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 = 50.0 kOhm, R_2 = 185 kOhm, and C = 10.0 muF. (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. (Use the following as necessary: t. Express your answer in units of muA.) The circuit in the figure below has been connected for a long time. Let R_1 = 8.80 Ohm and R_2 = 4.20 Ohm. (a)What is the potential difference across the capacitor? (b)If the battery is disconnected from the circuit, over what time interval does the capacitor discharge to one-eighth its initial voltage?

Explanation / Answer

a)

here

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 Topen is

Topen = R * C

Topen = (185 * 10^3 + 50 * 10^3 ) * 10 * 10^-6 = 2.35 sec

b)

when the switch is closed ,the capacitor discharges through 185 kohm resistor, so the time constant Tclosed is

Tclosed = 185 * 10^3 * 10 * 10^-6 = 1.85 sec

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

I = delta Vb / R1 + I0 * e^(-t/Tclosed)

I = (10 / 50 * 10^3) + I0 * e^(-t/1.85)

I = 200uA + I0 * e^(-t/1.85)

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