NEET (UG) · Physics, Class 12
Current Electricity: NEET PYQ, 2018–2026
Current Electricity (Physics, Class 12) has been asked 24 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 3.4 questions a paper, 7% of all Physics questions. It is the 3rd most-asked of 20 Physics chapters. NEET 2026 had 1 from it. Last checked 28 October 2026.
Questions per paper
Year by year.
| Year | Questions |
|---|---|
| 2018 | 3 |
| 2020 | 4 |
| 2021 | 4 |
| 2023 | 5 |
| 2024 | 4 |
| 2025 | 3 |
| 2026 | 1 |
From 2021 to 2024 each paper had 200 questions (a Section B with a choice); the others had 180.
Inside the chapter
Topics asked most.
- Resistivity6
- Ohm's law5
- Cells, emf and internal resistance4
- Kirchhoff's rules4
- Drift velocity and mobility2
- Wheatstone bridge2
- Metre bridge1
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10 Current Electricity questions to practise.
Our own questions, written in the NEET 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
An electric heater is rated 1000 W at 220 V. What is its resistance while working?
Answer: (B)
P = V²/R, so R = V²/P = 220² / 1000 = 48400 / 1000 = 48.4 Ω.
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
Two cells, one of emf 6 V and internal resistance 2 Ω and the other of emf 3 V and internal resistance 1 Ω, are joined in parallel (positive to positive) and connected across an external resistor of 2 Ω. What is the current in the external resistor?
Answer: (D)
Let V be the potential difference across the external resistor. Junction rule at the common positive terminal: (6 − V)/2 + (3 − V)/1 = V/2. So 3 − V/2 + 3 − V = V/2, giving 2V = 6 and V = 3 V. The current in the 2 Ω resistor is 3/2 = 1.5 A. (Here the 3 V cell happens to carry no current, since its emf equals V.)
Q10 · 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.
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All 24 real NEET 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 NEET.
How many questions come from Current Electricity in NEET?
Current Electricity (Physics, Class 12) has been asked 24 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 3.4 questions a paper, 7% of all Physics questions. It is the 3rd most-asked of 20 Physics chapters. NEET 2026 had 1 from it.
Which topics of Current Electricity are asked most in NEET?
Resistivity (6), Ohm's law (5), Cells, emf and internal resistance (4), Kirchhoff's rules (4), counted over the papers from 2018–2026.
Is Current Electricity important for NEET 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 3 of 20 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Current Electricity NEET previous year questions?
On Nikaalle, free: sign up and practise every past NEET 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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