NEET Current Electricity PYQs & Practice Questions

10 free NEET Physics questions on Current Electricity, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Parallel Circuits · Wheatstone Bridge · Wheatstone Bridge / Network Reduction · Series-Parallel Combination of Resistors · Bridge Circuit / Branch Current · Steady Current and Surface Charges · Equivalent EMF of Cells in Parallel · Drift Velocity and Current · Series-Parallel Combination of Wire Segments - Equivalent Resistance · Terminal Voltage of Battery with Internal Resistance · Wheatstone Bridge Balance with Diode in a Branch · Power Ratio - Series vs Parallel Combination of Heaters

1. Q1NEET 2026Temperature-Dependent Resistance in Series and Parallel Circuits

   Consider two circuits, (A) and (B), each having two resistors. In circuit (A), two resistors R₀(1−αΔT) and R₀(1+αΔT) are connected in series with a battery V, giving current I_A. In circuit (B), the same two resistors R₀(1−αΔT) and R₀(1+αΔT) are connected in parallel with a battery V, giving current I_B. As the temperature (ΔT) is increased, the correct option describing the variation of current in these circuits is:

   • A. I_A decreases while I_B increases
   • B. I_A increases while I_B decreases
   • C. both I_A and I_B remain constant
   • D. I_A remains constant while I_B increases
   Show answer

   Answer: C — both I_A and I_B remain constant

   Circuit A (series): total resistance = R₀(1-αΔT) + R₀(1+αΔT) = 2R₀, independent of ΔT (the temperature-dependent terms cancel). So I_A = V/(2R₀) stays constant. Circuit B (parallel): R_parallel = [R₀(1-αΔT)·R₀(1+αΔT)]/(2R₀) = R₀(1-α²ΔT²)/2, which decreases as ΔT increases. Since I_B = V/R_parallel, a decreasing resistance means I_B increases with ΔT. So I_A remains constant while I_B increases.

2. Q2NEET 2026Power dissipation with voltage drop

   A room heater is rated 1000 W, 220 V. If the supply voltage drops to 200 V, what will be the power consumed (approximately)?

   • A. 1000 W
   • B. 121 W
   • C. 826 W
   • D. 200 W
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   Answer: B — 121 W

   R=V²/P=220²/1000=48.4Ω. At 200V, P=200²/48.4≈826 W.

3. Q3NEET 2026Terminal voltage of a battery

   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:

   • A. 12 V
   • B. 1.2 V
   • C. 10 V
   • D. 10.8 V
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   Answer: C — 10 V

   Terminal voltage = EMF - I×r = 12 - 0.6×2 = 10.8 V.

4. Q4NEET 2026Meter bridge, interchanging cell and galvanometer

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

   • A. Only the right-sided deflection
   • B. Only the left-sided deflection
   • C. There will be no deflection irrespective of the position of the jockey
   • D. Both right-sided and left-sided deflection, and at balance point, no deflection
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   Answer: C — There will be no deflection irrespective of the position of the jockey

   By reciprocity of a Wheatstone bridge, interchanging cell and galvanometer leaves the balance condition unchanged; deflection on either side is possible away from balance, and none at balance.

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5. Q5NEET 2026Ammeter shunt resistance

   A galvanometer of resistance 100 Ω gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0–10 A. The shunt required is:

   • A. 0.001 Ω
   • B. 0.10 Ω
   • C. 1.0 Ω
   • D. 0.01 Ω
   Show answer

   Answer: C — 1.0 Ω

   S=IgG/(I-Ig)=(10⁻³×100)/(10-10⁻³)≈0.01 Ω.

6. Q6NEET 2026Current in diode-resistor network

   The current I in the circuit shown below is: (All diodes are ideal and identical; resistors 4Ω, 3Ω, 5Ω; battery 10V)

   • A. 1/3 A
   • B. 15/2 A
   • C. 5/3 A
   • D. 5/9 A
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   Answer: A — 1/3 A

   With ideal diodes conducting appropriately, the equivalent circuit reduces to give I=15/2 A as per the official key.

7. Q7NEET 2026Current in a wire bent into a square loop

   A uniform metallic wire having resistance 4 Ω is bent to form a square loop ABCD. 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 in figure. The value of current (I) is:

   • A. 4 A
   • B. 4.5 A
   • C. 8 A
   • D. 2 A
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   Answer: C — 8 A

   By symmetry and parallel combination of the two arms of the square with the BD resistor, the net resistance gives I=2 A per the official key.

8. Q8NEET 2025Wheatstone Bridge / Network Reduction

   The current passing through the battery in the given circuit (a resistor network ABCDEF with resistances 5Ω, 1.5Ω, 2.5Ω, 6Ω, 3Ω, 1.5Ω, 1/3Ω and a 5.5Ω branch, driven by a 5V battery), is:

   • A. 2.0 A
   • B. 0.5 A
   • C. 2.5 A
   • D. 1.5 A
   Show answer

   Answer: A — 2.0 A

   Reducing the bridge network (parallel/series combination of the 6 arms plus the 5.5Ω and 1/3Ω branches) gives total resistance R = 16/6 + 1/3 + 3/2 + 11/2 = (16+2+9+33)/6 = 60/6 = 10Ω. Current I = V/R = 5/10 = 0.5 A.

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9. Q9NEET 2025Series-Parallel Combination of Resistors

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

   • A. R/64
   • B. R/32
   • C. R/16
   • D. R/8
   Show answer

   Answer: B — R/32

   Each piece has resistance R/8. Four in parallel: R1 = (R/8)/4 = R/32. Two such sets (R1 and R2, each R/32) in series: Req = R/32 + R/32 = R/16.

10. Q10NEET 2025Bridge Circuit / Branch Current

    A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure (a Wheatstone-like network ACBD with resistances AC=1Ω, CB=2Ω, AD=3Ω, DB=4Ω). The current through the branch CD of the circuit is :

    • A. 1.5 A
    • B. 2.0 A
    • C. 2.5 A
    • D. 3.0 A
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    Answer: A — 1.5 A

    R1 (1Ω & 3Ω parallel on one path analysis) and R2 give Req = 3/4+4/3 = 25/12 Ω. I = V/Req = 50×12/25 = 24A total. Branch currents work out to I1=18A, I2=6A, I3=16A, I4=8A, giving ICD = I1−I3 = 18−16 = 2 A.