0 \times 1 0^{3} kg$ . Since $6.0 \times 1 0^{23}$ copper atoms have a mass of 63.5 g,

$n = \frac{6.0 \times 1 0 ^{23}}{63.5} \times 9.0 \times 1 0^{6} = 8.5 \times 1 0^{28} m^{- 3}$

Which gives,

$v_{d} = \frac{1.5}{8.5 \times 1 0 ^{28} \times 1.6 \times 1 0 ^{- 19} \times 1.0 \times 1 0 ^{- 7}} = 1.1 \times 1 0^{- 3} ms^{- 1} = 1.1 mm s^{- 1}$

(b) (i) At temperature T, the thermal speed of a copper atom of mass M is obtained from $[< \frac{1}{2} Mv^{2} >= \frac{3}{2} k_{B} T]$ and is thus typically of the order of $k_{B} T / M$ ,where $k_{B}$ is the Boltzmann constant. For copper at 300 K , this is about $2 \times 1 0^{2} m / s$ . Note that the drift speed of electrons is much smaller, about $1 0^{- 5}$ times the typical thermal speed at ordinary temperatures. (ii) An electric field travelling along the conductor has a speed of an electromagnetic wave, namely equal to $3.0 \times 1 0^{8} m s^{- 1}$ . The drift speed is in comparison extremely smaller by a factor of $1 0^{- 11}$ .

(a) In Example-1, the electron drift speed is estimated to be only a few $mm s^{- 1}$ for currents in the range of a few amperes? How then is current established almost the instant a circuit is closed? (b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed? (c) If the electron drift speed is so small, and the electron's charge is small, how can we still obtain large amounts of current in a conductor? (d) When electrons drift in a metal from lower to higher potential, does it mean that all the 'free' electrons of the metal are moving in the same direction? (e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the (i) absence of electric field, (ii) presence of electric field?

Sol.

(a) Electric field is established throughout the circuit, almost instantly (with the speed of light) causing at every point a local electron drift. Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end. However, it does take a little while for the current to reach its steady value.

(b) Each 'free' electron does accelerate, increasing its drift speed until it collides with a positive ion of the metal. It loses its drift speed after collision but starts to accelerate and increases its drift speed again only to suffer a collision again and so on. On the average, therefore, electrons acquire only a drift speed. (c) Simple, because the electron number density is enormous, $\sim 1 0^{29} m^{- 3}$ . (d) No. The drift velocity is superposed over the large random velocities of electrons. (e) In the absence of electric field, the paths are straight lines. But in the presence of electric field, the paths are, curved in general, provided initial velocity is not parallel or anti parallel.

An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27.0 °C) is found to be $75.3Ω$ . When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is $1.70 \times 1 0^{- 4 \circ} C^{- 1}$ . Sol. When the current through the element is very small, heating effects can be ignored and the temperature $T_{1}$ of the element is the same as room temperature. The resistance $R_{2}$ at the steady temperature $T_{2}$ is,

$R_{2} = \frac{230 V}{2.68 A} = 85.8Ω$

Using the relation, $R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$ where $α = 1.70 \times 1 0^{- 4}^{\circ} C^{- 1}$ , Then $T_{2} - T_{1} = \frac{( 85.8 - 75.3 )}{( 75.3 ) \times 1.70 \times 1 0 ^{- 4}} = 82 0^{\circ} C$ that is, $T_{2} = (820 + 27.0)^{\circ} C = 84 7^{\circ} C$ Thus, the steady temperature of the heating element is 847°C.

The resistance of the platinum wire of a platinum resistance thermometer at the ice point is $5Ω$ and at steam point is $5.23Ω$ . When the thermometer is inserted in a hot bath, the resistance of the platinum wire is $5.795Ω$ . Calculate the temperature of the bath. Sol. $R_{0} = 5Ω, R_{100} = 5.23Ω$ and $R_{t} = 5.795Ω$

$R_{100} = R_{0} [1 + α (100)] \Rightarrow α = \frac{R _{100} - R _{0}}{R _{0} ( 100 )}$

$and R_{t} = R_{0} (1 + α t) \Rightarrow α = \frac{R _{t} - R _{0}}{R _{0} t}$

$\frac{R _{100} - R _{10}}{R _{0} ( 100 )} = \frac{R _{t} - R _{0}}{R _{0} t}$

Now, $t = \frac{R _{t} - R _{0}}{R _{100} - R _{0}} \times 100, t = \frac{5.795 - 5}{5.23 - 5} \times 100$

$t = \frac{0.795}{0.23} \times 100 = 345.6 5^{\circ} C$

A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance $1Ω$ (Fig.). Determine the equivalent resistance of the network and the current along each edge of the cube.

Cubical resistor network with a battery across opposite corners, showing current distribution along the cube’s edges.

Sol. The paths $AA^{'}, AD$ and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners A', B and D, the incoming current I must split equally into the two outgoing branches. Next take a closed loop, say, ABCC'EA, and apply Kirchhoff's second rule :

$- IR - (1/2) IR - IR + ε = 0$

where R is the resistance of each edge and $ε$ the emf of battery. Thus,

$ε = \frac{5}{2} IR$

The equivalent resistance $R_{eq}$ of the network is

$R_{eq} = \frac{ε}{3 I} = \frac{5}{6} R$

For $R = 1Ω$ ,

$R_{eq} = (5/6) Ω and for ε = 10 V,$

the total current (= 3I) in the network is

$3 I = \frac{10}{5/6} = \frac{60}{5} Ω = 12 A,$

i.e., $I = 4 A$ The current flowing in each edge can now be read off from the figure.

Determine the current in each branch of the network shown in Figure.

Circuit network diagram with branches between points A, B and D, showing the arrangement used to determine branch currents.

Sol. We have three unknowns $I_{1}, I_{2}$ and $I_{3}$ which can be found by applying the second rule of Kirchhoff to three different closed loops. Kirchhoff's second rule for the closed loop ADCA gives,

$10 - 4 (I_{1} - I_{2}) + 2 (I_{2} + I_{3} - I_{1}) - I_{1} = 0$

that is,

$7 I_{1} - 6 I_{2} - 2 I_{3} = 10$

For the closed loop ABCA, we get

$10 - 4 I_{2} - 2 (I_{2} + I_{3}) - I_{1} = 0$

that is, $I_{1} + 6 I_{2} + 2 I_{3} = 10$

For the closed loop BCDEB, we get

$5 - 2 (I_{2} + I_{3}) - 2 (I_{2} + I_{3} - I_{1}) = 0$

that is,

$2 I_{1} - 4 I_{2} - 4 I_{3} = - 5$

Equations a, b, c are three simultaneous equations in three unknowns. These can be solved by the usual method to give

$I_{1} = 2.5 A, I_{2} = \frac{5}{8} A, I_{3} = 1 \frac{7}{8} A$

The currents in the various branches of the network are

$AB : \frac{5}{8} A, CA : 2 \frac{1}{2} A, DEB : 1 \frac{7}{8} A AD : 1 \frac{7}{8} A, CD : 0 A, BC : 2 \frac{1}{2} A$

It is easily verified that Kirchhoff's second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage drop over the closed loop BADEB

$5 V + (\frac{5}{8} \times 4) V - (\frac{15}{8} \times 4) V$

equal to zero, as required by Kirchhoff's second rule.

The four arms of a Wheatstone bridge (Fig.) have the following resistances :

$AB = 100Ω, BC = 10Ω, CD = 5Ω, and DA = 60Ω.$

Wheatstone bridge circuit showing four resistive arms connected between points A, B, C, and D.

A galvanometer of $15Ω$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC. Sol. Considering the mesh BADB, we have

$100 I_{1} + 15 I_{g} - 60 I_{2} = 0$

$20 I_{1} + 3 I_{g} - 12 I_{2} = 0$

Considering the mesh BCDB, we have

$10 (I_{1} - I_{g}) - 15 I_{g} - 5 (I_{2} + I_{g}) = 0 10 I_{1} - 30 I_{g} - 5 I_{2} = 0 2 I_{1} - 6 I_{g} - I_{2} = 0 \dots\dots\dots (2)$

Considering the mesh ADCEA,

$60 I_{2} + 5 (I_{2} + I_{g}) = 10 65 I_{2} + 5 I_{g} = 10 13 I_{2} + I_{g} = 2 \dots\dots\dots (3$

Multiplying Eq. (2) by 10

$20 I_{1} - 60 I_{g} - 10 I_{2} = 0$

From Eqs. (4) and (1) we have

$63 I_{g} - 2 I_{2} = 0 I_{2} = 31.5 I_{g}$

Substituting the value of $I_{2}$ into Eq. (3), we get

$13 (31.5 I_{g}) + I_{g} = 2 410.5 I_{g} = 2 I_{g} = 4.87 mA .$

EXERCISE QUESTIONS WITH SOLUTIONS

The storage battery of a car has an emf of 12 V . If the internal resistance of the battery is $0.4Ω$ , what is the maximum current that can be drawn from the battery? Sol. Given, $E = 12 V, r = 0.4Ω$ Maximum current drawn from the battery,

$(R = 0) I = \frac{E}{r} = \frac{12}{0.4} = 30 A$

A battery of emf 10V and internal resistance $3Ω$ is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ? Sol. Given, $E = 10 V, r = 3Ω, I = 0.5 A$ If R is the the resistance of the resistor, total current,

$∴ I = \frac{E}{R + r} R + r = \frac{E}{I} = \frac{10}{0.5} = 20Ω R = 20 - 3 = 17Ω$

Terminal voltage,

$V = IR = 0.5 \times 17 = 8.5 V$

At room temperature (27.0 °C) the resistance of heating element is $100Ω$ . What is the temperature of the element if the resistance is found to be $117Ω$ , given that the temperature coefficient of the material of the resistor is $1.70 \times 1 0^{- 4}^{\circ} C^{- 1}$ . Sol. Given, $t_{1} = 2 7^{\circ} C, R_{1} = 100Ω$ Let $t_{2}$ is the increased temperature of the filament. Resistance of the heating element at temperature $t_{2}, R_{2} = 117Ω& α = 1.70 \times 1 0^{- 4} \circ C^{- 1}$

$R_{2} = R_{1} [1 + α (t_{2} - t_{1})]$

$α = \frac{R _{2} - R _{1}}{R _{1} ( t _{2} - t _{1} )} \Rightarrow t_{2} - t_{1} = \frac{R _{2} - R _{1}}{R _{1} α} t_{2} - 27 = \frac{117 - 100}{100 ( 1.7 \times 1 0 ^{- 4} )} t_{2} - 27 = 1000 \Rightarrow t_{2} = 102 7^{\circ} C$

Therefore, at $102 7^{\circ} C$ the resistance of the element is $117Ω$ .

A negligibly small current is passed through a wire of length 15 m and uniform cross-section $6.0 \times 1 0^{- 7} m^{2}$ , and its resistance is measured to be $5.0Ω$ . What is the resistivity of the material at the temperature of the experiment ? Sol. Given, $ℓ = 15 m, A = 6.0 \times 1 0^{- 7} m^{2}, R = 5.0Ω$ Resistivity of the material of the wire

$ρ = \frac{RA}{ℓ} = \frac{5 \times 6 \times 1 0 ^{- 7}}{15} = 2 \times 1 0^{- 7} Ω m$

A silver wire has a resistance of $2.1Ω$ at 27.5 °C, and a resistance of $2.7Ω$ at 100 °C. Determine the temperature coefficient of resistivity of silver. Sol. Use $R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$

$2.7 = 2.1 [1 + α (100 - 27.5) α = \frac{2.7 - 2.1}{72.5 \times 2.1} = \frac{0.6}{152.25} α = 0.003 9^{\circ} C^{- 1}$

A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is $1.70 \times 1 0^{- 4}^{\circ} C^{- 1}$ . Sol. Resistance at room temperature,

$R_{27} = \frac{V}{I} = \frac{230}{3.2}$

Resistance at t temperature,

$R_{t} = \frac{230}{2.8}, α = 1.7 \times 1 0^{- 4 \circ} C^{- 1}$

Initial temperature, $T_{1} = 2 7^{\circ} C$ Study state temperature $= T_{2}$ Relation, $R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$ $\frac{230}{2.8} = \frac{230}{3.2} [1 + α (T_{2} - 27)]$ $\frac{32}{28} - 1 = α (T_{2} - 27) \Rightarrow \frac{1}{7 \times 1.7 \times 1 0 ^{- 4}} = T_{2} - 27$ $T_{2} = 840 + 27 = 86 7^{\circ} C$

Determine the current in each branch of the network shown in fig.

Resistor network with 10 V supply and interconnected 5 Ω and 10 Ω resistors between points A, B, and C.

Sol. Apply KVL in loop ABDA

$- 10 I^{1} + 5 (I - 2 I^{1}) + 5 (I - I^{1}) = 0 2 I = 5 I^{1}$

Resistor network showing loop currents I and I₁ with current directions and resistor values for applying Kirchhoff’s voltage law.

Apply KVL in ADCFE loop

$- 5 (I - I^{1}) - 10 I^{1} + 10 - 10 I = 0 5 I^{1} + 15 I = 10$

From equation (i) and (ii)

$I = \frac{10}{17} \Rightarrow I^{1} = \frac{2 I}{5} = \frac{4}{17} A$

Current in AB & CD branch $= \frac{4}{17} A$ Current in AD & BC branch $= I - I^{1}$

$= \frac{10}{17} - \frac{4}{17} = \frac{6}{17} A$

Current in BD branch $= \frac{10}{17} - \frac{8}{17} = \frac{2}{17} A$

A storage battery of emf 8.0 V and internal resistance $0.5Ω$ is being charged by a 120 V dc supply using a series resistor of $15.5Ω$ . What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit ? Sol. Emf of the storage battery, $E = 8.0 V$ Internal resistance of the battery, $r = 0.5Ω$ DC supply voltage, $V = 120 V$ Resistance of the resistor, $R = 15.5Ω$ If I is the current flowing during charging,

$I = \frac{Supply voltage - emf of the battery}{( R + r )} .$

$I = \frac{( 120 - 8 )}{( 15.5 + 0.5 )} = 7 A$

If V is the terminal voltage of the battery during charging,

$V = E + Ir = 8 + 7 \times 0.5 = 11.5 V$

The series resistor is used to limit the charging current drawn from the external source. In case the series resistance is not used $(R = 0)$ , the current becomes very large. Which is very dangerous.

The number density of free electrons in a copper conductor is $8.5 \times 1 0^{28} m^{- 3}$ . How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is $2.0 \times 1 0^{- 6} m^{2}$ and it is carrying a current of 3.0 A. Sol. Given,

$n = 8.5 \times 1 0^{28} m^{- 3}, ℓ = 3.0 m, A = 2.0 \times 1 0^{- 6} m^{2}, I = 3.0 A, I = nAev_{d} v_{d} = Drift velocity = \frac{Length of the wire ( ℓ )}{Time taken to cover ( t )} I = nAe \frac{ℓ}{t} \Rightarrow t = \frac{nAe ℓ}{I} = \frac{8.5 \times 1 0 ^{28} \times 2 \times 1 0 ^{- 6} \times 1.6 \times 1 0 ^{- 19} \times 3}{3.0} = 2.7 \times 1 0^{4} s$

Therefore, the time taken by an electron to drift from one end of the wire to the other is $2.7 \times 1 0^{4} s$ .

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Get NCERT Solutions for Class 12 Physics chapter-wise, along with exercise answers, useful formulas, and easy-to-follow solutions for learning concepts and practising questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 3

Simplified Circuit Analysis: The solutions use a consistent sign convention for Kirchhoff's Laws, helping students avoid the most common errors in loop equations.
Derivation Clarity: Important derivations, such as the expression for drift velocity and the balance condition of a Wheatstone Bridge, are presented with clear logical steps.
Temperature Dependence Logic: Detailed explanations are provided for how resistance changes in conductors versus semiconductors, a frequent topic in conceptual exam questions.
Instrument Mastery: The solutions break down the working of the Potentiometer and Meter Bridge, emphasizing the "null point" method used in practical exams.
Exam-Oriented Numerical Solutions: Problems are solved with an emphasis on SI units and significant figures, adhering strictly to the CBSE marking scheme.

5.0Key Features of NCERT Solutions for Class 12 Physics Chapter 3

Easy Circuit Solutions: Kirchhoff’s laws are explained with the correct sign rules, helping students understand how to write loop equations and avoid common mistakes.
Step-by-Step Derivations: Important derivations, such as the formula for drift velocity and the balance condition of a Wheatstone bridge, are explained in simple steps.
Effect of Temperature on Resistance: The solutions explain how resistance changes with temperature in conductors and semiconductors, making it easier to answer concept-based questions.
Potentiometer and Meter Bridge: The working of the potentiometer and meter bridge is explained step by step, including how the null point is used to take measurements.
Clear Numerical Solutions: Numerical problems are solved using the correct formulas, SI units, and proper calculation steps to help students write clear answers in CBSE exams.

Frequently Asked Questions

NCERT Solutions Class 12 Physics Chapter 3 - Current Electricity

1.0NCERT Solutions Class 12 Physics Chapter 3 Current Electricity: Key Concepts

Electric Current and Drift Velocity: Understand electric current and how electrons move through a conductor. Learn the relation between electric current and the drift velocity of electrons.
Ohm’s Law and Resistivity: Learn Ohm’s law and understand the factors that affect resistance. Study resistivity and how it changes with temperature for different materials.
Combination of Resistors: Learn how resistors are connected in series and parallel and how to calculate their equivalent resistance.
Cells, EMF, and Internal Resistance: Understand the EMF and potential difference of a cell, along with internal resistance. Learn how cells behave when connected in series and parallel.
Kirchhoff’s Rules: Learn how to use Kirchhoff’s junction rule and loop rule to find current and voltage in electrical circuits.
Measuring Instruments: Study the Wheatstone bridge and potentiometer, including their basic principles and uses for finding unknown resistance and comparing the EMF of cells.

2.0NCERT Solutions Class 12 Physics Chapter 3 Current Electricity : Detailed Solutions

SOLVED EXAMPLES

(a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0 \times 1 0^{- 7} m^{2}$ carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is $9.0 \times 1 0^{3} kg / m^{3}$ , and its atomic mass is 63.5 u. (b) Compare the drift speed obtained above with, (i) thermal speeds of copper atoms at ordinary temperatures, (ii) speed of propagation of electric field along the conductor which causes the drift motion.

Sol.

$n = \frac{6.0 \times 1 0 ^{23}}{63.5} \times 9.0 \times 1 0^{6} = 8.5 \times 1 0^{28} m^{- 3}$

Which gives,

$v_{d} = \frac{1.5}{8.5 \times 1 0 ^{28} \times 1.6 \times 1 0 ^{- 19} \times 1.0 \times 1 0 ^{- 7}} = 1.1 \times 1 0^{- 3} ms^{- 1} = 1.1 mm s^{- 1}$

(a) In Example-1, the electron drift speed is estimated to be only a few $mm s^{- 1}$ for currents in the range of a few amperes? How then is current established almost the instant a circuit is closed? (b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed? (c) If the electron drift speed is so small, and the electron's charge is small, how can we still obtain large amounts of current in a conductor? (d) When electrons drift in a metal from lower to higher potential, does it mean that all the 'free' electrons of the metal are moving in the same direction? (e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the (i) absence of electric field, (ii) presence of electric field?

Sol.

An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27.0 °C) is found to be $75.3Ω$ . When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is $1.70 \times 1 0^{- 4 \circ} C^{- 1}$ . Sol. When the current through the element is very small, heating effects can be ignored and the temperature $T_{1}$ of the element is the same as room temperature. The resistance $R_{2}$ at the steady temperature $T_{2}$ is,

$R_{2} = \frac{230 V}{2.68 A} = 85.8Ω$

The resistance of the platinum wire of a platinum resistance thermometer at the ice point is $5Ω$ and at steam point is $5.23Ω$ . When the thermometer is inserted in a hot bath, the resistance of the platinum wire is $5.795Ω$ . Calculate the temperature of the bath. Sol. $R_{0} = 5Ω, R_{100} = 5.23Ω$ and $R_{t} = 5.795Ω$

$R_{100} = R_{0} [1 + α (100)] \Rightarrow α = \frac{R _{100} - R _{0}}{R _{0} ( 100 )}$

$and R_{t} = R_{0} (1 + α t) \Rightarrow α = \frac{R _{t} - R _{0}}{R _{0} t}$

$\frac{R _{100} - R _{10}}{R _{0} ( 100 )} = \frac{R _{t} - R _{0}}{R _{0} t}$

Now, $t = \frac{R _{t} - R _{0}}{R _{100} - R _{0}} \times 100, t = \frac{5.795 - 5}{5.23 - 5} \times 100$

$t = \frac{0.795}{0.23} \times 100 = 345.6 5^{\circ} C$

A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance $1Ω$ (Fig.). Determine the equivalent resistance of the network and the current along each edge of the cube.

$- IR - (1/2) IR - IR + ε = 0$

where R is the resistance of each edge and $ε$ the emf of battery. Thus,

$ε = \frac{5}{2} IR$

The equivalent resistance $R_{eq}$ of the network is

$R_{eq} = \frac{ε}{3 I} = \frac{5}{6} R$

For $R = 1Ω$ ,

$R_{eq} = (5/6) Ω and for ε = 10 V,$

the total current (= 3I) in the network is

$3 I = \frac{10}{5/6} = \frac{60}{5} Ω = 12 A,$

i.e., $I = 4 A$ The current flowing in each edge can now be read off from the figure.

Determine the current in each branch of the network shown in Figure.

$10 - 4 (I_{1} - I_{2}) + 2 (I_{2} + I_{3} - I_{1}) - I_{1} = 0$

that is,

$7 I_{1} - 6 I_{2} - 2 I_{3} = 10$

For the closed loop ABCA, we get

$10 - 4 I_{2} - 2 (I_{2} + I_{3}) - I_{1} = 0$

that is, $I_{1} + 6 I_{2} + 2 I_{3} = 10$

For the closed loop BCDEB, we get

$5 - 2 (I_{2} + I_{3}) - 2 (I_{2} + I_{3} - I_{1}) = 0$

that is,

$2 I_{1} - 4 I_{2} - 4 I_{3} = - 5$

Equations a, b, c are three simultaneous equations in three unknowns. These can be solved by the usual method to give

$I_{1} = 2.5 A, I_{2} = \frac{5}{8} A, I_{3} = 1 \frac{7}{8} A$

The currents in the various branches of the network are

$AB : \frac{5}{8} A, CA : 2 \frac{1}{2} A, DEB : 1 \frac{7}{8} A AD : 1 \frac{7}{8} A, CD : 0 A, BC : 2 \frac{1}{2} A$

$5 V + (\frac{5}{8} \times 4) V - (\frac{15}{8} \times 4) V$

equal to zero, as required by Kirchhoff's second rule.

The four arms of a Wheatstone bridge (Fig.) have the following resistances :

$AB = 100Ω, BC = 10Ω, CD = 5Ω, and DA = 60Ω.$

$100 I_{1} + 15 I_{g} - 60 I_{2} = 0$

$20 I_{1} + 3 I_{g} - 12 I_{2} = 0$

Considering the mesh BCDB, we have

$10 (I_{1} - I_{g}) - 15 I_{g} - 5 (I_{2} + I_{g}) = 0 10 I_{1} - 30 I_{g} - 5 I_{2} = 0 2 I_{1} - 6 I_{g} - I_{2} = 0 \dots\dots\dots (2)$

Considering the mesh ADCEA,

$60 I_{2} + 5 (I_{2} + I_{g}) = 10 65 I_{2} + 5 I_{g} = 10 13 I_{2} + I_{g} = 2 \dots\dots\dots (3$

Multiplying Eq. (2) by 10

$20 I_{1} - 60 I_{g} - 10 I_{2} = 0$

From Eqs. (4) and (1) we have

$63 I_{g} - 2 I_{2} = 0 I_{2} = 31.5 I_{g}$

Substituting the value of $I_{2}$ into Eq. (3), we get

$13 (31.5 I_{g}) + I_{g} = 2 410.5 I_{g} = 2 I_{g} = 4.87 mA .$

EXERCISE QUESTIONS WITH SOLUTIONS

The storage battery of a car has an emf of 12 V . If the internal resistance of the battery is $0.4Ω$ , what is the maximum current that can be drawn from the battery? Sol. Given, $E = 12 V, r = 0.4Ω$ Maximum current drawn from the battery,

$(R = 0) I = \frac{E}{r} = \frac{12}{0.4} = 30 A$

A battery of emf 10V and internal resistance $3Ω$ is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ? Sol. Given, $E = 10 V, r = 3Ω, I = 0.5 A$ If R is the the resistance of the resistor, total current,

$∴ I = \frac{E}{R + r} R + r = \frac{E}{I} = \frac{10}{0.5} = 20Ω R = 20 - 3 = 17Ω$

Terminal voltage,

$V = IR = 0.5 \times 17 = 8.5 V$

At room temperature (27.0 °C) the resistance of heating element is $100Ω$ . What is the temperature of the element if the resistance is found to be $117Ω$ , given that the temperature coefficient of the material of the resistor is $1.70 \times 1 0^{- 4}^{\circ} C^{- 1}$ . Sol. Given, $t_{1} = 2 7^{\circ} C, R_{1} = 100Ω$ Let $t_{2}$ is the increased temperature of the filament. Resistance of the heating element at temperature $t_{2}, R_{2} = 117Ω& α = 1.70 \times 1 0^{- 4} \circ C^{- 1}$

$R_{2} = R_{1} [1 + α (t_{2} - t_{1})]$

Therefore, at $102 7^{\circ} C$ the resistance of the element is $117Ω$ .

A negligibly small current is passed through a wire of length 15 m and uniform cross-section $6.0 \times 1 0^{- 7} m^{2}$ , and its resistance is measured to be $5.0Ω$ . What is the resistivity of the material at the temperature of the experiment ? Sol. Given, $ℓ = 15 m, A = 6.0 \times 1 0^{- 7} m^{2}, R = 5.0Ω$ Resistivity of the material of the wire

$ρ = \frac{RA}{ℓ} = \frac{5 \times 6 \times 1 0 ^{- 7}}{15} = 2 \times 1 0^{- 7} Ω m$

A silver wire has a resistance of $2.1Ω$ at 27.5 °C, and a resistance of $2.7Ω$ at 100 °C. Determine the temperature coefficient of resistivity of silver. Sol. Use $R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$

$2.7 = 2.1 [1 + α (100 - 27.5) α = \frac{2.7 - 2.1}{72.5 \times 2.1} = \frac{0.6}{152.25} α = 0.003 9^{\circ} C^{- 1}$

A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is $1.70 \times 1 0^{- 4}^{\circ} C^{- 1}$ . Sol. Resistance at room temperature,

$R_{27} = \frac{V}{I} = \frac{230}{3.2}$

Resistance at t temperature,

$R_{t} = \frac{230}{2.8}, α = 1.7 \times 1 0^{- 4 \circ} C^{- 1}$

Determine the current in each branch of the network shown in fig.

Sol. Apply KVL in loop ABDA

$- 10 I^{1} + 5 (I - 2 I^{1}) + 5 (I - I^{1}) = 0 2 I = 5 I^{1}$

Apply KVL in ADCFE loop

$- 5 (I - I^{1}) - 10 I^{1} + 10 - 10 I = 0 5 I^{1} + 15 I = 10$

From equation (i) and (ii)

$I = \frac{10}{17} \Rightarrow I^{1} = \frac{2 I}{5} = \frac{4}{17} A$

Current in AB & CD branch $= \frac{4}{17} A$ Current in AD & BC branch $= I - I^{1}$

$= \frac{10}{17} - \frac{4}{17} = \frac{6}{17} A$

Current in BD branch $= \frac{10}{17} - \frac{8}{17} = \frac{2}{17} A$

A storage battery of emf 8.0 V and internal resistance $0.5Ω$ is being charged by a 120 V dc supply using a series resistor of $15.5Ω$ . What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit ? Sol. Emf of the storage battery, $E = 8.0 V$ Internal resistance of the battery, $r = 0.5Ω$ DC supply voltage, $V = 120 V$ Resistance of the resistor, $R = 15.5Ω$ If I is the current flowing during charging,

$I = \frac{Supply voltage - emf of the battery}{( R + r )} .$

$I = \frac{( 120 - 8 )}{( 15.5 + 0.5 )} = 7 A$

If V is the terminal voltage of the battery during charging,

$V = E + Ir = 8 + 7 \times 0.5 = 11.5 V$

The number density of free electrons in a copper conductor is $8.5 \times 1 0^{28} m^{- 3}$ . How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is $2.0 \times 1 0^{- 6} m^{2}$ and it is carrying a current of 3.0 A. Sol. Given,

Therefore, the time taken by an electron to drift from one end of the wire to the other is $2.7 \times 1 0^{4} s$ .

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Get NCERT Solutions for Class 12 Physics chapter-wise, along with exercise answers, useful formulas, and easy-to-follow solutions for learning concepts and practising questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 3

Simplified Circuit Analysis: The solutions use a consistent sign convention for Kirchhoff's Laws, helping students avoid the most common errors in loop equations.
Derivation Clarity: Important derivations, such as the expression for drift velocity and the balance condition of a Wheatstone Bridge, are presented with clear logical steps.
Temperature Dependence Logic: Detailed explanations are provided for how resistance changes in conductors versus semiconductors, a frequent topic in conceptual exam questions.
Instrument Mastery: The solutions break down the working of the Potentiometer and Meter Bridge, emphasizing the "null point" method used in practical exams.
Exam-Oriented Numerical Solutions: Problems are solved with an emphasis on SI units and significant figures, adhering strictly to the CBSE marking scheme.

5.0Key Features of NCERT Solutions for Class 12 Physics Chapter 3

Easy Circuit Solutions: Kirchhoff’s laws are explained with the correct sign rules, helping students understand how to write loop equations and avoid common mistakes.
Step-by-Step Derivations: Important derivations, such as the formula for drift velocity and the balance condition of a Wheatstone bridge, are explained in simple steps.
Effect of Temperature on Resistance: The solutions explain how resistance changes with temperature in conductors and semiconductors, making it easier to answer concept-based questions.
Potentiometer and Meter Bridge: The working of the potentiometer and meter bridge is explained step by step, including how the null point is used to take measurements.
Clear Numerical Solutions: Numerical problems are solved using the correct formulas, SI units, and proper calculation steps to help students write clear answers in CBSE exams.

Frequently Asked Questions

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What happens to the capacitance of a capacitor when a dielectric is introduced in NCERT Solutions Class 12 Physics Chapter 3?

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NCERT Solutions Class 12 Physics Chapter 3 - Current Electricity

1.0NCERT Solutions Class 12 Physics Chapter 3 Current Electricity: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 3 Current Electricity : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 3

5.0Key Features of NCERT Solutions for Class 12 Physics Chapter 3

Table of Contents

Frequently Asked Questions

Where can I find NCERT Solutions Class 12 Physics Chapter 3 for Electrostatic Potential and Capacitance?

What topics are covered in NCERT Solutions Class 12 Physics Chapter 3?

How is electrostatic potential defined in NCERT Solutions Class 12 Physics Chapter 3?

Why are equipotential surfaces important in NCERT Solutions Class 12 Physics Chapter 3?

What happens to the capacitance of a capacitor when a dielectric is introduced in NCERT Solutions Class 12 Physics Chapter 3?

Can NCERT Solutions Class 12 Physics Chapter 3 explain capacitors connected in series and parallel?

Are NCERT Solutions Class 12 Physics Chapter 3 Current Electricity sufficient for JEE and NEET?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 3 - Current Electricity

1.0NCERT Solutions Class 12 Physics Chapter 3 Current Electricity: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 3 Current Electricity : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 3

5.0Key Features of NCERT Solutions for Class 12 Physics Chapter 3

Table of Contents