JEE Physics Current Electricity MCQs Set 30

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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.

Chapter-wise Objective Questions: Current Electricity

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: A current of 5 A passes through a copper conductor (resistivity \( 1.7\times10^{-8}\ \Omega \)-m) of radius of cross-section 5 mm. Find the mobility of the charges if their drift velocity is \( 1.1\times10^{-3} \) m/s.
(a) 1.5 m\(^2\)/V-s
(b) 1.0 m\(^2\)/V-s
(c) 1.3 m\(^2\)/V-s
(d) 1.8 m\(^2\)/V-s
Show Answer & Explanation

Answer: (b) 1.0 m\(^2\)/V-s

Explanation:
1. \( E=\rho J=\rho\dfrac I{\pi r^2}=\dfrac{1.7\times10^{-8}\times5}{\pi(5\times10^{-3})^2}=1.08\times10^{-3} \) V/m.
2. Mobility \( \mu=\dfrac{v_d}E=\dfrac{1.1\times10^{-3}}{1.08\times10^{-3}}\approx1.0 \) m\(^2\)/V-s.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 2: A metal wire of resistance 3 \(\Omega\) is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle \( 60^\circ \) at the centre, the equivalent resistance between these two points will be
(a) \( \dfrac72\ \Omega \)
(b) \( \dfrac52\ \Omega \)
(c) \( \dfrac53\ \Omega \)
(d) \( \dfrac{12}5\ \Omega \)
Show Answer & Explanation

Answer: (c) \( \dfrac53\ \Omega \)

Explanation:
1. Doubling the length at constant volume quadruples the resistance: \( R=12\ \Omega \).
2. The \( 60^\circ \) arc has \( 2\ \Omega \) and the rest \( 10\ \Omega \), in parallel: \( \dfrac{2\times10}{12}=\dfrac53\ \Omega \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 3: A wire of uniform resistance 18 \(\Omega\) is bent into an equilateral triangle. The resistance between any two vertices of the triangle is
(a) \( 4\ \Omega \)
(b) \( 12\ \Omega \)
(c) \( 8\ \Omega \)
(d) \( 2\ \Omega \)
Show Answer & Explanation

Answer: (a) \( 4\ \Omega \)

Explanation:
1. Each side is \( 6\ \Omega \). Between two vertices, one side (6) is in parallel with two sides in series (12).
2. \( \dfrac{6\times12}{18}=4\ \Omega \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 4: A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance, if its volume remains unchanged, is
(a) 2.0%
(b) 1.0%
(c) 0.5%
(d) 2.5%
Show Answer & Explanation

Answer: (b) 1.0%

Explanation:
1. At constant volume \( R\propto L^2 \), so \( \dfrac{\Delta R}R=2\dfrac{\Delta L}L \).
2. \( 2\times0.5\%=1.0\% \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 5: The drift speed of electrons, when 1.5 A of current flows in a copper wire of cross-section 5 mm\(^2\), is v. If the electron density in copper is \( 9\times10^{28}\ \text{m}^{-3} \), the value of v in mm/s is close to (charge of electron \( =1.6\times10^{-19} \) C)
(a) 0.2
(b) 2
(c) 3
(d) 0.02
Show Answer & Explanation

Answer: (d) 0.02

Explanation:
1. \( v=\dfrac I{neA}=\dfrac{1.5}{9\times10^{28}\times1.6\times10^{-19}\times5\times10^{-6}} \).
2. \( =2.08\times10^{-5} \) m/s \( \approx0.02 \) mm/s.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 6: When a 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is \( 2.5\times10^{-4}\ \text{m s}^{-1} \). If the electron density in the wire is \( 8\times10^{28}\ \text{m}^{-3} \), the resistivity of the material is close to
(a) \( 1.6\times10^{-5}\ \Omega\text{m} \)
(b) \( 1.6\times10^{-8}\ \Omega\text{m} \)
(c) \( 1.6\times10^{-7}\ \Omega\text{m} \)
(d) \( 1.6\times10^{-6}\ \Omega\text{m} \)
Show Answer & Explanation

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

Explanation:
1. \( E=\dfrac Vl=50 \) V/m and \( J=nev_d \).
2. \( \rho=\dfrac EJ=\dfrac{50}{8\times10^{28}\times1.6\times10^{-19}\times2.5\times10^{-4}}=1.6\times10^{-5}\ \Omega \)m.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 7: A steady current flows in a metallic conductor of non-uniform cross-section. The quantity or quantities constant along the length of the conductor is/are
(a) current, electric field and drift speed
(b) drift speed only
(c) current only
(d) current and drift speed
Show Answer & Explanation

Answer: (c) current only

Explanation:
1. In steady state the same charge crosses every section per second, so the current is constant.
2. \( E=\dfrac{I\rho}A \) and \( v_d=\dfrac I{neA} \) both vary with the cross-sectional area A.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 8: A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of
(a) each of them increases
(b) each of them decreases
(c) copper increases and germanium decreases
(d) copper decreases and germanium increases
Show Answer & Explanation

Answer: (d) copper decreases and germanium increases

Explanation:
1. Copper is a metal: lower temperature means fewer collisions, so its resistance decreases.
2. Germanium is a semiconductor: fewer carriers are thermally excited at low temperature, so its resistance increases.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 9: A copper wire is stretched to make it 0.1% longer. The percentage change in its resistance is
(a) 0.1% increase
(b) 0.2% increase
(c) 0.05% increase
(d) 0.4% increase
Show Answer & Explanation

Answer: (b) 0.2% increase

Explanation:
1. \( R\propto L^2 \) at constant volume.
2. \( \dfrac{\Delta R}R=2\times0.1\%=0.2\% \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 10: In a circuit, a 18 V cell of negligible internal resistance drives a current through \( R_1=400\ \Omega \) in series with a parallel combination of \( R_2 \) and the series pair \( R_3=100\ \Omega,R_4=500\ \Omega \). The reading of an ideal voltmeter across \( R_4 \) is 5 V. The value of \( R_2 \) is
(a) \( 550\ \Omega \)
(b) \( 230\ \Omega \)
(c) \( 300\ \Omega \)
(d) \( 450\ \Omega \)
Show Answer & Explanation

Answer: (c) \( 300\ \Omega \)

Explanation:
1. Current in the \( R_3\text{-}R_4 \) branch \( =\dfrac5{500}=0.01 \) A, so the voltage across the parallel part \( =0.01\times600=6 \) V.
2. The drop across \( R_1 \) is \( 12 \) V, so \( I=\dfrac{12}{400}=0.03 \) A, and \( I_2=0.03-0.01=0.02 \) A. \( R_2=\dfrac6{0.02}=300\ \Omega \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 11: In a circuit two points A and B are joined by 2 \(\Omega\) to a node C. A 20 V source is connected through a 2 \(\Omega\) resistor from A to the node C, and a 10 V source through 4 \(\Omega\) from B to C; the node C is connected to the ground through a 2 \(\Omega\) resistor and a closed switch (both sources have their other terminals at zero potential). The current i through the 2 \(\Omega\) resistor to the ground is
(a) 5 A
(b) 4 A
(c) 3 A
(d) 2 A
Show Answer & Explanation

Answer: (a) 5 A

Explanation:
1. Let the node potential be V: \( \dfrac{20-V}2+\dfrac{10-V}4=\dfrac V2 \).
2. \( 40-2V+10-V=2V\Rightarrow V=10 \) V, so \( i=\dfrac{10}2=5 \) A.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 12: A 2 V battery is connected across one 30 \(\Omega\) resistor in parallel with two 30 \(\Omega\) resistors in series (three equal 30 \(\Omega\) resistors forming a triangle, the battery across two of its vertices). The current i drawn from the battery is
(a) \( \dfrac1{45}\ \text{A} \)
(b) \( \dfrac1{10}\ \text{A} \)
(c) \( \dfrac1{15}\ \text{A} \)
(d) \( \dfrac15\ \text{A} \)
Show Answer & Explanation

Answer: (b) \( \dfrac1{10}\ \text{A} \)

Explanation:
1. One branch of \( 30\ \Omega \) and another of \( 60\ \Omega \) are in parallel: \( R=\dfrac{30\times60}{90}=20\ \Omega \).
2. \( i=\dfrac2{20}=\dfrac1{10} \) A.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 13: An infinite ladder network of resistances is built from 1 \(\Omega\) series resistors and 2 \(\Omega\) shunt resistors (each section has a 1 \(\Omega\) in the top line followed by a 2 \(\Omega\) connecting the two lines). A 6 V battery of negligible internal resistance is connected at the start. The current in the 2 \(\Omega\) resistor nearest the battery is
(a) 3 A
(b) 1 A
(c) 0.75 A
(d) 1.5 A
Show Answer & Explanation

Answer: (d) 1.5 A

Explanation:
1. Effective resistance R of the ladder: \( R=1+\dfrac{2R}{2+R}\Rightarrow R^2-R-2=0\Rightarrow R=2\ \Omega \). Battery current \( =\dfrac62=3 \) A.
2. The first 1 \(\Omega\) drops 3 V, so the first 2 \(\Omega\) has 3 V across it: \( \dfrac32=1.5 \) A.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 14: In a Wheatstone bridge (see figure of the standard bridge with arms P, Q, R and X), resistances P and Q are approximately equal. When \( R=400\ \Omega \) the bridge is balanced. On interchanging P and Q, the value of R for balance is \( 405\ \Omega \). The value of X is close to
(a) \( 402.5\ \Omega \)
(b) \( 404.5\ \Omega \)
(c) \( 401.5\ \Omega \)
(d) \( 403.5\ \Omega \)
Show Answer & Explanation

Answer: (a) \( 402.5\ \Omega \)

Explanation:
1. First balance: \( \dfrac PQ=\dfrac RX=\dfrac{400}X \). After interchange: \( \dfrac QP=\dfrac{405}X \).
2. Multiply: \( 1=\dfrac{400\times405}{X^2}\Rightarrow X=\sqrt{162000}=402.5\ \Omega \).

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 15: A 16 V source drives a series chain made of: two resistors 4R and 4R in parallel, then a resistor R, then two resistors 6R and 12R in parallel, then another resistor R. The power consumed by the network is 4 W. The value of R is
(a) \( 6\ \Omega \)
(b) \( 1\ \Omega \)
(c) \( 8\ \Omega \)
(d) \( 16\ \Omega \)
Show Answer & Explanation

Answer: (c) \( 8\ \Omega \)

Explanation:
1. \( 4R\|4R=2R \) and \( 6R\|12R=4R \). Total \( =2R+R+4R+R=8R \).
2. \( P=\dfrac{V^2}{8R}=\dfrac{256}{8R}=4\Rightarrow R=8\ \Omega \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 16: A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when
(a) \( R=2r \)
(b) \( R=0.001r \)
(c) \( R=1000r \)
(d) \( R=r \)
Show Answer & Explanation

Answer: (d) \( R=r \)

Explanation:
1. \( P=I^2R=\dfrac{E^2R}{(R+r)^2} \).
2. \( \dfrac{dP}{dR}=0\Rightarrow R=r \) (maximum power transfer).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 17: Two equal resistances when connected in series to a battery consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be
(a) 60 W
(b) 240 W
(c) 30 W
(d) 120 W
Show Answer & Explanation

Answer: (b) 240 W

Explanation:
1. \( P=\dfrac{V^2}{R_{eq}} \). Series \( R_{eq}=2R \), parallel \( R_{eq}=\dfrac R2 \): four times smaller resistance.
2. \( P_{par}=4\times60=240 \) W.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 18: A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. The power dissipated when it is connected to an ideal power supply of 11 V is
(a) \( 11\times10^{-4}\ \text{W} \)
(b) \( 11\times10^{5}\ \text{W} \)
(c) \( 11\times10^{-5}\ \text{W} \)
(d) \( 11\times10^{-3}\ \text{W} \)
Show Answer & Explanation

Answer: (c) \( 11\times10^{-5}\ \text{W} \)

Explanation:
1. \( R=\dfrac P{I^2}=\dfrac{4.4}{(2\times10^{-3})^2}=1.1\times10^6\ \Omega \).
2. \( P'=\dfrac{11^2}{1.1\times10^6}=1.1\times10^{-4}=11\times10^{-5} \) W.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 19: A wire of length L and 3 identical cells of negligible internal resistance are connected in series. Due to the current, the temperature of the wire is raised by \( \Delta T \) in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section of length 2L. The temperature of the wire is raised by the same amount \( \Delta T \) in the same time t. The value of N is
(a) 6
(b) 4
(c) 8
(d) 9
Show Answer & Explanation

Answer: (a) 6

Explanation:
1. Case 1: \( H=\dfrac{(3V)^2t}R=ms\Delta T \). Case 2: R and the mass both double, so \( \dfrac{(NV)^2t}{2R}=2ms\Delta T \).
2. \( \dfrac{N^2}{2}=2\times9\Rightarrow N^2=36\Rightarrow N=6 \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 20: A resistor of 5 \(\Omega\) is in parallel with a series pair of 4 \(\Omega\) and 6 \(\Omega\) resistors. The heat produced in the 5 \(\Omega\) resistor is 10 cal/s. The heat generated in the 4 \(\Omega\) resistor is
(a) 1 cal/s
(b) 2 cal/s
(c) 3 cal/s
(d) 4 cal/s
Show Answer & Explanation

Answer: (b) 2 cal/s

Explanation:
1. The voltage across the parallel combination satisfies \( \dfrac{V^2}5=10\Rightarrow V^2=50 \).
2. The series branch carries \( I=\dfrac V{10} \): heat in 4 \(\Omega\) \( =\dfrac{V^2}{100}\times4=2 \) cal/s.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 21: An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb delivers 500 W, is
(a) \( 10\ \Omega \)
(b) \( 5\ \Omega \)
(c) \( 20\ \Omega \)
(d) \( 40\ \Omega \)
Show Answer & Explanation

Answer: (c) \( 20\ \Omega \)

Explanation:
1. Bulb: \( I=\dfrac{500}{100}=5 \) A and \( R_b=20\ \Omega \).
2. Total resistance \( =\dfrac{200}5=40\ \Omega \), so \( R=40-20=20\ \Omega \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

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