Topic: Self and mutual inductance· Electromagnetic Induction and Alternating Currents

AB is a part of an electrical circuit (see figure). The potential difference "V_A - V_B", at the instant when current i = 2 A and is increasing at a rate of 1 amp / second is:

AB is a part of an electrical circuit (see figure). The potential difference "$V_A - V_B$", at the instant when current i = 2 A and is increasing at a rate of 1 amp / second is:

A to B: inductor 1 H, battery 5 V, resistor 2 ohm in series, current i from A

1. 5 volt

2. 6 volt

3. 9 volt

4. 10 volt

Doubtly Asked 30 Oct 2026 NEET UG 2025 Practice test neet Save WhatsApp Report

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Answer: 4. 10 volt

Key idea: going from A to B, add the potential change across each element.

The current flows from A to B and is increasing, so the inductor has $L\,\dfrac{di}{dt}=1\times1=1$ V with A at the higher potential. The resistor drops $iR=2\times2=4$ V. The battery is connected with its positive terminal towards A, so it contributes 5 V in the same sense.

$$V_A-V_B=L\frac{di}{dt}+iR+5=1+4+5=10\ \mathrm V$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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AB is a part of an electrical circuit (see figure). The potential difference "V_A - V_B", at the instant when current i = 2 A and is increasing at a rate of 1 amp / second is:

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