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A capacitor is fully charged and connected in an RC circuit. The switch is closed at t = 0. After how much time does the current from the capacitor fall to 1/4 of its initial maximum value?

1. 2RC*ln(2)
2. 7RC/4 * ln(2)
3. RC*ln(2)
4. 7RC/2 * ln(2)

Correct answer: 2RC*ln(2)

Solution

For a capacitor discharging through resistance R with time constant tau = RC, I(t) = I_max * exp(-t/RC). Setting I = I_max/4: exp(-t/RC) = 1/4, so t = RC * ln(4) = 2RC * ln(2).

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