In the circuit of the figure below, the switch S has been open for a long time.

Question

In the circuit of the figure below, the switch S has been open for a long time. It is then suddenly closed. Take epsilon = 10.0 V, R _1 = 49.0 k Ohm, R_2 = 175 k Ohm, and C = 12.0 mu F. (a) Determine the time constant before the switch is closed. (b) Determine the time constant after the switch is closed. (c) Let the switch be closed at t = 0. Determine the current in the switch as a function of time. (Assume I is in A and t is in s. Do not enter units in your expression. Use the following as necessary: t.)

Explanation / Answer

(a) While the switch is open, the battery charges the capacitor. The current used to charge
the battery is determined by the two resistors in series with the battery. So, the time
constant open is:

open = RC = (49.0 k + 175 k) × 12µF = 2.68 s

(b) When the switch is closed, the capacitor discharges through 100 k resistor, so the time
constant closed is:

closed = 175 k × 12 µF = 2.04 s

(c) The current through the closed switch will produced by the battery Ib and while the
capacitor is discharging a current Ic, the total current trough the switch is then

I = Ib + Ic =Vb/R + Ie^t/closed =10/49.0 × 10^3 + Ie^t/1.00 = 204 µA + Ie^t/1.00

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