JEE Physics Current Electricity MCQs Set 19

Physics Objective Questions and Answers: Current Electricity

Review structured MCQ sets for JEE Physics Current Electricity. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Download Current Electricity MCQs with Answers

Access the complete set of multiple-choice questions for Current Electricity below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: If the specific resistance of a wire is ρ, its volume is 3 m³ and its resistance is 3 ohm, then its length will be:
(a) \( \sqrt{\frac1\rho} \)
(b) \( \frac{3}{\sqrt\rho} \)
(c) \( \frac{1}{\rho}\sqrt3 \)
(d) \( \rho\sqrt{\frac13} \)
Show Answer & Explanation

Answer: (b) \( \frac{3}{\sqrt\rho} \)

Explanation:
1. \( R=\frac{\rho l}{A}=\frac{\rho l^2}{V} \)
2. \( l^2=\frac{RV}{\rho}=\frac{3\times3}{\rho}=\frac9\rho \)
3. \( l=\frac{3}{\sqrt\rho} \)

Teacher's Note:
1. Eliminate the area using volume = A × l.

Question 2: A set of n identical resistors, each of resistance R ohm, when connected in series has an effective resistance of x ohm. When the resistors are connected in parallel, the effective resistance is y ohm. The relation between R, x and y is:
(a) \( R=\frac{xy}{x+y} \)
(b) \( R=y-x \)
(c) \( R=\sqrt{xy} \)
(d) \( R=x+y \)
Show Answer & Explanation

Answer: (c) \( R=\sqrt{xy} \)

Explanation:
1. \( x=nR,\quad y=\frac Rn \)
2. \( xy=R^2\Rightarrow R=\sqrt{xy} \)

Teacher's Note:
1. The product of the series and parallel resistances of n equal resistors is R².

Question 3: The supply voltage to a room is 120 V. The resistance of the lead wire is 6 Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb when a 240 W heater is switched on in parallel to the bulb? (Ratings are for 120 V.)
(a) Zero
(b) 2.9 V
(c) 13.3 V
(d) 10.4 V
Show Answer & Explanation

Answer: (d) 10.4 V

Explanation:
1. Bulb resistance \( \frac{120^2}{60}=240\ \Omega \); heater resistance \( \frac{120^2}{240}=60\ \Omega \).
2. Bulb alone: \( V_1=120\times\frac{240}{240+6}=117.07\ \text{V} \).
3. With heater: parallel resistance \( \frac{240\times60}{300}=48\ \Omega \), so \( V_2=120\times\frac{48}{54}=106.67\ \text{V} \).
4. Decrease \( =117.07-106.67\approx10.4\ \text{V} \)

Teacher's Note:
1. Lead-wire resistance acts as a series resistance, so extra load lowers the voltage across the bulb.

Question 4: Two batteries of emf ε1 and ε2 (ε2 greater than ε1) and internal resistances r1 and r2 are connected in parallel with the same polarity. Which statement about the equivalent emf εeq of the combination is correct?
(a) It lies between ε1 and ε2, i.e. ε1 < εeq < ε2
(b) It is smaller than ε1
(c) It is given by ε1 + ε2 always
(d) It is independent of the internal resistances r1 and r2
Show Answer & Explanation

Answer: (a) It lies between ε1 and ε2, i.e. ε1 < εeq < ε2

Explanation:
1. \( \varepsilon_{eq}=\frac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2} \), a weighted average of the two emfs.
2. A weighted average always lies between the two values, and it depends on r1 and r2.

Teacher's Note:
1. Parallel cells give a weighted average of emfs, weighted by the conductances 1/r.

Question 5: Two identical cells, each of emf Em and internal resistance r, are connected in parallel (same polarity) across a resistor R. The current through the resistor R is:
(a) \( \frac{E_m}{2R+r} \)
(b) \( \frac{E_m}{2r+R} \)
(c) \( \frac{2E_m}{R+2r} \)
(d) \( \frac{2E_m}{2R+r} \)
Show Answer & Explanation

Answer: (d) \( \frac{2E_m}{2R+r} \)

Explanation:
1. Equivalent cell: emf Em and internal resistance r/2.
2. \( I=\frac{E_m}{R+r/2}=\frac{2E_m}{2R+r} \)

Teacher's Note:
1. Identical cells in parallel keep the emf and halve the internal resistance.

Question 6: Which of the following characteristics of electrons determines the current in a conductor?
(a) Drift velocity alone
(b) Thermal velocity alone
(c) Both drift velocity and thermal velocity
(d) Neither drift nor thermal velocity
Show Answer & Explanation

Answer: (a) Drift velocity alone

Explanation:
1. Thermal motion is random and averages to zero over all electrons, so it gives no net charge flow.
2. Current \( I=neAv_d \) depends only on the drift velocity.

Teacher's Note:
1. Net flow of charge needs a net average velocity, which is the drift velocity.

Question 7: If the emf in a thermocouple is ε = αT + βT², then the neutral temperature of the thermocouple is:
(a) \( -\frac{\beta}{2\alpha} \)
(b) \( -\frac{2\beta}{\alpha} \)
(c) \( -\frac{\alpha}{2\beta} \)
(d) \( -\frac{2\alpha}{\beta} \)
Show Answer & Explanation

Answer: (c) \( -\frac{\alpha}{2\beta} \)

Explanation:
1. Neutral temperature is where \( \frac{d\varepsilon}{dT}=0 \).
2. \( \alpha+2\beta T=0\Rightarrow T_n=-\frac{\alpha}{2\beta} \)

Teacher's Note:
1. At the neutral temperature the thermo-emf is maximum, so its derivative vanishes.

Question 8: In a meter bridge experiment, the unknown resistance R = 20 Ω is in the left gap and S = 40 Ω in the right gap. The galvanometer shows no deflection when the jockey touches the wire at a point D, at a distance L1 from the left end A of the 100 cm wire. Then L1 is equal to:
(a) \( \frac{100}{3}\ \text{cm} \)
(b) \( \frac{50}{3}\ \text{cm} \)
(c) 25 cm
(d) \( \frac{25}{2}\ \text{cm} \)
Show Answer & Explanation

Answer: (a) \( \frac{100}{3}\ \text{cm} \)

Explanation:
1. Balance condition: \( \frac RS=\frac{L_1}{100-L_1} \).
2. \( \frac{20}{40}=\frac{L_1}{100-L_1}\Rightarrow 100-L_1=2L_1 \), so \( L_1=\frac{100}{3}\ \text{cm} \)

Teacher's Note:
1. In a balanced meter bridge the resistances are in the ratio of the wire lengths on either side of the jockey.

Question 9: A meter bridge has an unknown resistance R in the left gap and a resistance of 80 Ω in the right gap. The galvanometer shows zero current when the jockey divides the wire so that AC = 20 cm and CB = 80 cm. The resistance R is:
(a) 2 Ω
(b) 8 Ω
(c) 20 Ω
(d) 40 Ω
Show Answer & Explanation

Answer: (c) 20 Ω

Explanation:
1. \( \frac{R}{80}=\frac{20}{80}\Rightarrow R=20\ \Omega \)

Teacher's Note:
1. Balanced bridge: R/S = AC/CB.

Question 10: A battery of negligible internal resistance is connected in series with a resistor R. This combination is connected across a parallel arrangement of a second resistor R and a variable resistor r. If the heat generated in r is maximum when r = fR, then f is:
(a) 1/2
(b) 1
(c) 1/4
(d) 3/4
Show Answer & Explanation

Answer: (a) 1/2

Explanation:
1. Seen from the terminals of r, the battery is replaced by a short, so the Thevenin resistance of the rest is \( R\parallel R=\frac R2 \).
2. Power in the variable resistance is maximum when r equals the Thevenin resistance: \( r=\frac R2 \), so f = 1/2.

Teacher's Note:
1. Maximum power transfer occurs when the load resistance equals the source resistance seen by the load.

Question 11: The resistance of the series combination of two resistances is S. When they are joined in parallel, the total resistance is P. If S = nP, then the minimum possible value of n is:
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. \( S=R_1+R_2,\quad P=\frac{R_1R_2}{R_1+R_2} \)
2. \( n=\frac SP=\frac{(R_1+R_2)^2}{R_1R_2}\ge4 \) since \( (R_1+R_2)^2\ge4R_1R_2 \), with equality when R1 = R2.

Teacher's Note:
1. By the AM-GM inequality the ratio (R1 + R2)²/(R1R2) is at least 4.

Question 12: The resistance of an electric toaster has a temperature dependence R(T) = R0[1 + α(T − T0)] in its range of operation. At T0 = 300 K, R = 100 Ω and at T = 500 K, R = 120 Ω. The toaster is connected to a voltage source of 200 V and its temperature is raised at a constant rate from 300 K to 500 K in 30 s. The total work done in raising the temperature is:
(a) \( 6\times10^4\ln1.2\ \text{J} \) (about 1.1 × 10⁴ J)
(b) \( 400\ln\frac56\ \text{J} \)
(c) \( 200\ln\frac23\ \text{J} \)
(d) 300 J
Show Answer & Explanation

Answer: (a) \( 6\times10^4\ln1.2\ \text{J} \) (about 1.1 × 10⁴ J)

Explanation:
1. From the data \( 120=100[1+200\alpha]\Rightarrow\alpha=10^{-3}\ \text{K}^{-1} \). Temperature rate \( \frac{200}{30} \) K/s.
2. \( W=\int_0^{30}\frac{V^2}{R}dt=\frac{V^2}{R_0}\int_0^{30}\frac{dt}{1+\alpha\frac{200}{30}t}=\frac{V^2}{R_0}\cdot\frac{30}{200\alpha}\ln1.2 \).
3. \( W=400\times150\ln1.2=6\times10^4\ln1.2\approx1.1\times10^4\ \text{J} \)

Teacher's Note:
1. Resistance changes with time, so integrate the power V²/R over the 30 s interval.

Question 13: When a 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10⁻⁴ m/s. If the electron density in the wire is 8 × 10²⁸ m⁻³, the resistivity of the material is close to:
(a) \( 1.6\times10^{-6}\ \Omega\text{m} \)
(b) \( 1.6\times10^{-5}\ \Omega\text{m} \)
(c) \( 1.6\times10^{-8}\ \Omega\text{m} \)
(d) \( 1.6\times10^{-7}\ \Omega\text{m} \)
Show Answer & Explanation

Answer: (b) \( 1.6\times10^{-5}\ \Omega\text{m} \)

Explanation:
1. Field \( E=\frac Vl=\frac{5}{0.1}=50\ \text{V/m} \).
2. \( \rho=\frac{E}{J}=\frac{E}{nev_d}=\frac{50}{(8\times10^{28})(1.6\times10^{-19})(2.5\times10^{-4})}\approx1.6\times10^{-5}\ \Omega\text{m} \)

Teacher's Note:
1. Resistivity is E/J, and J = n e vd.

Question 14: The resistance of a wire is 5 Ω at 50°C and 6 Ω at 100°C. The resistance of the wire at 0°C will be:
(a) 2 Ω
(b) 1 Ω
(c) 4 Ω
(d) 3 Ω
Show Answer & Explanation

Answer: (c) 4 Ω

Explanation:
1. Resistance varies linearly with temperature: slope \( \frac{6-5}{100-50}=0.02\ \Omega/^\circ\text{C} \).
2. \( R_0=5-0.02\times50=4\ \Omega \)

Teacher's Note:
1. A linear R-T relation can be extrapolated back using the slope.

Question 15: Two cylindrical conductors of diameters 2d and d, lengths L and 2L, and different materials of resistivity ρ1 and ρ2 are connected in series with a battery. If the heat generated per unit time in the two conductors is equal, then ρ1/ρ2 is:
(a) 2
(b) 4
(c) 8
(d) 16
Show Answer & Explanation

Answer: (c) 8

Explanation:
1. Same current flows, so equal heating means equal resistances.
2. \( \frac{\rho_1L}{\pi d^2}=\frac{\rho_2(2L)}{\pi d^2/4}=\frac{8\rho_2L}{\pi d^2} \), so \( \frac{\rho_1}{\rho_2}=8 \)

Teacher's Note:
1. In series, P = I²R, so equal power means equal resistance.

Question 16: A battery of unknown emf and unknown internal resistance is connected in series with a variable resistance R. An ideal voltmeter is connected across the terminals of the battery. For R = 3 Ω the voltmeter reads 8 V, and for R = 6 Ω it reads 9.6 V. The emf of the battery is:
(a) 10 V
(b) 12 V
(c) 14 V
(d) 16 V
Show Answer & Explanation

Answer: (b) 12 V

Explanation:
1. \( V=\frac{\varepsilon R}{R+r}:\quad 8=\frac{3\varepsilon}{3+r},\quad 9.6=\frac{6\varepsilon}{6+r} \)
2. Equate the expressions for ε: \( \frac{8(3+r)}{3}=1.6(6+r)\Rightarrow r=1.5\ \Omega \).
3. \( \varepsilon=\frac{8(4.5)}{3}=12\ \text{V} \)

Teacher's Note:
1. Terminal voltage equals the emf times R/(R + r); two readings give two equations.

Question 17: A galvanometer has a range of 0-10 mA and internal resistance 15 Ω. To convert it into a voltmeter of range 0-3 V, a resistance R must be connected in series with it. The value of R is:
(a) 135 Ω
(b) 285 Ω
(c) 300 Ω
(d) 315 Ω
Show Answer & Explanation

Answer: (b) 285 Ω

Explanation:
1. \( V=I_g(G+R)\Rightarrow 3=0.01(15+R) \)
2. \( 15+R=300\Rightarrow R=285\ \Omega \)

Teacher's Note:
1. A voltmeter needs a large resistance in series, not in parallel.

Question 18: A wire of length l = 2 m is made of a material whose charge carrier density is n = 8 × 10²⁸ m⁻³ and resistivity is ρ = 1.5 × 10⁻⁸ Ω m. If a potential difference of 96 V is applied across the ends of the wire, the drift velocity of the charge carriers is (e = 1.6 × 10⁻¹⁹ C):
(a) 25 mm/s
(b) 250 mm/s
(c) 2.5 mm/s
(d) 25 cm/s
Show Answer & Explanation

Answer: (b) 250 mm/s

Explanation:
1. Field \( E=\frac{96}{2}=48\ \text{V/m} \).
2. \( v_d=\frac{E}{ne\rho}=\frac{48}{(8\times10^{28})(1.6\times10^{-19})(1.5\times10^{-8})}=0.25\ \text{m/s} \) = 250 mm/s.

Teacher's Note:
1. From J = σE = n e vd, the drift velocity is E/(neρ).

Question 19: The resistance of a wire is R1 = 20.00 Ω at 300 K and R2 = 20.07 Ω at 310 K. The temperature coefficient of resistance of the wire is α = ______ × 10⁻³ per K:
(a) 0.035
(b) 0.35
(c) 3.5
(d) 35
Show Answer & Explanation

Answer: (b) 0.35

Explanation:
1. \( \alpha=\frac{R_2-R_1}{R_1\,\Delta T}=\frac{0.07}{20\times10}=3.5\times10^{-4}=0.35\times10^{-3}\ \text{K}^{-1} \)

Teacher's Note:
1. Use the definition of α with the lower temperature as reference.

Question 20: Two resistances R1 and R2 are connected in series with a battery. The time taken for the resistances to produce a total heat H is t1. When they are connected in parallel to the same battery, the time taken to produce the same total heat H is t2. If t2/t1 = 12/169 and R2 is greater than R1, the value of R2/R1 is:
(a) 1/12
(b) 12
(c) 13
(d) 169/12
Show Answer & Explanation

Answer: (b) 12

Explanation:
1. Series: \( t_1=\frac{H(R_1+R_2)}{V^2} \); parallel: \( t_2=\frac{HR_1R_2}{V^2(R_1+R_2)} \).
2. Let \( x=R_2/R_1 \): \( \frac{t_2}{t_1}=\frac{x}{(1+x)^2}=\frac{12}{169}\Rightarrow12x^2-145x+12=0 \).
3. Roots: x = 12 or 1/12; with R2 greater than R1, x = 12.

Teacher's Note:
1. Heat equals power times time, with power V²/R for each arrangement.

Question 21: Two cylindrical conductors A and B of the same diameter, lengths L and 2L, and resistivities ρ and 5ρ/4 respectively are connected in series with a battery of emf 28 V. If L = 25 cm, the magnitude of the electric field in conductor A is:
(a) 8 N/C
(b) 16 N/C
(c) 32 N/C
(d) 64 N/C
Show Answer & Explanation

Answer: (c) 32 N/C

Explanation:
1. With area A: \( R_A=\frac{\rho L}{A},\ R_B=\frac{5\rho}{4}\cdot\frac{2L}{A}=\frac{2.5\rho L}{A} \).
2. Current \( I=\frac{28A}{3.5\rho L}=\frac{8A}{\rho L} \).
3. \( E_A=\rho J=\rho\frac IA=\frac8L=\frac{8}{0.25}=32\ \text{N/C} \)

Teacher's Note:
1. In series, the same current flows, and E = ρJ in each conductor.

Question 22: In a potentiometer experiment a driver cell of emf 40 V in series with a 2 Ω resistor is connected across the ends of a uniform wire AB of length 1.2 m. A cell of emf 32 V in series with a 1 Ω resistor and a galvanometer is connected between A and the jockey P, opposing the driver cell. The galvanometer shows zero current when AP = 1.0 m. The resistance per unit length of the wire AB is:
(a) 20 Ω/m
(b) 30 Ω/m
(c) 40 Ω/m
(d) 50 Ω/m
Show Answer & Explanation

Answer: (c) 40 Ω/m

Explanation:
1. Let the wire resistance be r per metre, so \( R_{AB}=1.2r \) and the driver current is \( I=\frac{40}{2+1.2r} \).
2. At balance \( 32=I\cdot(1.0\,r)=\frac{40r}{2+1.2r} \).
3. \( 64+38.4r=40r\Rightarrow r=40\ \Omega/\text{m} \)

Teacher's Note:
1. At balance the potentiometer wire drop across AP equals the emf of the test cell, since no current flows in that branch.

Question 23: A bulb A has voltage rating 220 V and power rating 40 W, and a bulb B has voltage rating 220 V and power rating 120 W. If the two bulbs are connected in series with a DC source of 220 V, the potential difference across bulb B is:
(a) 55 V
(b) 110 V
(c) 165 V
(d) 220 V
Show Answer & Explanation

Answer: (a) 55 V

Explanation:
1. \( R=\frac{V^2}{P}:\ R_A=1210\ \Omega,\ R_B=403.3\ \Omega \)
2. \( V_B=220\times\frac{403.3}{1210+403.3}=55\ \text{V} \)

Teacher's Note:
1. In series the voltage divides in proportion to resistance, and the lower-power bulb has the larger resistance.

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Current Electricity Objective Questions & Solutions for JEE Physics

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