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An ideal inductor of 10 H is connected in series with a 5 ohm resistor and a 5 V battery. The current flowing in the circuit 2 seconds after the circuit is switched on is (in amperes):

1. 1 - e⁻¹
2. e⁻¹
3. 1 - e
4. e

Correct answer: 1 - e⁻¹

Solution

For a series LR circuit, I(t) = (V/R)(1 - e^(-Rt/L)). Here V/R = 5/5 = 1 A and tau = L/R = 10/5 = 2 s. At t = 2 s, t/tau = 1, so I = 1*(1 - e⁻¹) = (1 - e⁻¹) A.

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