NEET 2025 Physics Question 11 Solution
Question
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:
Dynamic 180-question full-length simulator mixed dynamically from all 6 years for 2027 & 2026 aspirants.
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:
V_A - L(di/dt) - E - iR = V_B ⟹ V_A - V_B = (1 × 1) + 5 + (2 × 2) = 1 + 5 + 4 = 10 V!
Writing the potential step equation starting from point A and proceeding to point B:
Substitute the given values into the potential difference expression:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Incorrect. Considers only battery EMF while ignoring inductive and resistive drops.
Incorrect. Subtracts resistive drop: 5 + 1 - 4 = 2, or misses inductor.
Incorrect. Forgets inductor back EMF (5 + 4 = 9 V).
100% correct. V_A - V_B = L(di/dt) + E + iR = 1 + 5 + 4 = 10 V.
Students often assume an inductor behaves like a plain wire (0 V drop) because current is 2 A. But current is INCREASING at 1 A/s, creating an active induced back EMF of L di/dt = 1 V!
"Potential drop across an inductor carrying increasing current is L di/dt opposing current flow."
Test your speed and accuracy in the authentic 180-minute timed NTA simulator.
Spotted a mistake in this solution? Report an error ([email protected])
Cover all 1,140 official NEET questions in 285 days at your own pace. Progress is saved on your device. No login needed.