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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 - e⁻¹
- e⁻¹
- 1 - e
- 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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