An inductor ( L = 100 mH ), a resistor ( R = 100 Ω ) and a battery ( E = 100 V ) are connected in series. After a long time, the battery is disconnect - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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An inductor (L = 100 mH), a resistor (R = 100 Ω) and a battery (E = 100 V) are connected in series. After a long time, the battery is disconnected after short-circuiting. The current in the circuit 1 ms after the short-circuit is:

Solution & Step-by-step Explanation

Initial steady-state current I_{0} = E / R = 100/100 = 1 A .After short-circuiting, the current decays according to I = I_{0} e^{- Rt / L} .Time t = 1 ms = 1 0^{- 3} s .Inductance L = 100 mH = 0.1 H .Resistance R = 100 Ω .Exponent term \frac{Rt}{L} = \frac{100 \times 1 0 ^{- 3}}{0.1} = \frac{0.1}{0.1} = 1 . I = 1 \cdot e^{- 1} = 1/ e A

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An inductor (L = 100 mH), a resistor (R = 100 Ω) and a battery (E = 100 V) are connected in series. After a long time, the battery is disconnected after short-circuiting. The current in the circuit 1 ms after the short-circuit is:

1 A

1/ e A

e A

0.1 A

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