Current Electricity NEET PYQs (2005–2026) with Solutions

10 solved Current Electricity PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasySeries and parallel combinations of resistors; Ideal diode circuits

   The current (I) in the circuit shown below is: (All diodes are ideal and identical)

   1. AOption A: 5/3 A
   2. BOption B: 5/9 A
   3. COption C: 1/3 A
   4. DOption D: 15/2 A

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   Correct answer: (D) 15/2 A

   Explanation

   For ideal diodes: forward resistance = 0 and reverse resistance = $\infty$. Only the branches containing $4\Omega$ and $2\Omega$ conduct. Hence the equivalent circuit has two parallel resistors across $10V$. Current: $$I = \frac{10}{4} + \frac{10}{2} = \frac{5}{2} + 5 = \frac{15}{2}\,A$$ Therefore, the correct answer is $\boxed{\frac{15}{2}\,A}$.

2. Question 2 (NEET 2026)

   NEET 2026EasyInternal resistance, potential difference and emf of a cell

   A resistor is connected to a battery of 12 V emf and internal resistance 2 Ω. If the current in the circuit is 0.6 A, the terminal voltage of the battery is:

   1. AOption A: 10 V
   2. BOption B: 1.2 V
   3. COption C: 12 V
   4. DOption D: 10.8 V

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   Correct answer: (D) 10.8 V

   Explanation

   Terminal voltage of a battery is: V = E - ir Substituting values: V = 12 - (0.6 × 2) V = 12 - 1.2 = 10.8 V Hence, option (D) is correct.

3. Question 3 (NEET 2026)

   NEET 2026MediumWheatstone & Metre Bridge

   A uniform metallic wire having resistance 4 Ω is bent to form a square loop (ABCD). A resistance of 2 Ω is connected between points B and D and a battery of 2 V is connected across points A and C as shown. The value of current I is:

   1. AOption A: 2 A
   2. BOption B: 8 A
   3. COption C: 4.5 A
   4. DOption D: 4 A

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   Correct answer: (A) 2 A

   Explanation

   Total wire resistance = 4 Ω, so each side has 1 Ω. The network forms a balanced Wheatstone bridge, hence no current flows through the 2 Ω diagonal resistor. Equivalent resistance across A and C becomes 1 Ω. Thus current I = V/R = 2/1 = 2 A. Therefore option (A) is correct.

4. Question 4 (NEET 2026)

   NEET 2026EasyElectrical Energy & Power

   A room heater is rated 400 W, 220 V. If the supply voltage drops to 200 V, what will be the power consumed (approximately)?

   1. AOption A: 200 W
   2. BOption B: 400 W
   3. COption C: 331 W
   4. DOption D: 121 W

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   Correct answer: (C) 331 W

   Explanation

   For a heater of constant resistance: P ∝ V². Therefore, P=(200/220)²×400 ≈ 331 W. Hence option (C) is correct.

5. Question 5 (NEET 2026)

   NEET 2026MediumTemperature dependence of resistance

   Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, $+\alpha$, while the other one has a negative temperature of coefficient, $-\alpha$, as shown in the figure. The current through these circuits are denoted by $I_A$ and $I_B$. At initial temperature, the resistance of the two resistors is $R_0$. As the temperature is increased, the correct option that describes the variation of current in these circuits is:

   1. AOption A: $I_A$ remains constant while $I_B$ increases
   2. BOption B: $I_A$ decreases while $I_B$ increases
   3. COption C: $I_A$ increases while $I_B$ decreases
   4. DOption D: Both $I_A$ and $I_B$ remain constant

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   Correct answer: (A) $I_A$ remains constant while $I_B$ increases

   Explanation

   Circuit (A) has the two resistors in series: $R_A = R_0(1 - \alpha\Delta T) + R_0(1 + \alpha\Delta T) = 2R_0$, independent of temperature. So $I_A = \frac{V}{2R_0}$ remains constant. Circuit (B) has them in parallel: $R_B = \frac{R_0^2(1 - \alpha\Delta T)(1 + \alpha\Delta T)}{2R_0} = \frac{R_0}{2}(1 - \alpha^2\Delta T^2)$, which decreases as $\Delta T$ increases. So $I_B = \frac{V}{R_B}$ increases. Hence option (A).

6. Question 6 (NEET 2026)

   NEET 2026MediumWheatstone & Metre Bridge

   In a metre bridge experiment (see figure), the positions of the cell E and galvanometer G are interchanged. We shall observe in the galvanometer:

   1. AOption A: Only the left-sided deflection
   2. BOption B: There will be no deflection irrespective of the position of the jockey
   3. COption C: Only the right-sided deflection
   4. DOption D: Both right-sided and left-sided deflection and at balance point, no deflection

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   Correct answer: (D) Both right-sided and left-sided deflection and at balance point, no deflection

   Explanation

   Interchanging the cell and galvanometer does not affect the balance condition of the metre bridge. At balance point, galvanometer shows no deflection; otherwise, deflection can be either side. Hence option (D) is correct.

7. Question 7 (NEET 2025)

   NEET 2025HardOhm's Law & Resistivity

   The current passing through the battery in the given circuit, is:

   1. AOption A: 2.0 A
   2. BOption B: 0.5 A
   3. COption C: 2.5 A
   4. DOption D: 1.5 A

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   Correct answer: (B) 0.5 A

   Explanation

   Reducing the resistor network using series and parallel combinations gives an equivalent resistance of 10 Ω across the 5 V battery. Hence current through the battery is I = V/R = 5/10 = 0.5 A.

8. Question 8 (NEET 2025)

   NEET 2025MediumOhm's Law & Resistivity

   A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these pieces together in parallel. Then these two sets are added in series. The effective resistance of the combination is :

   1. AOption A: R/64
   2. BOption B: R/32
   3. COption C: R/16
   4. DOption D: R/8

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   Correct answer: (C) R/16

   Explanation

   Each small piece has resistance: $$r = \frac{R}{8}$$ Four such resistors in parallel give: $$R_p = \frac{r}{4} = \frac{R}{32}$$ Two such combinations in series: $$R_{eq} = \frac{R}{32} + \frac{R}{32} = \frac{R}{16}$$

9. Question 9 (NEET 2024)

   NEET 2024EasyCells & Internal Resistance

   The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω as shown in the figure is:

   1. AOption A: 8 V
   2. BOption B: 10 V
   3. COption C: 4 V
   4. DOption D: 6 V

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   Correct answer: (A) 8 V

   Explanation

   Given: $$E = 10\,V, \quad r = 1\,\Omega, \quad R = 4\,\Omega$$ Current in the circuit: $$I = \frac{E}{R+r} = \frac{10}{4+1} = 2\,A$$ Terminal voltage: $$V = E - Ir$$ $$V = 10 - (2)(1) = 8\,V$$ Hence, option (A) is correct.

10. Question 10 (NEET 2024)

    NEET 2024EasyDisplacement Current

    A parallel plate capacitor is charged by connecting it to a battery through a resistor. If I is the current in the circuit, then in the gap between the plates:

    1. AOption A: displacement current of magnitude equal to I flows in a direction opposite to that of I.
    2. BOption B: displacement current of magnitude greater than I flows but can be in any direction.
    3. COption C: there is no current.
    4. DOption D: displacement current of magnitude equal to I flows in the same direction as I.

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    Correct answer: (D) displacement current of magnitude equal to I flows in the same direction as I.

    Explanation

    According to Maxwell's theory, during charging of a capacitor, displacement current exists between the plates. The displacement current has the same magnitude as the conduction current in the circuit and flows in the same direction. Hence, option (D) is correct.