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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.)
- 4 μC
- 5 μC
- 6 μC
- 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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