In the circuit of the figure below the switch S has been ope

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 = 50.0 kOhm, R_2 = 210 kOhm, and C = 10.0 muF. Determine the time constant before the switch is closed. Determine the time constant after the switch is closed. Let the switch be closed at t = 0. Determine the current in the switch as a function of time. (Use the following as necessary: t. Express your answer in units of (muA.)

Solution

The current used to charge the battery is determined by the two resistors in series with the battery. So, the time constant

T_open = RC = (50 + 210) kohms * 10 * 10^-6

T_open = 2.6 s

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b) When the switch is closed, the capacitor discharges through 210 kohm resistor, so the time constant

T_closed = RC = 210 kohms * 10 * 10^-6

T_closed = 2.1 s

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c) I = I_b+ I_c= delta V_b/ R + I₀ e^-t/Tclosed

= 10 / (50 * 10³) + I₀ e^-t/2.1

= (200 * 10^-6) + I₀ e^-t/2.1

I₀ = 10 / 210 kohms = 47.6 * 10^-6 A

Total current becomes I = (200 * 10^-6) + (47.6 * 10^-6) I₀ e^-t/2.1

after the switch is closed the capacitor is fully discharged and the current in the switch will be mainly from the battery