Class 12 Physics Chapter 3 Important Questions (2026-27)

Class 12 Physics Chapter 3 Current Electricity: important questions with answers on Kirchhoff's rules and Ohm's law for CBSE 2026-27.

Important Questions for Class 12 Physics Chapter 3 – Current Electricity

Looking for Class 12 Physics Chapter 1 important questions? This page covers Electric Charges and Fields with exam-focused questions and step-by-step answers, arranged by marks: 1 mark, 2–3 marks and 5 marks. Written for the CBSE 2026–27 syllabus, it focuses on Gauss's law, electric flux and the electric dipole. Attempt them after finishing the NCERT chapter.

Very Short Answer Questions (1 Mark)

Cells, EMF and Internal Resistance

Question 1

The plot of the variation of potential difference across a combination of three identical cells in series, versus current is as shown. What is the emf of each cell?

Answer: Total emf of three cells in series = p.d. corresponding to zero current = 6 V.

So the emf of each cell \(= \dfrac{6}{3} =\) 2 V

Question 2

The emf of a cell is always greater than its terminal voltage. Why? Give reason.

Answer: Emf is the p.d. when no current is drawn. When current is drawn, there is a potential drop across the internal resistance of the cell. So the terminal voltage is less than the emf.

Question 3

A cell of emf \(E\) and internal resistance \(r\) draws a current \(I\). Write the relation between terminal voltage \(V\) in terms of \(E\), \(I\) and \(r\).

Answer: \(V = E - Ir\)

Question 4

Two identical cells, each of emf \(E\), having negligible internal resistance, are connected in parallel with each other across an external resistance \(R\). What is the current through this resistance?

Answer: When cells are connected in parallel, the emf remains unchanged. So the current is \(I = \dfrac{E}{R}\).

Question 5

Why is the terminal voltage of a cell less than its emf?

Answer: The terminal voltage is less than the emf because some current, however small, is drawn to measure the terminal voltage due to the internal resistance of the cell.

Question 6

A resistance \(R\) is connected across a cell of emf \(\varepsilon\) and internal resistance \(r\). A potentiometer now measures the potential difference between the terminals of the cell as \(V\). Write the expression for \(r\) in terms of \(\varepsilon\), \(V\) and \(R\).

Answer: Current in the circuit \(I = \dfrac{\varepsilon}{R + r}\). Terminal voltage \(V = \varepsilon - Ir\).

Solving: \(r = R\left(\dfrac{\varepsilon - V}{V}\right)\)

Question 7

Why is a potentiometer preferred over a voltmeter for determining the emf of a cell?

Answer: A potentiometer does not draw any (net) current from the cell, while a voltmeter draws some current from the cell when connected across it, hence it measures terminal voltage. That is why a potentiometer is preferred over a voltmeter to measure emf.

Drift Velocity and Current

Question 8

Two conducting wires X and Y of same diameter are connected across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires.

Answer: Since the wires are in series, current \(I\) is the same. Using \(I = neAv_d\) and same area: \(v_d \propto \dfrac{1}{n}\).

So \(\dfrac{v_{dX}}{v_{dY}} = \dfrac{n_Y}{n_X} = \dfrac{1}{2}\). Ratio = 1 : 2

Question 9

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?

Answer: No, only the drift velocities of the electrons are superposed over their random (haphazard) thermal velocities.

Question 10

Define the term 'Mobility' of charge carriers in a conductor. Write its SI unit.

Answer: Mobility of charge carriers is defined as the magnitude of the drift velocity per unit electric field \(E\): \(\mu = \dfrac{v_d}{E}\). SI unit: \(\text{m}^2\,\text{V}^{-1}\,\text{s}^{-1}\).

Question 11

Define the term 'drift velocity' of charge carriers in a conductor and write its relationship with the current flowing through it.

Answer: Drift velocity is the velocity with which a free electron in a conductor gets drifted under the influence of the applied external electric field.

Relationship: \(I = neAv_d\), where \(n\) is number density, \(e\) is electronic charge, \(A\) is cross-sectional area.

Question 12

How does the random motion of free electrons in a conductor get affected when a potential difference is applied across its ends?

Answer: The random motion of free electrons gets directed towards the point at a higher potential.

Question 13

Write the expression for the drift velocity of charge carriers in a conductor of length \(l\) across which a potential difference \(V\) is applied.

Answer: \(v_d = \dfrac{eE\tau}{m} = \dfrac{eV\tau}{ml}\), where \(\tau\) is the relaxation time.

Resistivity and Conductivity

Question 14

Show on a graph the variation of resistivity with temperature for a typical semiconductor.

Answer: The resistivity of a semiconductor decreases rapidly (exponentially) with an increase in temperature.

Question 15

Two wires of equal length, one of copper and the other of manganin, have the same resistance. Which wire is thicker?

Answer: For both wires \(R\) and \(l\) are the same, and \(\rho_{\text{copper}} < \rho_{\text{manganin}}\). So \(A_{\text{copper}} < A_{\text{manganin}}\), i.e. the manganin wire is thicker.

Question 16

Show variation of resistivity of copper as a function of temperature in a graph.

Answer: For a metal like copper, resistivity increases almost linearly with temperature.

Question 17

Define the term 'electrical conductivity' of a metallic wire. Write its SI unit.

Answer: Conductivity is the reciprocal of the resistivity of the material of a conductor: \(\sigma = \dfrac{1}{\rho}\). SI unit: \(\Omega^{-1}\,\text{m}^{-1}\) (or \(\text{S}\,\text{m}^{-1}\)).

Question 18

How does one explain the increase in resistivity of a metal with increase of temperature?

Answer: With an increase in temperature, the relaxation time (average time between successive collisions) decreases, and hence resistivity \(\rho = \dfrac{m}{ne^2\tau}\) increases.

Question 19

Define the conductivity of a conductor. Write its SI unit.

Answer: Conductivity is the reciprocal of resistivity, \(\sigma = \dfrac{1}{\rho}\). SI unit: \(\text{S}\,\text{m}^{-1}\) (Siemens per metre).

Circuits and Kirchhoff's Laws

Question 20

A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 Ω as shown. Find the value of the current in the circuit.

Answer: Applying Kirchhoff's rule: net emf \(= 200 - 10 = 190\ \text{V}\).

Current \(I = \dfrac{190}{38} =\) 5 A

Question 21

A 5 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 39 Ω as shown. Find the value of the current.

Answer: Net emf \(= 200 - 5 = 195\ \text{V}\). Current \(I = \dfrac{195}{39} =\) 5 A.

Potentiometer and Meter Bridge

Question 22

State the underlying principle of a potentiometer.

Answer: When a constant current flows through a wire of uniform cross-section and uniform composition, the potential difference across any length of wire is directly proportional to its length: \(V \propto l\).

Galvanometer and Sensitivity

Question 23

Define current sensitivity and voltage sensitivity of a galvanometer. Increasing the current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer. Justify.

Answer: Current sensitivity is the deflection produced in the galvanometer when unit current is passed through its coil. Voltage sensitivity is the deflection produced when unit voltage is applied across the coil.

Since \(V_s = \dfrac{I_s}{R}\), an increase in current sensitivity may not necessarily increase voltage sensitivity — it may be affected by the resistance used.

Question 24

Define the current sensitivity of a galvanometer. Write its SI unit.

Answer: Current sensitivity of a moving coil galvanometer is defined as the deflection produced in the galvanometer per unit current passed through it. SI unit: \(\text{rad}\,\text{A}^{-1}\).

Short Answer Questions (2–3 Marks)

Drift Velocity and Current Density

Question 25

Two metallic wires of the same material have the same length but cross-sectional area in the ratio 1 : 2. They are connected (i) in series and (ii) in parallel. Compare the drift velocities of electrons in the two wires in both cases.

Answer: \(I = neAv_d\).

(i) In series, \(I\) is the same. So \(v_d \propto \dfrac{1}{A}\). Hence \(v_{d1} : v_{d2} = A_2 : A_1 =\) 2 : 1.

(ii) In parallel, potential difference \(V\) is the same. Since \(v_d = \dfrac{eV\tau}{ml}\), drift velocity is independent of area. So \(v_{d1} : v_{d2} =\) 1 : 1.

Question 26

Derive an expression for the resistivity of a good conductor in terms of the relaxation time of electrons.

Answer: Drift speed \(v_d = \dfrac{eE\tau}{m}\). Current \(I = neAv_d = \dfrac{ne^2EA\tau}{m}\).

Current density \(J = \dfrac{I}{A} = \dfrac{ne^2E\tau}{m}\).

Using \(J = \sigma E\), conductivity \(\sigma = \dfrac{ne^2\tau}{m}\).

Resistivity: \(\rho = \dfrac{1}{\sigma} = \dfrac{m}{ne^2\tau}\)

Question 27

Using the mathematical expression for the conductivity of a material, explain how it varies with temperature for (i) semiconductors, (ii) good conductors.

Answer: Conductivity \(\sigma = \dfrac{ne^2\tau}{m}\).

(i) Semiconductors: With an increase in temperature, conductivity increases. This is because the increase in number density \(n\) dominates the decrease in \(\tau\).

(ii) Good conductors: With an increase in temperature, conductivity decreases. This is because the relaxation time \(\tau\) decreases; the change in \(n\) is negligible.

Question 28

Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time.

Answer: In the absence of an electric field, the electron motion is random and the net velocity is zero. In the presence of field \(E\), they experience acceleration \(a = \dfrac{eE}{m}\).

The average time for this acceleration is \(\tau\). The drift velocity: \(v_d = \dfrac{eE\tau}{m}\)

Question 29

Explain the term 'drift velocity' of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of drift velocity.

Answer: Drift velocity is the velocity with which free electrons in a conductor get drifted in a direction opposite to the applied field. Its unit is \(\text{m}\,\text{s}^{-1}\).

Expression: Field \(E = \dfrac{V}{l}\). Let \(n\) be the number density. Total free electrons in the conductor \(= nAl\), so total charge \(Q = nAle\).

Time to cross the conductor: \(t = \dfrac{l}{v_d}\). Current: \(I = \dfrac{Q}{t} = neAv_d\)

Question 30

Write a relation between current and drift velocity of electrons in a conductor. Use this relation to explain how the resistance of a conductor changes with the rise in temperature.

Answer: \(I = neAv_d\).

On increasing temperature, the resistance of the conductor increases due to an increase in frequent collisions of electrons with each other, resulting in a decrease in drift velocity.

Question 31

Define mobility of a charge carrier. Write the relation expressing mobility in terms of relaxation time. Give its SI unit.

Answer: Mobility \(\mu\) is the magnitude of drift velocity per unit electric field: \(\mu = \dfrac{v_d}{E}\).

Since \(v_d = \dfrac{eE\tau}{m}\), we get \(\mu = \dfrac{e\tau}{m}\). SI unit: \(\text{m}^2\,\text{V}^{-1}\,\text{s}^{-1}\).

Question 32

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area \(1.0 \times 10^{-7}\ \text{m}^2\) carrying a current of 1.5 A. Assume the density of conduction electrons to be \(9 \times 10^{28}\ \text{m}^{-3}\).

Answer: \(v_d = \dfrac{I}{neA} = \dfrac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-7}}\)

\(= \dfrac{1.5}{1.44 \times 10^{3}} =\) \(1.04 \times 10^{-3}\ \text{m/s}\)

Question 33

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area \(2.5 \times 10^{-7}\ \text{m}^2\) carrying a current of 1.8 A. Assume the density of conduction electrons to be \(9 \times 10^{28}\ \text{m}^{-3}\).

Answer: \(v_d = \dfrac{I}{neA} = \dfrac{1.8}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.5 \times 10^{-7}}\)

\(= \dfrac{1.8}{3.6 \times 10^{3}} =\) \(5 \times 10^{-4}\ \text{m/s}\)

Question 34

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area \(2.5 \times 10^{-7}\ \text{m}^2\) carrying a current of 2.7 A. Assume the density of conduction electrons to be \(9 \times 10^{28}\ \text{m}^{-3}\).

Answer: \(v_d = \dfrac{I}{neA} = \dfrac{2.7}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.5 \times 10^{-7}}\)

\(= \dfrac{2.7}{3.6 \times 10^{3}} =\) \(7.5 \times 10^{-4}\ \text{m/s}\)

Question 35

Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.

Answer: Suppose a potential difference \(V\) is applied across a conductor of length \(l\) and uniform cross-section \(A\). The field \(E = \dfrac{V}{l}\).

Number of electrons per unit volume = \(n\). Charge on an electron = \(e\). Number of electrons in length \(l\) = \(nAl\). Total charge \(q = enAl\).

Time for electrons to cross the conductor: \(t = \dfrac{l}{v_d}\). Current \(I = \dfrac{q}{t} = neAv_d\).

Current density \(J = \dfrac{I}{A} = nev_d\). Thus \(J \propto v_d\).

Question 36

Derive the expression for the current density of a conductor in terms of the conductivity and applied electric field. Explain, with reason, how the mobility of electrons in a conductor changes when the potential difference applied is doubled, keeping the temperature of the conductor constant.

Answer: \(v_d = \dfrac{eE\tau}{m}\), \(J = nev_d = \dfrac{ne^2E\tau}{m} = \sigma E\), where \(\sigma = \dfrac{ne^2\tau}{m}\).

Mobility \(\mu = \dfrac{v_d}{E} = \dfrac{e\tau}{m}\). Since \(\tau\) depends only on temperature (not on \(V\) or \(E\)), doubling the potential difference does not change mobility. Mobility remains unchanged.

Cells, EMF and Internal Resistance

Question 37

A cell of emf \(E\) and internal resistance \(r\) is connected to two external resistances \(R_1\) and \(R_2\) and a perfect ammeter. The current in the circuit is measured in four different situations: (i) without any external resistance (ii) with resistance \(R_2\) only (iii) with \(R_1\) and \(R_2\) in series (iv) with \(R_1\) and \(R_2\) in parallel. The currents measured are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order. Identify the currents corresponding to the four cases.

Answer: The current is maximum when external resistance is minimum and minimum when external resistance is maximum.

(i) Without any external resistance: \(I = \dfrac{E}{r}\) → 4.2 A (largest)
(ii) With \(R_2\) only: 1.4 A
(iii) \(R_1\) and \(R_2\) in series (max external resistance): 0.42 A (smallest)
(iv) \(R_1\) and \(R_2\) in parallel: 1.05 A

Question 38

A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, find (i) the resistance of the resistor; (ii) the terminal voltage of the battery.

Answer: (i) \(I = \dfrac{E}{R + r} \Rightarrow 0.5 = \dfrac{10}{R + 3} \Rightarrow R + 3 = 20 \Rightarrow R =\) 17 Ω

(ii) \(V = E - Ir = 10 - (0.5)(3) =\) 8.5 V

Question 39

A battery of emf 6 V and internal resistance 2 Ω is connected to a resistor. If the current in the circuit is 0.25 A, find (i) the resistance of the resistor; (ii) the terminal voltage of the battery.

Answer: (i) \(0.25 = \dfrac{6}{R + 2} \Rightarrow R + 2 = 24 \Rightarrow R =\) 22 Ω

(ii) \(V = 6 - 0.25 \times 2 =\) 5.5 V

Question 40

Two cells of emfs 1.5 V and 2.0 V having internal resistances 0.2 Ω and 0.3 Ω respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.

Answer: For parallel combination:

\(E_{eq} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2} = \dfrac{1.5 \times 0.3 + 2.0 \times 0.2}{0.2 + 0.3} = \dfrac{0.45 + 0.4}{0.5} =\) 1.7 V

\(r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2} = \dfrac{0.2 \times 0.3}{0.5} =\) 0.12 Ω

Question 41

A battery of emf \(E\) and internal resistance \(r\) when connected across an external resistance of 12 Ω produces a current of 0.5 A. When connected across a resistance of 25 Ω, it produces a current of 0.25 A. Determine (i) the emf and (ii) the internal resistance of the cell.

Answer: \(0.5 = \dfrac{E}{12 + r}\) ...(i), \(0.25 = \dfrac{E}{25 + r}\) ...(ii)

From (i): \(E = 0.5(12 + r) = 6 + 0.5r\)
From (ii): \(E = 0.25(25 + r) = 6.25 + 0.25r\)

\(6 + 0.5r = 6.25 + 0.25r \Rightarrow 0.25r = 0.25 \Rightarrow r =\) 1 Ω

\(E = 6 + 0.5(1) =\) 6.5 V

Question 42

A battery of emf 12 V and internal resistance 2 Ω is connected to a 4 Ω resistor. (a) Show that a voltmeter placed across the cell and across the resistor, in turn, gives the same reading. (b) To record the voltage and current in the circuit, why is the voltmeter placed in parallel and ammeter in series?

Answer: (a) \(I = \dfrac{12}{4 + 2} = 2\ \text{A}\). Terminal voltage \(V = 12 - 2 \times 2 = 8\ \text{V}\).

Voltage across the resistor: \(V = IR = 2 \times 4 = 8\ \text{V}\). Both give 8 V. ✓

(b) An ammeter is connected in series because the same current flows through the circuit, while a voltmeter is connected in parallel to measure the potential difference across two points.

Question 43

A cell of emf 4 V and internal resistance 1 Ω is connected to a d.c. source of 10 V through a resistor of 5 Ω. Calculate the terminal voltage across the cell during charging.

Answer: During charging, net emf \(= 10 - 4 = 6\ \text{V}\). Total resistance \(= 1 + 5 = 6\ \Omega\).

Current \(I = \dfrac{6}{6} = 1\ \text{A}\).

Terminal voltage across the cell: \(V = E + Ir = 4 + 1 \times 1 =\) 5 V

Question 44

Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.2 V. When the terminals of the cell are also connected to a resistance of 5 Ω as shown, the voltmeter reading drops to 1.8 V. Find the internal resistance of the cell.

Answer: (a) Internal resistance depends on: (i) nature of electrolyte, (ii) separation between electrodes (also electrode area).

(b) Given \(\varepsilon = 2.2\ \text{V}\) (emf), \(V = 1.8\ \text{V}\), \(R = 5\ \Omega\).

\(r = R\left(\dfrac{\varepsilon - V}{V}\right) = 5 \times \left(\dfrac{2.2 - 1.8}{1.8}\right) = 5 \times \dfrac{0.4}{1.8} =\) 1.11 Ω

Question 45

Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2.0 V. When the terminals of the cell are also connected to a resistance of 3 Ω as shown, the voltmeter reading drops to 1.5 V. Find the internal resistance of the cell.

Answer: Factors: (i) nature of electrolyte, (ii) separation between electrodes.

\(\varepsilon = 2.0\ \text{V}\), \(V = 1.5\ \text{V}\), \(R = 3\ \Omega\).

\(r = 3 \times \left(\dfrac{2.0 - 1.5}{1.5}\right) = 3 \times \dfrac{0.5}{1.5} =\) 1 Ω

Question 46

Distinguish between emf (\(\varepsilon\)) and terminal voltage (\(V\)) of a cell having internal resistance \(r\). Draw a plot showing the variation of terminal voltage \(V\) vs the current \(I\) drawn from the cell. Using this plot, how does one determine the internal resistance of the cell?

Answer: Emf is the potential difference between the two terminals of the cell when no current is drawn from the cell. Terminal voltage is the potential difference between the two terminals when current passes through it.

Plot: \(V\) vs \(I\) is a straight line with intercept \(E\) on the \(V\)-axis and slope \(-r\). The negative slope gives the internal resistance.

Resistivity, Resistance and Colour Code

Question 47

Two students A and B were asked to pick a resistor of 15 kΩ from a collection of carbon resistors. A picked a resistor with bands of colours: brown, green, orange while B chose a resistor with bands of black, green, red. Who picked the correct resistor?

Answer: Brown = 1, Green = 5, Orange = ×10³ → 15 × 10³ = 15 kΩ.

Black = 0, Green = 5, Red = ×10² → 05 × 10² = 500 Ω.

Student A picked the correct resistor.

Question 48

A wire of 15 Ω resistance is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 3.0 V battery. Find the current drawn from the battery.

Answer: On stretching, \(R \propto l^2\) and volume is constant. Doubling length: \(R_1 = 15 \times 4 = 60\ \Omega\).

Each half: \(30\ \Omega\). In parallel: \(R_{eq} = 15\ \Omega\).

Current \(I = \dfrac{3}{15} =\) 0.2 A

Question 49

A wire of 20 Ω resistance is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 4.0 V battery. Find the current drawn from the battery.

Answer: Doubling length: \(R_1 = 20 \times 4 = 80\ \Omega\). Each half: \(40\ \Omega\). In parallel: \(R_{eq} = 20\ \Omega\).

Current \(I = \dfrac{4}{20} =\) 0.2 A

Question 50

Nichrome and copper wires of same length and same radius are connected in series. Current \(I\) is passed through them. Which wire gets heated up more? Justify.

Answer: Nichrome wire gets heated more because its resistivity is higher than that of copper. Since \(H = I^2Rt\) and \(R \propto \rho\), nichrome dissipates more heat.

Question 51

Draw a plot showing the variation of resistivity of a (i) conductor and (ii) semiconductor, with increase in temperature. How does one explain this behaviour in terms of number density of charge carriers and the relaxation time?

Answer: (i) Conductor: \(\rho\) increases with temperature (roughly linear). (ii) Semiconductor: \(\rho\) decreases with temperature (exponentially).

Explanation: In conductors, average relaxation time decreases with increase in temperature, so \(\rho\) increases. In semiconductors, the increase in number density \(n\) with temperature is more than the decrease in relaxation time, so \(\rho\) decreases.

Question 52

Draw a graph showing the variation of resistivity with temperature for nichrome. Which property of nichrome is used to make standard resistance coils?

Answer: For nichrome, resistivity is roughly constant (very slightly increasing) over a wide temperature range. The temperature coefficient of resistivity \(\alpha\) is very small for nichrome, which is why it's used to make standard resistance coils.

Question 53

Define resistivity of a conductor. Plot a graph showing the variation of resistivity with temperature for a metallic conductor. How does one explain such a behaviour, using the mathematical expression of the resistivity of a material?

Answer: Resistivity is the resistance of a conductor of unit length and unit area of cross-section.

For a metallic conductor, resistivity increases with temperature (linear plot). Mathematical expression: \(\rho = \dfrac{m}{ne^2\tau}\). On increasing temperature, the average speed of electrons increases, so collisions are more frequent. The average relaxation time \(\tau\) decreases, hence \(\rho\) increases.

Question 54

(a) The potential difference applied across a given resistor is altered so that the heat produced per second increases by a factor of 9. By what factor does the applied potential difference change? (b) In the figure, an ammeter A and a resistor of 4 Ω are connected to the terminals of the source. The emf of the source is 12 V with internal resistance 2 Ω. Calculate the voltmeter and ammeter readings.

Answer: (a) \(H = \dfrac{V^2}{R}t\), so \(H \propto V^2\). For \(H\) to become 9 times, \(V\) must become \(\sqrt{9} =\) 3 times.

(b) \(I = \dfrac{12}{4 + 2} = 2\ \text{A}\) (ammeter reading).

Voltmeter reading (across 4 Ω): \(V = IR = 2 \times 4 =\) 8 V

Kirchhoff's Rules

Question 55

State Kirchhoff's rules. Explain briefly how these rules are justified.

Answer:

Kirchhoff's junction rule: At any junction, the sum of the currents entering the junction equals the sum of currents leaving the junction. (Justified by conservation of charge.)
Kirchhoff's loop rule: The algebraic sum of changes in potential in any closed loop involving resistors and cells is zero. (Justified by conservation of energy.)

Question 56

In the given circuit, assuming point A to be at zero potential, use Kirchhoff's rules to determine the potential at point B.

Answer: Applying Kirchhoff's loop rule along ACDE and using \(V_A = 0\), we can write:

\(V_A - I R_1 + E_1 - I R_2 + E_2 - I R_3 = V_B\)

Solving gives the potential at B relative to A. (Specific numerical value depends on circuit parameters.)

Question 57

Use Kirchhoff's rules to determine the value of the current \(I_1\) flowing in the circuit shown.

Answer: Using Kirchhoff's junction rule at junction E: \(I_3 = I_1 + I_2\).

In loop ABCDA (second law): \(80 - 20I_2 + 30I_1 = 0\).

Solving with the junction equation gives \(I_1\). A negative sign of current indicates that the direction of current is opposite to that shown in the circuit diagram.

Question 58

State Kirchhoff's rules. Use these rules to write the expressions for the currents \(I_1\), \(I_2\) and \(I_3\) in the circuit diagram shown.

Answer: Kirchhoff's rules:

(i) Junction rule: At any junction, sum of currents entering equals sum of currents leaving.
(ii) Loop rule: The algebraic sum of potential changes in any closed loop is zero.

From the junction rule: \(I_1 + I_2 = I_3\). Applying loop rule to the two loops and solving the resulting equations gives the three currents.

Question 59

State Kirchhoff's rules. Apply Kirchhoff's rules to the loops ACBPA and ACBQA to write the expressions for the currents \(I_1\), \(I_2\) and \(I_3\) in the network.

Answer: Junction rule: \(I_3 = I_1 + I_2\).

Loop ACBPA: \(10 = 12I_3 + 1I_2\) ...(i)
Loop ACBQA: \(12I_3 + 1I_2 = 10\) ...(ii)

Solving the simultaneous equations gives \(I_1\), \(I_2\) and \(I_3\).

Potentiometer and Meter Bridge

Question 60

State the principle of a potentiometer. Describe briefly, with the help of a circuit diagram, how this device is used to compare the emfs of two cells.

Answer: A potentiometer is a device used to measure potential difference.

Principle: When a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length: \(V = Kl\), where \(K\) is the potential gradient.

Method to compare emfs: Close key \(K_1\) so a constant current flows through the potentiometer wire. With \(K_2\) open, move the jockey along AB until it balances the emf \(\varepsilon_1\) of the first cell. Let the balancing length be \(l_1\). Then \(\varepsilon_1 = Kl_1\).

Now introduce the second cell and find its balancing length \(l_2\). Then \(\varepsilon_2 = Kl_2\). Ratio: \(\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2}\).

Question 61

Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.

Answer: The cell (emf \(\varepsilon\)) whose internal resistance \(r\) is to be determined is connected across a resistance box through key \(K_2\).

With \(K_2\) open, balance is obtained at length \(l_1\): \(\varepsilon = Kl_1\).

When \(K_2\) is closed, the cell sends current \(I\) through the resistance box \(R\). If \(V\) is the terminal potential difference and balance is obtained at \(l_2\): \(V = Kl_2\).

Since \(\varepsilon = I(R + r)\) and \(V = IR\), we get \(\dfrac{\varepsilon}{V} = \dfrac{R + r}{R} = \dfrac{l_1}{l_2}\).

Solving: \(r = R\left(\dfrac{l_1}{l_2} - 1\right)\)

Question 62

In a meter bridge experiment, balance point was observed at J with AJ = \(l\). (i) The values of R and X were doubled and then interchanged. What would be the new position of the balance point? (ii) If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected?

Answer: (i) Balance condition: \(\dfrac{R}{X} = \dfrac{l}{100 - l}\). Doubling both R and X doesn't change the ratio. Interchanging them gives \(\dfrac{X}{R} = \dfrac{l'}{100 - l'}\). So \(l' = 100 - l\). The new balance point is at distance \(100 - l\) from A.

(ii) From the Wheatstone bridge principle, the condition for a balanced bridge is still satisfied when the galvanometer and cell are interchanged. So the galvanometer will not show any deflection and the balance point remains the same.

Question 63

Write the principle on which the working of a meter bridge is based. In an experiment on a meter bridge, a student obtains the balance point at J such that AJ = 40 cm. The values of R and X are both doubled and then interchanged. Find the new position of the balance point. If the galvanometer and battery are also interchanged, how will the position of the balance point be affected?

Answer: The working of a meter bridge is based on the Wheatstone bridge principle.

Initial: \(\dfrac{R}{X} = \dfrac{40}{60} = \dfrac{2}{3}\). After doubling and interchanging: \(\dfrac{X}{R} = \dfrac{3}{2}\). Let the new balance point be at length \(l\) from A: \(\dfrac{3}{2} = \dfrac{l}{100 - l} \Rightarrow 3(100 - l) = 2l \Rightarrow l = 60\ \text{cm}\). New position: AJ = 60 cm.

If galvanometer and battery are also interchanged, the balance condition is still satisfied and the balance point position remains at 60 cm.

Question 64

Write the principle on which the working of a meter bridge is based. In an experiment on a meter bridge, a student obtains the balance point at J such that AJ = 60 cm. The values of R and X are both doubled and then interchanged. Find the new position of the balance point. If the galvanometer and battery are also interchanged, how will the position of balance point be affected?

Answer: Principle: Wheatstone bridge.

Initial: \(\dfrac{R}{X} = \dfrac{60}{40} = \dfrac{3}{2}\). After doubling and interchanging: \(\dfrac{X}{R} = \dfrac{2}{3}\). Then \(\dfrac{2}{3} = \dfrac{l}{100 - l} \Rightarrow 2(100 - l) = 3l \Rightarrow l = 40\ \text{cm}\). New position: AJ = 40 cm.

If galvanometer and battery are interchanged, the balance point position is unaffected (remains at 40 cm).

Question 65

In a meter bridge, the null point is found at a distance of 40 cm from A. If a resistance of 12 Ω is connected in parallel with S, the null point occurs at 50.0 cm from A. Determine the values of R and S.

Answer: From \(\dfrac{R}{S} = \dfrac{40}{60} = \dfrac{2}{3}\) → \(S = \dfrac{3R}{2}\).

With 12 Ω in parallel with S, effective \(S' = \dfrac{12S}{12 + S}\). Balance at 50 cm: \(\dfrac{R}{S'} = \dfrac{50}{50} = 1\), so \(R = S' = \dfrac{12S}{12 + S}\).

Substitute \(S = \dfrac{3R}{2}\): \(R = \dfrac{12 \times (3R/2)}{12 + 3R/2} = \dfrac{18R}{12 + 1.5R}\). Cancelling R: \(1 = \dfrac{18}{12 + 1.5R} \Rightarrow 12 + 1.5R = 18 \Rightarrow R = 4\ \Omega\).

Then \(S = \dfrac{3 \times 4}{2} =\) 6 Ω. So R = 4 Ω, S = 6 Ω.

Question 66

In a meter bridge, the null point is found at a distance of 60.0 cm from A. If now a resistance of 5 Ω is connected in series with S, the null point occurs at 50 cm. Determine the values of R and S.

Answer: Initially \(\dfrac{R}{S} = \dfrac{60}{40} = \dfrac{3}{2}\).

After adding 5 Ω to S, \(\dfrac{R}{S + 5} = \dfrac{50}{50} = 1\), so \(R = S + 5\).

Substituting \(R = \dfrac{3S}{2}\): \(\dfrac{3S}{2} = S + 5 \Rightarrow S = 10\ \Omega\), \(R = 15\ \Omega\).

R = 15 Ω, S = 10 Ω

Question 67

In a meter bridge, the null point is found at a distance of \(l_1\) cm from A. If now a resistance of X is connected in parallel with S, the null point occurs at \(l_2\). Obtain a formula for X in terms of \(l_1\), \(l_2\) and S.

Answer: Initially: \(\dfrac{R}{S} = \dfrac{l_1}{100 - l_1}\).

With X in parallel with S, the effective resistance of the right arm is \(S' = \dfrac{SX}{S + X}\). New balance condition: \(\dfrac{R}{S'} = \dfrac{l_2}{100 - l_2}\).

Divide the two equations: \(\dfrac{S'}{S} = \dfrac{l_1}{100 - l_1} \cdot \dfrac{100 - l_2}{l_2} = k\) (say). Then \(S' = kS\).

From \(S' = \dfrac{SX}{S + X}\): \(kS = \dfrac{SX}{S + X} \Rightarrow k = \dfrac{X}{S + X} \Rightarrow X = \dfrac{kS}{1 - k}\).

Substituting \(k\): \(X = \dfrac{l_1(100 - l_2) \cdot S}{(100 - l_1)l_2 - l_1(100 - l_2)}\)

Question 68

Write the principle of working of a potentiometer. Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a given cell.

Answer: Principle: When a constant current flows through a wire of uniform cross-section and composition, the potential drop across any length of the wire is directly proportional to that length: \(V = Kl\).

Method to determine internal resistance: The cell of emf \(\varepsilon\) is connected across a resistance box R through a key \(K_2\).

With \(K_2\) open, find the balance length \(l_1\) (for emf \(\varepsilon\)): \(\varepsilon = Kl_1\).

With \(K_2\) closed, find the balance length \(l_2\) (for terminal voltage \(V\)): \(V = Kl_2\).

Using \(\varepsilon = I(R + r)\) and \(V = IR\): \(\dfrac{\varepsilon}{V} = \dfrac{R + r}{R} = \dfrac{l_1}{l_2} \Rightarrow r = R\left(\dfrac{l_1}{l_2} - 1\right)\).

Question 69

State the underlying principle of a potentiometer. Write two factors on which the sensitivity of a potentiometer depends. In the potentiometer circuit shown, the balance point is at X. State, giving reason, how the balance point is shifted when (i) resistance R is increased? (ii) resistance S is increased, keeping R constant?

Answer: Principle: When a current flows through a wire of uniform thickness, the potential difference between its two points is directly proportional to the length of the wire between these two points: \(V \propto l\).

Two factors: (a) Potential gradient (b) Length of potentiometer wire.

(i) When R is increased, the current through the potentiometer wire decreases, so the potential gradient decreases. Therefore, the same emf needs a longer length to balance. Balance point shifts towards B.

(ii) If S is increased keeping R constant, the main circuit current is unchanged, so the balance point does not change.

Question 70

A potentiometer wire of length 1 m has a resistance of 10 Ω. It is connected to a 6 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 40 cm.

Answer: Current through wire: \(I = \dfrac{6}{10 + 5} = 0.4\ \text{A}\).

Potential drop across full length (1 m): \(V = IR = 0.4 \times 10 = 4\ \text{V}\).

Potential gradient \(K = 4\ \text{V/m}\). For balancing length 40 cm = 0.4 m: \(\varepsilon = K \times l = 4 \times 0.4 =\) 1.6 V

Question 71

A potentiometer wire of length 1.0 m has a resistance of 15 Ω. It is connected to a 5 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.

Answer: \(I = \dfrac{5}{15 + 5} = 0.25\ \text{A}\).

Potential drop across 1 m: \(V = 0.25 \times 15 = 3.75\ \text{V}\).

Potential gradient \(K = 3.75\ \text{V/m}\). For 60 cm = 0.6 m: \(\varepsilon = 3.75 \times 0.6 =\) 2.25 V

Question 72

A potentiometer wire of length 1 m has a resistance of 5 Ω. It is connected to an 8 V battery in series with a resistance of 15 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.

Answer: \(I = \dfrac{8}{5 + 15} = 0.4\ \text{A}\).

Potential drop across 1 m: \(V = 0.4 \times 5 = 2\ \text{V}\).

\(K = 2\ \text{V/m}\). For 0.6 m: \(\varepsilon = 2 \times 0.6 =\) 1.2 V

Question 73

A potentiometer wire of length 1 m is connected to a driver cell of emf 3 V. When a cell of 1.5 V emf is used in the secondary circuit, the balance point is found at 60 cm. On replacing this cell with a cell of unknown emf, the balance point shifts to 80 cm. (i) Calculate the unknown emf. (ii) Explain with reason whether the circuit works if the driver cell is replaced with a cell of emf 1 V. (iii) Does the high resistance R, used in the secondary circuit, affect the balance point? Justify.

Answer: (i) \(\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2} \Rightarrow \dfrac{1.5}{\varepsilon_2} = \dfrac{60}{80} \Rightarrow \varepsilon_2 = 1.5 \times \dfrac{80}{60} =\) 2.0 V

(ii) The circuit will not work. If the emf of the auxiliary battery is less than that of the cell in the secondary circuit (1.5 V), there will be no balance point on the wire. The max p.d. across the wire would be 1 V, which cannot balance 1.5 V.

(iii) No, the balance point is not affected by the high resistance R because no current flows through the cell at the balance point.

Question 74

In the meter bridge experiment, the balance point was observed at J with AJ = l. The values of R and X were doubled and then interchanged. What would be the new position of the balance point? If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected?

Answer: Same as Question 62. (i) New balance point at \(100 - l\). (ii) Unchanged when galvanometer and battery are interchanged.

Question 75

Draw a circuit diagram of a potentiometer. State its working principle. Derive the necessary formula to describe how it is used to compare the emfs of two cells.

Answer: Same as Question 60. Working principle: \(V \propto l\) for a uniform wire carrying constant current.

For comparison: balance length \(l_1\) for cell with emf \(\varepsilon_1\), and \(l_2\) for \(\varepsilon_2\). Then \(\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2}\).

Causes for one-sided deflection: (i) potential difference between ends of wire < emf of the cell in the secondary circuit; (ii) positive side of the driving cell connected to the negative terminal of the cell in the secondary circuit.

Question 76

With the help of the circuit diagram, explain the working principle of a meter bridge. How is it used to determine the unknown resistance of a given wire? Write the necessary precautions to minimize the error in the result.

Answer: Meter bridge is a special case of the Wheatstone bridge. It is used to determine the unknown resistance of a wire.

Principle: When the bridge is balanced, \(\dfrac{P}{Q} = \dfrac{R}{S}\).

Circuit: The unknown resistance S is connected in the gap between C and D. Resistance box R is connected between A and D. A cell, rheostat and key K are connected between A and C. The wire AC is divided into two parts by the jockey, which act as P and Q.

Method: Close key K. Choose a resistance R such that when the jockey is pressed at both ends A and C, galvanometer deflection is on both sides. Slide the jockey to a null point. Measure the length AB = l cm and BC = (100 - l) cm.

Then \(\dfrac{R}{S} = \dfrac{l}{100 - l} \Rightarrow S = R \cdot \dfrac{100 - l}{l}\).

Precautions: (i) Very small resistances cannot be measured accurately (end resistances not accounted for). (ii) Current should not flow for a long time, otherwise the wire gets hot and its resistance changes.

Mixed Numerical Problems

Question 77

Calculate the current drawn from the battery in the given network.

Answer: Simplify the network by identifying series and parallel combinations, find the equivalent resistance, then use \(I = \dfrac{V}{R_{eq}}\). The current drawn from the battery is the total current through the external circuit.

Question 78

The network PQRS has batteries of 4 V and 5 V and negligible internal resistance. A milliammeter of 20 Ω resistance is connected between P and R. Calculate the reading in the milliammeter.

Answer: Apply loop rule to loop PQRP: \(-4 = 60(I - I_1) - 20I_1\) ... (i)

Apply loop rule to loop PRSP: \(-5 + 200I + 20I_1 = 0\) ... (ii)

Solving gives \(I\) through the milliammeter. The reading is 0.064 A.

Question 79

The network PQRS has batteries of 5 V and 10 V and negligible internal resistance. A milliammeter of 50 Ω resistance is connected between P and R. Calculate the reading in the milliammeter.

Answer: Loop PSRP: \(-10 + 100I + 50I_1 = 0 \Rightarrow 10I + 5I_1 = 1\) ... (i)

Loop PRQP: solving gives \(I_1\). The reading is 0.08 A.

Question 80

An ammeter of resistance 1 Ω can measure current up to 1.0 A. (i) What must be the value of the shunt resistance to enable the ammeter to measure up to 5.0 A? (ii) What is the combined resistance of the ammeter and the shunt?

Answer: Let shunt resistance be \(R_s\). Max current \(I = 5\ \text{A}\), ammeter resistance \(G = 1\ \Omega\), ammeter current \(I_g = 1\ \text{A}\).

(i) \(I_g G = (I - I_g) R_s \Rightarrow 1 \times 1 = 4 R_s \Rightarrow R_s =\) 0.25 Ω

(ii) \(R_{eq} = \dfrac{G \cdot R_s}{G + R_s} = \dfrac{1 \times 0.25}{1.25} =\) 0.2 Ω

Question 81

Using Kirchhoff's rules, determine the potential difference between points A and D when no current flows in arm BE of the electric network shown.

Answer: Considering loop ABED and using Kirchhoff's second law, we can write the potential changes around the loop. Setting the current through BE to zero, the loops reduce to a simpler network. Solving gives the potential difference \(V_A - V_D\).

Question 82

Calculate the current drawn from the battery by the network of resistors shown.

Answer: Rearranging the given circuit, it forms a Wheatstone's bridge. Resistances \(P = 1\ \Omega\), \(R = 2\ \Omega\) in series; similarly \(Q = 2\ \Omega\) and \(S\) in series. If it's a balanced bridge, no current flows through BD. Then find the equivalent resistance and current using \(I = \dfrac{V}{R_{eq}}\).

Question 83

Use Kirchhoff's rules to determine the value of unknown resistance R in the circuit so that no current flows through the 4 Ω resistance. Also find the potential difference between A and D.

Answer: If no current flows through the 4 Ω resistance, it acts like an open circuit. Apply Kirchhoff's laws to the remaining loops to find the current and hence R, then calculate the potential difference \(V_A - V_D\).

Question 84

Calculate the value of the resistance R in the circuit shown so that the current in the circuit is 0.2 A. What would be the potential difference between points B and E?

Answer: Using \(I = 0.2\ \text{A}\) and the total emf in the circuit, find the total resistance required. Then subtract the known resistances to find R.

\(R_{total} = \dfrac{V_{total}}{I}\). Then \(R = R_{total} - R_{other}\)

The potential difference between B and E is the p.d. across the part of the circuit between these points, calculated using Ohm's law.

Question 85

Calculate the value of the resistance R in the circuit shown so that the current is 0.2 A. What would be the potential difference between points A and B?

Answer: Same approach as Question 84 but for the p.d. between A and B.

Question 86

In the circuit shown, \(R_1 = 4\ \Omega\), \(R_2 = R_3 = 15\ \Omega\), \(R_4 = 30\ \Omega\) and \(E = 10\ \text{V}\). Calculate the equivalent resistance of the circuit and the current in each resistor.

Answer: Identify series and parallel combinations: \(R_2\) and \(R_3\) in parallel, then in series with \(R_1\), then \(R_4\) in parallel. Find equivalent resistance, then total current, then branch currents using current division.

Question 87

In the circuit shown, \(R_1 = 4\ \Omega\), \(R_2 = R_3 = 5\ \Omega\), \(R_4 = 10\ \Omega\) and \(E = 6\ \text{V}\). Work out the equivalent resistance of the circuit and the current in each resistor.

Answer: Same method as Question 86: identify series and parallel combinations, then find \(R_{eq}\), total current, and branch currents.

Question 88

In the figure, a long uniform potentiometer wire AB has a constant potential gradient along its length. The null points for the two primary cells of emfs \(\varepsilon_1\) and \(\varepsilon_2\) connected in the manner shown are obtained at distances of 120 cm and 300 cm from end A. Find (i) \(\varepsilon_1/\varepsilon_2\) (ii) position of null point for cell \(\varepsilon_1\). How is the sensitivity of a potentiometer increased?

Answer: From the arrangement, the cells are in series-opposing configuration. Let \(\varepsilon_1 > \varepsilon_2\).

The null points correspond to \((\varepsilon_1 - \varepsilon_2)\) and \((\varepsilon_1 + \varepsilon_2)\): \((\varepsilon_1 - \varepsilon_2) \propto 120\), \((\varepsilon_1 + \varepsilon_2) \propto 300\).

Dividing: \(\dfrac{\varepsilon_1 - \varepsilon_2}{\varepsilon_1 + \varepsilon_2} = \dfrac{120}{300} = \dfrac{2}{5} \Rightarrow 5\varepsilon_1 - 5\varepsilon_2 = 2\varepsilon_1 + 2\varepsilon_2 \Rightarrow 3\varepsilon_1 = 7\varepsilon_2 \Rightarrow \varepsilon_1/\varepsilon_2 = 7/3\)

Balancing length for \(\varepsilon_1\): \(l_1 = \dfrac{300 + 120}{2} = 210\ \text{cm}\)

Sensitivity of a potentiometer can be increased by reducing the potential gradient.

Question 89

In the electric network shown, use Kirchhoff's rules to calculate the power consumed by the resistance \(R = 4\ \Omega\).

Answer: Using Kirchhoff's rules:

\(-12 + 2I_1 + 4(I_1 + I_2) = 0 \Rightarrow 3I_1 + 2I_2 = 6\) ... (i)
Loop ADFEA: \(-4(I_1 + I_2) + 6 = 0 \Rightarrow 2I_1 + 2I_2 = 3\) ... (ii)

Solving: \(I_1 = 3\ \text{A}\), \(I_2 = -1.5\ \text{A}\).

Current through 4 Ω: \(I = I_1 + I_2 = 1.5\ \text{A}\). Power \(P = I^2R = (1.5)^2 \times 4 =\) 9 W.

Question 90

In the electric network shown, use Kirchhoff's rules to calculate the power consumed by the resistance \(R = 8\ \Omega\).

Answer: Loop BCDA: \(4I_1 + 8(I_1 + I_2) = 12 \Rightarrow 3I_1 + 2I_2 = 3\) ... (i)

Loop ADFE: \(8(I_1 + I_2) = 8 \Rightarrow I_1 + I_2 = 1\) ... (ii)

Solving: \(I_1 = 1\ \text{A}\), \(I_2 = 0\ \text{A}\).

Current through 8 Ω: \(I = I_1 + I_2 = 1\ \text{A}\). Power \(P = I^2R = 1^2 \times 8 =\) 8 W.

Question 91

In the circuit shown, find the current through each resistor.

Answer: Total emf = \(8 - 4 = 4\ \text{V}\). Two resistors 3 Ω and 6 Ω in parallel: \(R_p = \dfrac{3 \times 6}{3 + 6} = 2\ \Omega\).

Total resistance = 0.5 + 4.5 + 2 + 1 = 8 Ω.

Total current \(I = \dfrac{4}{8} = 0.5\ \text{A}\). This flows through the 0.5 Ω, 4.5 Ω and 1 Ω resistors.

The current through 3 Ω and 6 Ω in parallel: \(I_3 = \dfrac{6}{9} \times 0.5 = 0.333\ \text{A}\), \(I_6 = \dfrac{3}{9} \times 0.5 = 0.167\ \text{A}\).

Question 92

Calculate the current density of a metallic conductor in terms of the drift speed of electrons.

Answer: Current \(I = neAv_d\). Current density \(J = \dfrac{I}{A} = nev_d\). So \(J \propto v_d\). The current density is directly proportional to the drift speed.

Question 93

Two cells of emf \(E_1\) and \(E_2\) with internal resistances \(r_1\) and \(r_2\) are connected such that their combination has a net internal resistance R. Draw a diagram of the setup and obtain an expression for the current flowing through the resistance R.

Answer: Since the net internal resistance is less than \(r_1\), the two cells must be connected in parallel.

For parallel cells, the equivalent emf and internal resistance are: \(E_{eq} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2}\), \(r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2} = R\)

Current through external resistance R (assumed same as \(r_{eq}\)): \(I = \dfrac{E_{eq}}{R_{ext} + r_{eq}}\).

Question 94

A resistance of R draws current from a potentiometer. The potentiometer wire, AB, has a total resistance of \(R_0\). A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer wire.

Answer: Let the total length be L. At the middle, the resistance of the left half AC is \(R_0/2\) and the right half CB is \(R_0/2\).

The left half (resistance \(R_0/2\)) is in parallel with R. Their equivalent: \(R_p = \dfrac{(R_0/2)R}{R_0/2 + R}\).

This is in series with the right half (\(R_0/2\)). Total resistance of the circuit: \(R_{total} = R_p + R_0/2\).

Total current from source: \(I = \dfrac{V}{R_{total}}\). Voltage across the parallel combination: \(V' = I R_p = \dfrac{V R_p}{R_p + R_0/2}\)

Question 95

Calculate the equivalent resistance between points A and B in the given electrical network. Also calculate the current through CD and ACB, if a 10 V d.c. source is connected between A and B, and the value of R is assumed as 2 Ω.

Answer: The given circuit simplifies to a balanced Wheatstone bridge.

(i) Since the bridge is balanced, no current flows through the CD arm. The equivalent resistance can be calculated by considering the two series-parallel paths.

(ii) With \(V = 10\ \text{V}\) and R = 2 Ω, find the total current using \(I = \dfrac{V}{R_{eq}}\). Current through CD is zero (balanced bridge). Current through ACB is the current through that arm.

Question 96

The figure shows an experimental setup of a meter bridge. When two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from end A. When a resistance of 10 Ω is connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the 10 Ω resistance is instead connected in series with resistance Y. Determine the values of resistances X and Y.

Answer: Initially: \(\dfrac{X}{Y} = \dfrac{40}{60} = \dfrac{2}{3}\). With 10 Ω in series with X: \(\dfrac{X + 10}{Y} = \dfrac{50}{50} = 1 \Rightarrow X + 10 = Y\).

Substitute \(Y = \dfrac{3X}{2}\): \(X + 10 = \dfrac{3X}{2} \Rightarrow 2X + 20 = 3X \Rightarrow X = 20\ \Omega\), \(Y = 30\ \Omega\).

When 10 Ω is in series with Y instead: \(\dfrac{X}{Y + 10} = \dfrac{20}{40} = \dfrac{1}{2}\), so balance point at 33.3 cm from A.

Question 97

State the principle of working of a meter bridge. In a meter bridge, balance point is found at distance \(l_1\) with resistances R and S. When an unknown resistance X is connected in parallel with S, the balance point shifts to \(l_2\). Find the expression for X in terms of \(l_1\), \(l_2\) and S.

Answer: Meter bridge works on the Wheatstone bridge principle: at balance, no current flows through the galvanometer.

Initial: \(\dfrac{R}{S} = \dfrac{l_1}{100 - l_1}\) ... (i)

With X in parallel with S: effective resistance \(S' = \dfrac{SX}{S + X}\). New balance: \(\dfrac{R}{S'} = \dfrac{l_2}{100 - l_2}\) ... (ii)

Dividing (ii) by (i): \(\dfrac{S}{S'} = \dfrac{l_2(100 - l_1)}{l_1(100 - l_2)}\). Let \(k = \dfrac{l_1(100 - l_2)}{l_2(100 - l_1)}\), so \(S' = kS\).

From \(S' = \dfrac{SX}{S + X}\): \(kS = \dfrac{SX}{S + X} \Rightarrow k(S + X) = X \Rightarrow X = \dfrac{kS}{1 - k}\).

Substituting \(k\): \(X = \dfrac{S \cdot l_1(100 - l_2)}{l_2(100 - l_1) - l_1(100 - l_2)}\)

Question 98

Use Kirchhoff's rules to obtain conditions for the balance condition in a Wheatstone bridge.

Answer: Applying Kirchhoff's loop rule to closed loop ADBA and to loop BDCB, and setting the current through the galvanometer (BD) to zero for balance, we get:

Loop ADBA: \(I_1 P - I_2 R = 0 \Rightarrow I_1 P = I_2 R\) ... (i)

Loop BDCB: \(I_1 Q - I_2 S = 0 \Rightarrow I_1 Q = I_2 S\) ... (ii)

Dividing (i) by (ii): \(\dfrac{P}{Q} = \dfrac{R}{S}\) — the balance condition.

Long Answer Questions (5 Marks)

Drift Velocity and Current Density

Question 99

(a) Derive the relation between current density \(\vec{J}\) and potential difference \(V\) across a current-carrying conductor of length \(l\), area of cross-section \(A\), and number density of free electrons \(n\). (b) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area \(1.0 \times 10^{-7}\ \text{m}^2\) carrying a current of 1.5 A. [Assume number density \(n = 9 \times 10^{28}\ \text{m}^{-3}\)]

Answer: (a) \(E = \dfrac{V}{l}\). Drift velocity \(v_d = \dfrac{eE\tau}{m} = \dfrac{eV\tau}{ml}\).

Current \(I = neAv_d = \dfrac{ne^2AV\tau}{ml}\).

Current density \(J = \dfrac{I}{A} = \dfrac{ne^2\tau}{m} \cdot \dfrac{V}{l} = \sigma E\), where \(\sigma = \dfrac{ne^2\tau}{m}\).

So \(J = \sigma \dfrac{V}{l}\).

(b) \(v_d = \dfrac{I}{neA} = \dfrac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-7}} =\) \(1.04 \times 10^{-3}\ \text{m/s}\)

Question 100

A number of identical cells \(n\), each of emf \(\varepsilon\), internal resistance \(r\), are connected in series and charged by a d.c. source of emf \(\varepsilon'\) using a resistor R. (i) Draw the circuit arrangement. (ii) Deduce the expressions for (a) the charging current and (b) the potential difference across the combination of the cells.

Answer: (i) The d.c. source \(\varepsilon'\) is connected in series with R and the combination of \(n\) cells (all opposing).

(ii) (a) Net emf = \(\varepsilon' - n\varepsilon\). Total resistance = \(R + nr\).

Charging current: \(I = \dfrac{\varepsilon' - n\varepsilon}{R + nr}\)

(b) Potential difference across the combination of cells: \(V = \varepsilon' - IR\) or equivalently the terminal voltage of the source.

Question 101

Define the terms (i) drift velocity (ii) relaxation time. A conductor of length L is connected to a d.c. source of emf \(\varepsilon\). If this conductor is replaced by another conductor of same material and same area of cross-section but of length 3L, how will the drift velocity change?

Answer: (i) Drift velocity: The average velocity gained by free electrons of a conductor in the direction opposite to the applied field.

(ii) Relaxation time: The average time that elapses between two successive collisions of an electron.

Field \(E = \dfrac{\varepsilon}{L}\). Drift velocity \(v_d = \dfrac{eE\tau}{m} = \dfrac{e\varepsilon\tau}{mL}\).

When length becomes 3L: \(E' = \dfrac{\varepsilon}{3L} = \dfrac{E}{3}\), so \(v_d' = \dfrac{v_d}{3}\). Drift velocity becomes one-third.

Question 102

Define relaxation time of the free electrons drifting in a conductor. How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material.

Answer: Relaxation time is the average time that elapses between two successive collisions of an electron.

Relation: \(v_d = \dfrac{eE\tau}{m}\).

Current \(I = neAv_d = \dfrac{ne^2EA\tau}{m}\). Also \(E = \dfrac{V}{l}\), so \(I = \dfrac{ne^2A\tau}{ml}V\).

From Ohm's law, \(V = IR = I \cdot \dfrac{\rho l}{A}\), so \(R = \dfrac{\rho l}{A}\). Comparing: \(\rho = \dfrac{m}{ne^2\tau}\).

Question 103

Find the relation between drift velocity and relaxation time of charge carriers in a conductor. A conductor of length L is connected to a d.c. source of emf E. If the length of the conductor is tripled by stretching it, keeping E constant, explain how its drift velocity would be affected.

Answer: Relation: \(v_d = \dfrac{eE\tau}{m}\) where \(E = \dfrac{V}{L}\).

When length is tripled (3L), \(E' = \dfrac{E}{3}\). Since \(v_d \propto E\), \(v_d' = \dfrac{v_d}{3}\). The drift velocity becomes one-third of its original value.

Question 104

(i) Derive an expression for drift velocity of free electrons. (ii) How does drift velocity of electrons in a metallic conductor vary with increase in temperature? Explain.

Answer: (i) \(v_d = \dfrac{eE\tau}{m}\). When a potential difference is applied, an electric field is produced. Free electrons experience a force \(-eE\) and accelerate, but keep colliding with each other, acquiring a constant average velocity \(v_d\).

(ii) The drift velocity decreases with increasing temperature because the relaxation time decreases (more frequent collisions).

Question 105

(a) Deduce the relation between current I flowing through a conductor and drift velocity \(\vec{v}_d\) of the electrons. (b) The figure shows a plot of current I flowing through the cross-section of a wire versus time t. Use the plot to find the charge flowing in 10 s through the wire.

Answer: (a) Consider a wire of length \(l\), area \(A\), with \(n\) free electrons per unit volume. Total charge in length \(l\): \(q = nAle\). Time to cross: \(t = \dfrac{l}{v_d}\). Current: \(I = \dfrac{q}{t} = neAv_d\). The negative sign shows that current is opposite to electron drift.

(b) Charge = area under I-t graph. For 0 to 5 s, average current = 2.5 A, so \(q_1 = 2.5 \times 5 = 12.5\ \text{C}\). For 5 to 10 s, current = 5 A, so \(q_2 = 5 \times 5 = 25\ \text{C}\).

Total charge \(q = 12.5 + 25 =\) 37.5 C

Key Formulas to Remember

Concept	Formula
Current	\(I = \dfrac{q}{t}\)
Ohm's law	\(V = IR\)
Drift velocity	\(v_d = \dfrac{eE\tau}{m}\)
Current in terms of drift velocity	\(I = neAv_d\)
Current density	\(J = \dfrac{I}{A} = \sigma E\)
Conductivity	\(\sigma = \dfrac{ne^2\tau}{m}\)
Resistivity	\(\rho = \dfrac{m}{ne^2\tau}\)
Mobility	\(\mu = \dfrac{v_d}{E} = \dfrac{e\tau}{m}\)
Resistance (depends on dimensions)	\(R = \dfrac{\rho l}{A}\)
Terminal voltage	\(V = E - Ir\)
Cells in series (n cells)	\(E_{eq} = nE\), \(r_{eq} = nr\)
Cells in parallel (2 cells)	\(E_{eq} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2}\), \(r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}\)
Wheatstone bridge balance	\(\dfrac{P}{Q} = \dfrac{R}{S}\)
Meter bridge balance	\(\dfrac{R}{S} = \dfrac{l}{100 - l}\)
Potentiometer (balance)	\(\varepsilon \propto l\), \(\varepsilon = Kl\)
Internal resistance (potentiometer)	\(r = R\left(\dfrac{l_1}{l_2} - 1\right)\)
Heat produced	\(H = I^2Rt = \dfrac{V^2}{R}t\)
Power	\(P = I^2R = \dfrac{V^2}{R} = VI\)

FAQs

Are these questions enough for CBSE Class 12 Physics Chapter 3?

Yes. These questions cover every major topic — drift velocity, resistivity and conductivity, internal resistance of cells, Kirchhoff's rules, Wheatstone bridge, meter bridge and potentiometer — that CBSE has asked in previous board exams. Combine them with NCERT exercises and you'll be well prepared.

Which topics are most important for board exams?

Drift velocity and its relation to current, resistivity in terms of relaxation time, internal resistance and terminal voltage, Kirchhoff's rules and their applications, Wheatstone bridge balance condition, meter bridge calculations, and potentiometer principle are consistently high-weightage topics.

How should I prepare for numericals in this chapter?

Practise drift velocity calculations, series-parallel resistance networks, Kirchhoff's loop problems, meter bridge unknown resistance calculations, potentiometer balance length problems, and internal resistance determination. Focus on understanding when to apply series/parallel formulas and when to use Kirchhoff's rules.