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In a series circuit containing a resistor, an inductor and a capacitor driven so that the current is increasing at the constant rate dI/dt = 4 A/s, take R = 1 ohm, L = 1 H and C = 1 microfarad. At the instant when the current is I = 2 A, find the charge on the capacitor. (The applied EMF equals the sum of the resistor, inductor and capacitor voltages; the capacitor voltage at this instant is the relevant unknown determined by the given data.)

1. 4 μC
2. 5 μC
3. 6 μC
4. none of these

Correct answer: 6 μC

Solution

For the series R-L-C branch the instantaneous voltages add. With R = 1 ohm and I = 2 A, V_R = 2 V; with L = 1 H and dI/dt = 4 A/s, V_L = 4 V. Their sum sets the capacitor voltage at this instant as V_C = V_R + V_L = 6 V. The charge is then q = C * V_C = 1 microfarad * 6 V = 6 microcoulombs.

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