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A 10 H ideal inductor is joined in series with a 5 Ω resistor and a 5 V battery. What is the current in the circuit 2 s after the circuit is closed?

1. (1 - e⁻¹) A
2. (1 - e⁻²) A
3. e⁻¹ A
4. e⁻¹/2 A

Correct answer: (1 - e⁻¹) A

Solution

I = (V/R)(1 - e^(-Rt/L)) = (5/5)(1 - e^(-5*2/10)) = (1 - e^(-1)) A.

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