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An inductor of inductance 5 H is connected to a 6 V cell through a 10 Ω resistance, with the circuit closed at t = 0. Find the magnitude of the emf induced across the inductor at time t = ln(sqrt(2)) seconds.

1. 3 V
2. 1.5 V
3. 0.75 V
4. 4.5 V

Correct answer: 4.5 V

Solution

τ = L/R = 5/10 = 0.5 s. The inductor voltage is V_L = E e^(-t/τ) = 6 e^(-t/0.5) = 6 e^(-2t). At t = ln(sqrt2) = (1/2)ln2, we get -2t = -ln2, so e^(-2t) = 1/2. Thus V_L = 6 * (1/2) = 3 V.

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