JEE (Main) · Physics, Class 12
Current Electricity: JEE Main PYQ, 2020–2025
Current Electricity (Physics, Class 12) has been asked 186 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.7 questions a paper, 6% of all Physics questions. It is the 7th most-asked of 20 Physics chapters. The 2025 shifts had 14 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 16in 14 papers |
| 2021 | 45in 26 papers |
| 2022 | 49in 24 papers |
| 2023 | 23in 12 papers |
| 2024 | 39in 18 papers |
| 2025 | 14in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- Ohm's law50
- Resistivity39
- Cells, emf and internal resistance37
- Kirchhoff's rules27
- Wheatstone bridge12
- Drift velocity and mobility11
- Metre bridge10
Try it
10 Current Electricity questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Current Electricity
A metallic wire carries a current of 25 mA when a potential difference of 5 V is applied across its ends. What is its resistance, in ohms?
Answer: (B)
Ohm's law in the form R = V/I gives the resistance directly. The main trap is units: the current is given in milliamperes, so it must be converted to amperes (25 mA = 0.025 A) before dividing. Then R = 5/0.025 = 200 Ω.
Q2 · 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.
Q3 · 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.
Q4 · Physics · Current Electricity
What is the SI unit of electrical resistance expressed in terms of SI base units (kg, m, s, A)?
Answer: (A)
Start from Ohm's law, R = V/I, so the unit of R is volt per ampere. A volt is a joule per coulomb. A joule is kg m² s⁻², a coulomb is A s, so a volt is kg m² s⁻³ A⁻¹. Dividing by the ampere gives kg m² s⁻³ A⁻².
Q5 · 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.
Q6 · 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.
Q7 · Physics · Current Electricity
Two cells, one of emf 3 V and one of emf 2 V, are connected in series so that their emfs oppose each other. What is the net emf of the combination?
Answer: (B)
In a series combination, the emfs add when the cells are arranged with their polarities in the same sense around the circuit. When one cell is reversed, its emf subtracts. So the net emf is 3 V − 2 V = 1 V, directed as the 3 V cell drives it. The same reasoning applies to internal resistances, which always add in series.
Q8 · 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.
Q9 · 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.
Q10 · Physics · Current Electricity
A metallic conductor of resistivity 4 × 10⁻⁸ Ω m carries a uniform electric field of 2 V/m inside it. What is the magnitude of the current density in the conductor, in A/m²?
Answer: (A)
Ohm's law can be written in terms of current density J and field E as J = σE, where σ = 1/ρ is the conductivity. Equivalently J = E/ρ. Substituting E = 2 V/m and ρ = 4 × 10⁻⁸ Ω m gives J = 2 / (4 × 10⁻⁸) = 0.5 × 10⁸ = 5 × 10⁷ A/m². Keep track of the powers of ten carefully.
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All 186 real JEE Main 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 Main.
How many questions come from Current Electricity in JEE Main?
Current Electricity (Physics, Class 12) has been asked 186 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.7 questions a paper, 6% of all Physics questions. It is the 7th most-asked of 20 Physics chapters. The 2025 shifts had 14 from it in all.
Which topics of Current Electricity are asked most in JEE Main?
Ohm's law (50), Resistivity (39), Cells, emf and internal resistance (37), Kirchhoff's rules (27), counted over the papers from 2020–2025.
Is Current Electricity important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 7 of 20 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Current Electricity JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main 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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