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An emf of 15 V is applied to a series circuit containing a 5 H inductor and a 10 ohm resistor. Find the ratio of the current at time t = infinity to the current at time t = 1 s.

1. e² / (e² - 1)
2. e^(1/2) / (e^(1/2) - 1)
3. 1 - e⁻¹
4. e⁻¹

Correct answer: e² / (e² - 1)

Solution

For an LR circuit, i(t) = i0(1 - e^(-t/tau)) with i0 = E/R and tau = L/R = 5/10 = 0.5 s. At t = infinity, i = i0. At t = 1 s, t/tau = 1/0.5 = 2, so i(1) = i0(1 - e⁻²). The ratio i(infinity)/i(1) = i0/[i0(1 - e⁻²)] = 1/(1 - e⁻²) = e²/(e² - 1).

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