The circuit in the figure has a capacitor connected to a battery, two switches,

Question

The circuit in the figure has a capacitor connected to a battery, two switches, and three resistors. Initially, the capacitor is uncharged and both of the switches are open. (a) Switch S1 is closed. What is the current flowing out of the battery immediately after switch S1 is closed? (b) After about 10 min, switch S2 is closed. What is the current flowing out of the battery immediately after switch S2 is closed? (c) What is the current flowing out of the battery about 10 min after switch S2 has been closed? (d) After another 10 min, switch S1 is opened. How long will it take until the current in the 200-? resistor is below 1 mA?

Explanation / Answer

a) With only S1 closed, we have a series circuit with a total resistance of 400

I = V/R = 6V/400 = 0.015A

b) immediately after the switch S2 is closed and the capacitor begins to charge, the 100 and 200 are in parallel and then in series with the 300.

1/R = 1/100 + 1/200 = 3/200

R = 66.67

Rtot = 66.67 + 300 = 366.67

So the current would be:

I = 6V / 366.67 = 0.01636A

c) After a while, the capacitor becomes charged and no more current flows in the 200. so the current will be 0.015A

d) For a discharging capacitor:

I = (V/R)e^(-t/RC)

The voltage across the capacitor is the same as that of the 100 when switch S1 is closed:

V = (0.015A)(100) = 1.5V

And it is discharging through a 100 and 200. So, if the current is 1mA,

1*10^-3A = (1.5V)/(300) *e^(-t/RC)

e^(-t/RC) = 0.2

-t/RC = Ln0.2

t = -(300)(4*10^-3F)Ln0.2 = 1.931s

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