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28 November, 15:52

Given that e = 9.0 v, r = 98 ω and c = 23 μf, how much charge is on the capacitor at time t = 4.0 ms

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  1. 28 November, 18:16
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    Let charge across the capacitor be Q, current through the circuit be I.

    Voltage difference across the resistor = rI

    Voltage difference across the capacitor = Q/c

    Loop rule: net voltage change through a loop must be zero, so

    9 = rI + Q/c. Since I = dQ/dt,

    r dQ/dt + Q/c = 9

    Solving, Q = 9c (1 - e^ (t/rc)). Plug in the numbers from the problem for the numerical answer.
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