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NEET Current Electricity MCQs with answers

381 questions on Current Electricity for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. A real cell has an internal resistance of 2.0 Ω. When it is connected to an external resistor of 8.0 Ω, the power dissipated in the external resistor is 8.0 W. What is the electromotive force (emf) of the cell?

  1. 10.0 V
  2. 8.0 V
  3. 12.0 V
  4. 16.0 V
Show answer and explanation

Answer: (A) 10.0 V

The power dissipated in the load resistor R is P = I²R. Thus, 8.0 = I²(8.0), giving I = 1.0 A. Using the circuit equation E = I(R + r), we get E = 1.0 × (8.0 + 2.0) = 10.0 V.

medium

2. A cell of emf 12 V and internal resistance 0.5 Ω is connected to an external 2 Ω resistor. The terminal potential difference is:

  1. 2.4 V
  2. 9.6 V
  3. 12 V
  4. 6 V
Show answer and explanation

Answer: (B) 9.6 V

I = E/(R + r) = 12/2.5 A. V = IR = 12 × 2/2.5 = 9.6 V.

hard

3. A copper wire has a circular cross-section whose radius tapers linearly from r₁ = 1.0 mm at one end to r₂ = 2.0 mm at the other end over a length L = 10 cm. When a constant current flows through it, the ratio of the drift speed of electrons at the narrower end to that at the wider end is:

  1. 1 : 2
  2. 2 : 1
  3. 4 : 1
  4. 1 : 4
Show answer and explanation

Answer: (C) 4 : 1

From I = n e A v_d, drift speed is inversely proportional to cross-sectional area: v_d ∝ 1/A ∝ 1/r². Thus, v_d1 / v_d2 = (r₂ / r₁)² = (2.0 / 1.0)² = 4 : 1.

medium

4. A mixed combination of 48 identical cells, each having an internal resistance of 3.0 Ω, is used to supply maximum current to an external resistor of 4.0 Ω. If there are n cells in series in each branch and m parallel branches, the values of n and m are respectively:

  1. n = 12, m = 4
  2. n = 6, m = 8
  3. n = 8, m = 6
  4. n = 16, m = 3
Show answer and explanation

Answer: (C) n = 8, m = 6

For maximum current in mixed grouping, the total internal resistance must equal the external resistance: nr / m = R. Given r = 3 Ω, R = 4 Ω, and nm = 48. Substituting m = 48/n gives n(3)/(48/n) = 4, so n²(3/48) = 4, which means n²/16 = 4, so n² = 64 and n = 8. Then m = 48/8 = 6.

easy

5. Which of the following statements is INCORRECT regarding Ohm's law and electrical conduction in materials?

  1. The microscopic form of Ohm's law is represented as J = σE, where σ is electrical conductivity.
  2. Ohm's law is a universal law obeyed by all conducting materials and devices under all conditions.
  3. In a semiconductor diode, the relation between current and voltage is non-linear, showing a failure of Ohm's law.
  4. For metallic conductors, Ohm's law holds well as long as the temperature and other physical conditions remain constant.
Show answer and explanation

Answer: (B) Ohm's law is a universal law obeyed by all conducting materials and devices under all conditions.

Ohm's law is not a fundamental or universal law of nature; it is an empirical relation satisfied by many materials (mostly metals) over limited ranges of electric field. Materials like semiconductors, electrolytes, and vacuum tubes do not obey Ohm's law.

hard

6. For a vacuum tube diode, the plate current I_p is related to the plate voltage V_p by Child's Law: I_p = k V_p^(3/2), where k is a constant. The dynamic resistance r_d of this diode relates to its static resistance R_s (V_p / I_p) as:

  1. r_d = (3/2) R_s
  2. r_d = (2/3) R_s
  3. r_d = (4/9) R_s
  4. r_d = R_s
Show answer and explanation

Answer: (B) r_d = (2/3) R_s

Static resistance is R_s = V_p / I_p = V_p / (k V_p^(3/2)) = 1 / (k V_p^(1/2)). Dynamic conductance is dI_p/dV_p = (3/2) k V_p^(1/2). Thus dynamic resistance r_d = dV_p/dI_p = 2 / (3 k V_p^(1/2)) = (2/3) R_s.

easy

7. Kirchhoff's loop rule, Σ ΔV = 0, directly reflects which fundamental law of physics?

  1. Conservation of electric charge
  2. Conservation of momentum
  3. Conservation of energy
  4. Conservation of angular momentum
Show answer and explanation

Answer: (C) Conservation of energy

Kirchhoff's loop rule states that the algebraic sum of potential differences around any closed loop is zero. Since electric potential difference is work done per unit charge in an electrostatic field (a conservative field), the net energy gained or lost by a charge completing a closed round trip must be zero, expressing the conservation of energy.

medium

8. A wire of resistance 9 Ω is stretched uniformly to 2 times its original length (volume unchanged). Its new resistance is:

  1. 36 Ω
  2. 4.5 Ω
  3. 18 Ω
  4. 72 Ω
Show answer and explanation

Answer: (A) 36 Ω

With volume fixed, A becomes A/2, so R ∝ L/A grows by 2² = 4. New R = 36 Ω.

medium

9. Consider the following two statements regarding electrical conduction in metals: Statement I: The path of an electron between two successive collisions in the presence of an electric field is a curved line. Statement II: In the absence of an external electric field, the average velocity of conduction electrons in a metal at room temperature is zero. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

In the absence of an electric field, thermal velocities are randomly oriented, so their vector average is zero. In the presence of an electric field, an acceleration is superimposed on thermal motion, bending the trajectory between collisions into a parabolic curve.

medium

10. Resistors of 6 Ω and 11 Ω are connected in parallel, and this combination is connected in series with a 7 Ω resistor. The equivalent resistance is:

  1. 24 Ω
  2. 3.88 Ω
  3. 10.9 Ω
  4. 4.96 Ω
Show answer and explanation

Answer: (C) 10.9 Ω

Parallel part = 6×11/17 = 3.88 Ω. Adding 7 Ω in series gives 10.9 Ω.

medium

11. Given below are two statements: Statement I: When two identical cells are connected in series aiding each other, the maximum current delivered to an external resistor of negligible resistance is double the short-circuit current of a single cell. Statement II: The maximum power transfer from a battery of fixed emf and internal resistance to an external load occurs when the load resistance equals the internal resistance of the battery. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are true.
  2. Both Statement I and Statement II are false.
  3. Statement I is true but Statement II is false.
  4. Statement I is false but Statement II is true.
Show answer and explanation

Answer: (D) Statement I is false but Statement II is true.

For Statement I: The short-circuit current of a single cell is I₁ = E/r. For two identical cells in series, total emf is 2E and total internal resistance is 2r, so the short-circuit current is I₂ = 2E/(2r) = E/r, which is equal to (not double) that of a single cell; hence Statement I is false. Statement II is the standard Maximum Power Transfer Theorem, which is true.

medium

12. A cell of emf E and internal resistance r is connected across an external variable resistor R. As the resistance R is gradually increased from a very small value towards infinity, how do the terminal potential difference V across the cell and the current I in the circuit change?

  1. V decreases towards zero, while I increases towards E/r
  2. V increases towards E, while I decreases towards zero
  3. Both V and I decrease continuously towards zero
  4. V remains constant at E, while I decreases towards zero
Show answer and explanation

Answer: (B) V increases towards E, while I decreases towards zero

The circuit current is I = E/(R + r), which decreases monotonically towards zero as R → ∞. The terminal potential difference is V = IR = E·R/(R + r) = E/(1 + r/R), which increases monotonically from 0 towards E as R increases towards infinity.

medium

13. In a closed circuit loop, a student traverses clockwise through three components: across a resistor of 4 Ω in the direction of a 2 A current, across an ideal battery from its positive to negative terminal of emf 10 V, and across another resistor R in the direction of the same 2 A current back to the starting point. If the total change in potential around the loop is zero, the resistance R is:

  1. 1 Ω
  2. 5 Ω
  3. This situation is physically impossible with positive R
  4. 9 Ω
Show answer and explanation

Answer: (C) This situation is physically impossible with positive R

Traversing in the direction of current gives potential drop -I·R, and traversing from positive to negative terminal of a battery gives -E. Thus, - (2 × 4) - 10 - (2 × R) = 0, leading to -18 - 2R = 0, or R = -9 Ω. Because a passive resistor cannot have a negative resistance, this loop cannot exist with the assumed directions and values.

medium

14. Given below are two statements: Statement I: The temperature coefficient of resistivity is positive for metals and negative for semiconductors. Statement II: In semiconductors, with an increase in temperature, the rapid decrease in relaxation time outweighs the increase in the number density of charge carriers.

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because metals show increasing resistivity with temperature (positive α), whereas semiconductors show decreasing resistivity with temperature (negative α). Statement II is incorrect because in semiconductors, the exponential increase in the number density (n) of charge carriers with temperature dominates over any decrease in relaxation time (τ), causing resistivity ρ = m/(n e² τ) to decrease.

medium

15. A cell of emf 9 V and internal resistance 1 Ω is connected to an external 6 Ω resistor. The terminal potential difference is:

  1. 1.29 V
  2. 7.71 V
  3. 9 V
  4. 1.5 V
Show answer and explanation

Answer: (B) 7.71 V

I = E/(R + r) = 9/7 A. V = IR = 9 × 6/7 = 7.71 V.

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