Physics · Current Electricity · PYQ set

Current Electricity · JEE PYQ Practice with Solutions

Twenty-six Current Electricity questions built on the shapes JEE actually repeats, sixteen at Main level and ten at Advanced, every one with a full worked solution and an answer key. Free to read or download.

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In short

Most JEE questions on Current Electricity come back to a handful of shapes: a stretched wire changing resistance, a meter bridge balance point, a potentiometer comparison, internal resistance under load, and combinations of cells. This set works through twenty-six of them at Main and Advanced level, with the full solution shown each time.

Contents

·How This Set Is Built
·JEE Main
·JEE Advanced
1Answer Key
2Solutions: JEE Main Tier
3Solutions: JEE Advanced Tier

How This Set Is Built

Tier 1, JEE Main: 13 single-correct plus 3 numerical answer questions.
Tier 2, JEE Advanced: 6 single-correct, 2 multiple-correct and 2 numerical.
Marking: Main +4 and −1. Advanced single-correct +3 and −1, multiple-correct +4 with partial credit and −2 for any wrong tick, numerical +3 with no negative.
Every assumption is written into the question, as a real Advanced paper does.

Think it through

What is in the JEE Main syllabus, and what is not

NTA removed the potentiometer, its principle and its applications, in the 2024 syllabus revision. It is still out for 2025 and 2026, so there are no potentiometer questions in this set.
Resistor colour codes were removed in the same revision.
The Wheatstone bridge and metre bridge remain, and so do Kirchhoff's laws, temperature coefficient and galvanometer conversion. The Main tier is weighted to those.
JEE Advanced is set by the IITs, not NTA, and its syllabus was not reduced in 2024. Check the current brochure before dropping anything for Advanced.

Trap alert

One honest note on the word PYQ

These are PYQ-pattern questions, matched to real JEE style, scope and difficulty.
They are not verbatim year-tagged papers reproduced from memory.
Every number here was computed and checked before printing, which is not true of most question sets you will find online.

TIER 1

JEE Main

Single-correct and numerical · Q1 to Q16

A galvanometer of resistance 99 Ω gives a full-scale deflection at 10 mA. To convert it into an ammeter reading up to 1 A, the shunt required is:

(A)1 Ω in parallel

(B)1 Ω in series

(D)9.9 Ω in parallel

A galvanometer of resistance 100 Ω gives full-scale deflection at 1 mA. The series resistance needed to convert it into a voltmeter reading up to 10 V is:

(A)10000 Ω

(B)1000 Ω

(D)100 Ω

In a metre bridge, a 10 Ω resistance in the left gap balances against an unknown S at 40 cm from the left end. The value of S is:

(A)6.67 Ω

(B)15 Ω

(D)40 Ω

In a Wheatstone bridge the arms are P = 2 Ω, Q = 3 Ω and R = 4 Ω. The bridge is balanced. The value of S, and the effect of interchanging the cell and the galvanometer, are:

(A)6 Ω, balance is lost

(B)8 Ω, balance is lost

(D)6 Ω, balance is unaffected

The resistance of a metal wire is 10 Ω at 20 °C and 15 Ω at 120 °C. Its temperature coefficient of resistance is:

(A)0.05 /°C

(B)0.005 /°C

(D)0.5 /°C

A cell of EMF 12 V and internal resistance 2 Ω delivers maximum power to an external resistance R. That maximum power is:

(A)72 W

(B)36 W

(D)18 W

A wire of resistance 5 Ω is stretched uniformly until its length becomes three times the original. Its new resistance is:

(A)15 Ω

(B)5/3 Ω

(D)9 Ω

Two bulbs rated 60 W and 100 W, both for 220 V, are connected in series across a 220 V supply. Which glows brighter, and what is the potential difference across the 60 W bulb?

(A)60 W bulb, 137.5 V

(B)100 W bulb, 82.5 V

(D)Both equally, 110 V

A 10 V cell of internal resistance 1 Ω is connected in opposition to a 4 V cell of internal resistance 2 Ω, through an external 3 Ω. The current in the circuit is:

(A)1 A

(B)2.33 A

(D)3.5 A

In the circuit of question 9, the terminal potential difference across the 4 V cell is:

(A)2 V

(B)4 V

(D)0 V

Four cells, each of EMF 2 V and internal resistance 0.5 Ω, are joined in series across a 6 Ω resistor. The current is:

(A)0.25 A

(B)1 A

(D)1.33 A

A copper wire of cross-section 2 mm² carries a current of 2 A. If n = 8.5 × 10²⁸ per m³, the drift velocity is about:

(A)1.5 × 10⁵ m/s

(B)7.4 × 10⁻² m/s

(D)7.4 × 10⁻⁵ m/s

A capacitor of 5 μF is charged through a 2 kΩ resistor. The time constant, and the fraction of final charge reached at t = τ, are:

(A)10 ms and 37%

(B)10 ms and 63%

(D)10 s and 63%

Numerical Answer Type

In a metre bridge a known resistance of 4 Ω is in the left gap. The balance point is found at 36.0 cm from the left end. Find the unknown resistance in the right gap, in ohm.

NUMERICAL ANSWER · Round off to TWO decimal places.

A metal wire has a resistance of 20.0 Ω at 0 °C. Its temperature coefficient of resistance is 0.004 /°C. Find its resistance in ohm at 100 °C.

NUMERICAL ANSWER · Round off to TWO decimal places.

A cell of EMF 12 V and internal resistance 2 Ω is connected to a 4 Ω resistor. Find the power dissipated in the external resistor, in watt.

NUMERICAL ANSWER · Round off to the nearest integer.

TIER 2

JEE Advanced

Multi-concept · single-correct, multiple-correct and numerical · Q17 to Q26

A conductor of resistivity ρ is a truncated cone of length L. Its end radii are a and b. Current flows along its axis. Assume the current density is uniform across every cross-section. Its resistance is:

(A)ρL/[π(a+b)²]

(B)ρL/(πab)

(D)ρL/[π(b²−a²)]

A rod of uniform cross-section A and length L has a resistivity that varies along its length as ρ(x) = ρ₀(1 + x/L). Its end-to-end resistance is:

(A)ρ₀L/A

(B)ρ₀L/(2A)

(D)3ρ₀L/(2A)

A charged capacitor discharges through a resistor R. Find the time for the energy stored in the capacitor to fall to half its initial value.

(A)RC ln 2

(B)(RC/2) ln 2

(D)RC ln 4

Two cells, 12 V with 2 Ω and 6 V with 1 Ω, are connected in parallel across a common 3 Ω resistor, with their positive terminals joined. Which is true?

(A)Both cells discharge

(B)The 12 V cell is being charged

(C)No current flows in the 3 Ω resistor

(D)The 6 V cell is being charged

Two rods of the same cross-section are joined end to end and carry the same steady current. Rod 1 has twice the free electron density of rod 2. The ratio of drift velocities v₁/v₂ and of the internal electric fields E₁/E₂ are:

(A)2 and 2

(B)1/2 and 2

(D)2 and 1/2

Multiple Correct Answer Type

An uncharged capacitor C is charged to a final voltage V through a resistor R by a battery of EMF V. Which of the following are correct? One or more options may be right.

(A) The battery supplies a total energy CV²
(B) The heat dissipated in R equals ½CV²
(C) The heat in R depends on the value of R
(D) Exactly half the energy supplied is stored in the capacitor

ONE OR MORE OPTIONS MAY BE CORRECT

A 4 μF capacitor charged to 10 V is connected through a resistor across an uncharged 6 μF capacitor. Which are correct? One or more options may be right.

(A) The final common voltage is 4 V
(B) The charge that flows through the connecting wire is 24 μC
(C) The energy lost is 120 μJ
(D) The energy lost can be made smaller by using a larger resistor

ONE OR MORE OPTIONS MAY BE CORRECT

Numerical Answer Type

A 12 V battery of negligible internal resistance is connected in series with a 2 Ω resistor. This then splits into two parallel branches: a 4 Ω resistor, and a 6 Ω resistor in series with a 5 μF capacitor. In the steady state, find the charge on the capacitor in microcoulomb.

NUMERICAL ANSWER · Round off to the nearest integer.

Two 500 Ω resistors are in series across a 10 V ideal supply. A voltmeter of resistance 1000 Ω is connected across one of them. Find the percentage error in the reading, relative to the true potential difference.

NUMERICAL ANSWER · Round off to the nearest integer.

Twenty-four cells, each of EMF 1.5 V and internal resistance 0.5 Ω, are arranged in m parallel rows of n cells each to drive an external 0.75 Ω. Find the maximum current in ampere that can be obtained.

NUMERICAL ANSWER · Round off to TWO decimal places.

Answer Key

Mark your paper first. Then read every solution, including the ones you got right.

10C

11B

12D

13B

147.11

1528.00

1616

17B

18D

19B

20D

21C

22A, B, D

23A, B, C

2440

2520

266.00

Solutions: JEE Main Tier

Each solution names the idea, shows the working, then names the exact mistake behind each wrong option.

1Answer AMedium · Galvanometer conversion

Core principleAn ammeter must carry a large current. So most of it is sent around the coil, through a small parallel shunt.

MathS = I_gG/(I − I_g) = (0.01)(99)/(1 − 0.01) = 99/99 = 1 Ω in parallel.

Trap(B) is the giveaway error: a series resistance makes a voltmeter, not an ammeter. Using I instead of (I − I_g) in the denominator gives 0.99 Ω, which is close enough to look right.

2Answer CMedium · Galvanometer conversion

Core principleA voltmeter must draw almost no current, so a large resistance is placed in series with the coil.

MathR = V/I_g − G = 10/0.001 − 100 = 10000 − 100 = 9900 Ω.

Trap(A) forgets to subtract the galvanometer's own resistance. It is a small correction here, but examiners set the options exactly 100 Ω apart to catch it.

3Answer BMedium · Metre bridge

Core principleA metre bridge is a Wheatstone bridge whose two lower arms are lengths of uniform wire. So resistance ratios equal length ratios.

MathR/S = l/(100 − l), so 10/S = 40/60, giving S = 10 × 60/40 = 15 Ω.

Trap(A) inverts the ratio, using 10 × 40/60. Always pair the known resistance with the length on its own side of the balance point.

4Answer DMedium · Wheatstone bridge

Core principleAt balance P/Q = R/S. The condition contains only the four arms, so neither the cell nor the galvanometer appears in it.

MathS = QR/P = (3)(4)/2 = 6 Ω. Since the balance condition has no EMF and no galvanometer resistance in it, swapping those two leaves the balance unchanged.

Trap(A) and (D) assume the source position matters. It does not. (C) inverts the ratio to PR/Q. Pair each arm with the one opposite it in the condition.

5Answer BMedium · Temperature coefficient

Core principleUse R_T = R₀[1 + αΔT], with R₀ the resistance at the lower reference temperature.

Math15 = 10[1 + α(100)], so 1.5 = 1 + 100α and α = 0.5/100 = 0.005 /°C.

TrapDividing by 15 instead of 10 gives 0.0033. The reference resistance is the one at the starting temperature, not the final one.

6Answer DMedium · Maximum power transfer

Core principleExternal power peaks when R equals the internal resistance r.

MathAt R = r = 2 Ω, P_max = E²/4r = 144/8 = 18 W.

Trap(B) uses E²/2r, and (A) uses E²/2r doubled. Note the efficiency at maximum power is only 50 percent, since an equal amount is wasted inside the cell.

7Answer CMedium · Resistivity

Core principleStretching conserves the volume. So if the length becomes n times, the area becomes A/n and R becomes n²R.

MathR′ = 3² × 5 = 45 Ω.

Trap(A) scales only the length and is the standard error. The area shrinks at the same time, which is where the second factor of 3 comes from.

8Answer AHard · Power in circuits

Core principleGet each resistance from R = V²/P. In series the current is common, so P = I²R and the larger resistance takes more.

MathR₆₀ = 807 Ω and R₁₀₀ = 484 Ω. So V₆₀ = 220 × 807/1291 = 137.5 V, and the 60 W bulb glows brighter.

Trap(A) and (C) follow the everyday intuition that 100 W is always brighter. That holds only in parallel, which is how houses are wired.

9Answer AMedium · Kirchhoff's laws

Core principleOpposing EMFs subtract, but every resistance in the loop still adds.

MathI = (10 − 4)/(1 + 2 + 3) = 6/6 = 1 A.

Trap(A) adds the EMFs instead of subtracting, giving 14/6. Read whether the cells help or oppose each other before writing the numerator.

10Answer CHard · Cells and terminal voltage

Core principleThe 4 V cell is driven backwards by the stronger cell, so it is being charged.

MathFor a cell being charged, V = E + Ir = 4 + (1)(2) = 6 V.

Trap(A) applies V = E − Ir, which is the discharging formula. A cell being charged always shows a terminal voltage greater than its EMF.

11Answer BEasy · Grouping of cells

Core principleIn series both the EMFs and the internal resistances add.

MathI = nE/(R + nr) = 8/(6 + 2) = 1 A.

TrapUsing r = 0.5 instead of nr = 2 gives 1.23 A. The internal resistances add just as the EMFs do.

12Answer DMedium · Drift velocity

Core principleUse I = nAev_d, converting the area to square metres first.

Mathv_d = 2/[(8.5×10²⁸)(2×10⁻⁶)(1.6×10⁻¹⁹)] = 7.4 × 10⁻⁵ m/s.

Trap(B) uses 2 mm² as 2 × 10⁻³ m². Squared units convert with the square of the factor, so 1 mm² = 10⁻⁶ m².

13Answer BMedium · RC circuits

Core principleThe time constant is τ = RC. Charging follows q = q₀(1 − e^−t/τ).

Mathτ = (2000)(5×10⁻⁶) = 0.01 s = 10 ms. At t = τ, q/q₀ = 1 − 1/e = 63%.

Trap(A) quotes 37%, which is the fraction remaining during discharge. Charging rises to 63%, discharging falls to 37%.

14Answer 7.11Medium · Metre bridge

Core principleResistance ratio equals the ratio of the two wire lengths on either side of the balance point.

MathS = R(100 − l)/l = 4(64)/36 = 7.11 Ω.

TrapInverting to 4(36)/64 gives 2.25 Ω. The unknown is in the right gap, so it pairs with the right-hand length of 64 cm.

15Answer 28.00Medium · Temperature coefficient

Core principleUse R_T = R₀(1 + αΔT), with R₀ measured at the reference temperature of 0 °C.

MathR = 20.0[1 + (0.004)(100)] = 20.0(1.4) = 28.00 Ω.

TrapAdding αΔT as an absolute value rather than a fraction gives 20.4 Ω. The bracket is a multiplying factor, not an addition in ohm.

16Answer 16Medium · Power

Core principleFind the current first, then use P = I²R for the external resistor only.

MathI = 12/(4 + 2) = 2 A. P = I²R = 4 × 4 = 16 W.

TrapUsing P = E²/R = 36 W ignores the internal resistance entirely. The total power from the cell is 24 W, of which 8 W is wasted inside it.

Solutions: JEE Advanced Tier

17Answer BHard · Non-uniform conductor

Core principleThe area changes along the length, so R = ρL/A cannot be used directly. Slice the cone, write dR for one slice, and integrate.

MathRadius at x is r = a + (b−a)x/L. dR = ρdx/πr². Integrating from 0 to L gives ρL/[π(b−a)](1/a − 1/b) = ρL/(πab).

Trap(C) uses the mean radius (a+b)/2 and treats the cone as a uniform cylinder. That is the standard shortcut, and it is wrong: the thin end dominates the resistance, so the true answer is the geometric mean ab, not the arithmetic one.

18Answer DHard · Non-uniform resistivity

Core principleResistance in series adds. So integrate dR = ρ(x)dx/A along the rod.

MathR = (1/A)∫₀^Lρ₀(1 + x/L)dx = (ρ₀/A)(L + L/2) = 3ρ₀L/(2A).

Trap(A) uses the resistivity at x = 0 only. Using the value at the midpoint happens to give the same answer here, but that shortcut fails for any non-linear ρ(x), so integrate.

19Answer BHard · RC transients

Core principleEnergy goes as the square of the charge, so it decays twice as fast as the charge does.

Mathq = q₀e^−t/RC, so U ∝ q² = U₀e^−2t/RC. Setting U = U₀/2 gives 2t/RC = ln 2, so t = (RC/2)ln 2.

Trap(A) is the time for the charge to halve, which is the answer most students give. Read whether the question asks about charge, current or energy: the three give different times.

20Answer DHard · Kirchhoff, multi-loop

Core principleLet the common node be at potential V and write one node equation. A negative branch current means that cell is absorbing energy.

Math(12−V)/2 + (6−V)/1 = V/3 gives V = 6.55 V. Then I₁ = 2.73 A but I₂ = −0.55 A. The node sits above 6 V, so current is pushed into the 6 V cell: it is being charged.

TrapMost students assume every cell in a circuit must discharge. Whenever the node potential exceeds a cell's EMF, that cell absorbs current instead, and its terminal voltage then exceeds its EMF.

21Answer CHard · Drift and current density

Core principleThe current is common, not the field. Use I = nAev_d for the first ratio, then E = ρJ with ρ ∝ 1/n for the second.

MathSame I and A, so v_d ∝ 1/n, giving v₁/v₂ = 1/2. Resistivity falls as n rises, and J is common, so E ∝ ρ ∝ 1/n, giving E₁/E₂ = 1/2.

Trap(C) and (D) come from assuming the field is common across the junction. In a series circuit the current is what is shared. The field jumps at the junction, which is what builds up surface charge there.

22Answer A, B, DHard · Energy in RC charging

Core principleDo not integrate i²R. Use an energy balance over the whole charging process instead.

MathCharge delivered is Q = CV, so the battery does work QV = CV². The capacitor stores ½CV². The difference, ½CV², must appear as heat in R.

Trap(C) is the trap and it is deeply counterintuitive. R sets how fast the charging happens, never how much heat is produced. A tiny R charges quickly with a huge current, and the total heat comes out identical.

23Answer A, B, CHard · Capacitor redistribution

Core principleCharge is conserved during the redistribution. Energy is not, and the loss does not depend on the resistor.

MathQ_total = 40 μC over 10 μF gives V = 4 V. The 6 μF ends with 6 × 4 = 24 μC, which is what flowed. Energy falls from 200 μJ to 80 μJ, a loss of 120 μJ, matching ½C₁C₂(V₁−V₂)²/(C₁+C₂).

Trap(D) is false for the same reason as in the previous question. The loss is fixed by the capacitances and the initial voltages alone. Even a superconducting wire loses the same energy, radiated away instead.

24Answer 40Hard · Steady state with a capacitor

Core principleIn the steady state no current flows into a capacitor. So the branch containing it carries zero current, and the resistor in that branch has no potential drop across it.

MathAll the current goes through the 2 Ω and 4 Ω: I = 12/6 = 2 A. The drop across the 4 Ω is 8 V. With no current in the 6 Ω, the full 8 V sits across the capacitor. Q = 5 × 8 = 40 μC.

TrapIncluding the 6 Ω in the parallel combination gives a different current and a wrong charge. In the steady state that branch is an open circuit, so the 6 Ω may as well not exist.

25Answer 20Hard · Non-ideal measurement

Core principleThe voltmeter is part of the circuit. It loads the resistor it measures, which lowers the very voltage being read.

Math500 ‖ 1000 = 333.3 Ω. I = 10/833.3 = 12 mA, so the reading is 12 × 333.3 = 4.00 V. The true value is 5.00 V, so the error is (5−4)/5 = 20 percent.

TrapAnswering 0 assumes an ideal voltmeter. This loading error is exactly why a good voltmeter is built with a very high resistance: the less current it draws, the less it disturbs what it is measuring.

26Answer 6.00Hard · Mixed grouping of cells

Core principleA row gives EMF nE and resistance nr, and m rows in parallel give nr/m. The current peaks when the external resistance matches the internal one.

MathSet 0.5n/m = 0.75 with mn = 24. This gives n = 1.5m, so 1.5m² = 24 and m = 4 rows of n = 6. Then I = nE/2R = 9/1.5 = 6.00 A.

TrapStacking all 24 in one series row gives only 2.77 A. Adding cells in series raises the EMF and the internal resistance at the same rate, so beyond the matching point you gain nothing.

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