JEE (Advanced) · Physics, Class 12
Current Electricity: JEE Advanced PYQ, 2007–2026
Current Electricity (Physics, Class 12) has been asked 32 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.8 questions a paper, 4% of all Physics questions. It is the 11th most-asked of 19 Physics chapters. The two 2026 papers had 1 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 1in 2 papers |
| 2008 | 2in 2 papers |
| 2009 | 1in 2 papers |
| 2010 | 5in 2 papers |
| 2011 | 2in 2 papers |
| 2012 | 1in 2 papers |
| 2013 | 1in 2 papers |
| 2014 | 3in 2 papers |
| 2015 | 2in 2 papers |
| 2016 | 3in 2 papers |
| 2017 | 0in 2 papers |
| 2018 | 0in 2 papers |
| 2019 | 1in 2 papers |
| 2020 | 2in 2 papers |
| 2021 | 3in 2 papers |
| 2022 | 3in 2 papers |
| 2023 | 1in 2 papers |
| 2024 | 0in 2 papers |
| 2025 | 0in 2 papers |
| 2026 | 1in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Kirchhoff's rules8
- Resistivity7
- Ohm's law6
- Metre bridge5
- Wheatstone bridge4
- Cells, emf and internal resistance2
Try it
10 Current Electricity questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Current Electricity
A copper resistor of 10 Ω is connected in series with a carbon resistor so that the combined resistance stays practically constant over a small range of temperature. The temperature coefficients of resistivity are +4.0 × 10⁻³ °C⁻¹ for copper and −5.0 × 10⁻⁴ °C⁻¹ for carbon, and each is referred to the common starting temperature. What is the resistance of the carbon resistor at that temperature?
Answer: (D)
For small temperature changes, the resistance change of each resistor is ΔR = R·α·ΔT. In series the changes add. For the total to stay constant, R_Cu·α_Cu·ΔT + R_C·α_C·ΔT = 0. This gives R_C = −R_Cu·α_Cu/α_C = 10 × (4.0 × 10⁻³)/(5.0 × 10⁻⁴) = 80 Ω. The carbon resistor has a negative α, so its resistance falls by the same amount as the copper's rises.
Q2 · Physics · Current Electricity
A metal wire carries a constant current I. Its temperature is raised so that its resistivity increases by 25%. Neglect any change in its dimensions and in the number density of free electrons. Which of the following describes correctly what happens to the drift speed v_d, the mobility μ and the electric field E inside the wire?
Answer: (A)
For constant current, v_d = I/(nAe) depends only on I, n and A, so it is unchanged. Resistivity is ρ = m/(ne²τ) = 1/(neμ). With n fixed, ρ is inversely proportional to μ, so μ becomes 1/1.25 = 0.8 times its original value. The field is E = ρJ, and J is constant, so E rises by 25%. This is consistent with v_d = μE, since 0.8 × 1.25 = 1.
Q3 · Physics · Current Electricity
Two cells, each of emf 2 V, have internal resistances of 1.5 Ω and 0.5 Ω. They are connected in series, with their emfs aiding, across an external resistor R. For what value of R is the potential difference across the terminals of the 1.5 Ω cell exactly zero?
Answer: (A)
The current in the series circuit is I = (2 + 2)/(R + 1.5 + 0.5) = 4/(R + 2). The terminal voltage of the first cell is V₁ = ε − Ir₁. For V₁ = 0 we need I = ε/r₁ = 2/1.5 = 4/3 A. Then R + 2 = 4/(4/3) = 3, so R = 1.0 Ω. In general, R = r₁ − r₂ for equal emfs.
Q4 · Physics · Current Electricity
The power delivered by a cell of emf 10 V and some fixed internal resistance to a variable external resistor is found to be exactly the same when the external resistance is set to 2 Ω and to 8 Ω. What is the maximum power that this cell can deliver to any external resistor?
Answer: (B)
Power in the load is P = E²R/(R+r)². Setting P(R1) = P(R2) gives R1(R2+r)² = R2(R1+r)². Expanding and simplifying gives r² = R1R2, so r = √(2×8) = 4 Ω. Check: at R = 2 Ω, I = 10/6 A and P = (100/36)×2 = 50/9 W. At R = 8 Ω, I = 10/12 A and P = (100/144)×8 = 50/9 W. The maximum power transfer theorem states that P is maximum when R = r, so Pmax = E²/(4r) = 100/16 = 6.25 W.
Q5 · Physics · Current Electricity
In a metre bridge, a 5 Ω resistor in the right gap balances an unknown resistance X in the left gap at 40 cm from the left end. A resistor P is now connected across the 5 Ω resistor so that the null point shifts to 50 cm. The value of P is:
Answer: (A)
First balance: X/5 = 40/60, so X = 10/3 Ω. At the 50 cm null point the two gap resistances are equal, so the right-gap resistance must be 10/3 Ω. A parallel resistor lowers the resistance below 5 Ω, which is consistent with this. Solve 5P/(5+P) = 10/3: 15P = 50 + 10P, so P = 10 Ω.
Q6 · Physics · Current Electricity
In a metre bridge, a uniform wire X is connected in the left gap and a resistor S, equal in resistance to X, in the right gap, so the null point is at 50.0 cm. The wire X is then drawn out uniformly to twice its original length, with its volume and resistivity unchanged, and reconnected in the left gap with S unchanged. The new null point, measured from the left end, is at:
Answer: (A)
R = ρL/A, and the volume AL is fixed. If L becomes 2L, then A becomes A/2, so R' = ρ(2L)/(A/2) = 4R. At balance, X'/S = l/(100 − l), so l/(100 − l) = 4 and l = 400/5 = 80 cm. The null point moves towards the end that has the smaller resistance in its gap, which is the right end here.
Q7 · Physics · Current Electricity
Six identical cells, each of emf 2 V and internal resistance 0.5 Ω, are joined in series, but one cell is connected with its terminals reversed. The combination is connected across an external resistor of 5 Ω. What current flows in the resistor?
Answer: (A)
In series the emfs add algebraically, and a reversed cell contributes −ε. Net emf = 5ε − ε = 4ε = 8 V. Internal resistances always add, whichever way the cell faces, so r_total = 6 × 0.5 = 3 Ω. Then I = 8/(5 + 3) = 1.00 A.
Q8 · Physics · Current Electricity
Two straight wires made of the same metal at the same temperature are connected in parallel across an ideal battery. Wire 1 has length L and radius r. Wire 2 has length 2L and radius 2r. If v₁ and v₂ are the drift speeds of the free electrons in wire 1 and wire 2 respectively, what is v₁ : v₂?
Answer: (C)
Drift speed is v = μE, where μ = eτ/m depends only on the material and temperature. Both wires are connected directly across the battery, so each has potential difference V. The field in each is E = V/L, so v = μV/L and the radius does not matter. Hence v₁/v₂ = L₂/L₁ = 2L/L = 2. Check with currents: I = nAev gives I₁/I₂ = (1/4)(2) = 1/2, which is consistent with R ∝ L/A, since R₁/R₂ = 2/4 = 1/2 and so I₁/I₂ = 2... Note that for a given V, I ∝ A/L = (r²/L), giving I₁/I₂ = (1/2)/(4/2)·... so use the direct result v ∝ 1/L, which does not need the currents.
Q9 · Physics · Current Electricity
A wire of length 50.0 cm and diameter 0.20 mm is placed in the left gap of a metre bridge, and a 4.0 Ω resistor is placed in the right gap. The balance point is 60.0 cm from the left end. Taking π = 3.14, the resistivity of the wire material is nearest to:
Answer: (A)
At balance X/S = l/(100 − l), so X = 4.0 × 60/40 = 6.0 Ω. The area is A = π(d/2)² = 3.14 × (1.0 × 10⁻⁴ m)² = 3.14 × 10⁻⁸ m². Then ρ = XA/L = (6.0 × 3.14 × 10⁻⁸)/0.50 ≈ 3.8 × 10⁻⁷ Ω m.
Q10 · Physics · Current Electricity
A copper wire of length 5 m has resistivity 1.7 × 10⁻⁸ Ω m and free-electron density 8.5 × 10²⁸ m⁻³. A potential difference of 0.20 V is maintained across its ends, so that a uniform field exists in it. Taking e = 1.6 × 10⁻¹⁹ C, what is the drift speed of the conduction electrons?
Answer: (A)
Resistivity is ρ = 1/(neμ), so μ = 1/(ρne). Here ρne = 1.7 × 10⁻⁸ × 8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ ≈ 231 (SI units), so μ ≈ 4.3 × 10⁻³ m² V⁻¹ s⁻¹. The field is E = V/L = 0.20/5 = 0.04 V m⁻¹. Then v_d = μE = 4.3 × 10⁻³ × 0.04 ≈ 1.7 × 10⁻⁴ m s⁻¹. Equivalently, v_d = E/(ρne).
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All 32 real JEE Advanced questions from Current Electricity.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Current Electricity in JEE Advanced.
How many questions come from Current Electricity in JEE Advanced?
Current Electricity (Physics, Class 12) has been asked 32 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.8 questions a paper, 4% of all Physics questions. It is the 11th most-asked of 19 Physics chapters. The two 2026 papers had 1 from it.
Which topics of Current Electricity are asked most in JEE Advanced?
Kirchhoff's rules (8), Resistivity (7), Ohm's law (6), Metre bridge (5), counted over the papers from 2007–2026.
Is Current Electricity important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 11 of 19 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Current Electricity JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced question from Current Electricity with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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