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Find the current flowing through the branch AC in the steady state as also the charge on the capacitor C. If the externally applied potentials are now withdrawn, how will the charge on the capacitor vary as a function of time? (R = 1 kOmega, C = 10 muF) |
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Answer» Solution :The current through `BA = (10V-5V)/(1) = 5 mA` Similarly current through `AC = 5 mA, & BC = 10 mA` steady state CHARGE on `C = 5V xx 10 muF = 50 muC`If the potential differences are withdrawn at time t = 0, the charge on the capacitor varies as a function of time as it discharges through the external resistance. The equivalent resistance of the circuit across AC is `(2R.R)/(2R+R)=(2)/(3)xxR=667Omega(approx)` The time CONSTANT `tau=(2//3R)XXC=6.67msec` The charge across the capacitor decreases exponentially`q(t) 50 muC xsx e^(-t//6.67) MS` |
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