An inductor of inductance L = 400 mH and resistors of resistances R 1 = 4Ω and R 2 = 2Ω…
Physics · JEE Advanced · NTA Exams — Electromagnetic Induction and Alternating Currents
An inductor of inductance L = 400 mH and resistors of resistances R1 = 4Ω and R2 = 2Ω are connected to battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0 . The potential drop across L as a function ot time is
- 6e–5t V
- \(\frac{12}{t} e^{-3 t} \mathrm{~V}\)
- 6 (1 – e–t/0.2)
- 12 e–5t V
Answer
(D) 12 e –5t V
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