76 The switch in FIGURE CP3176 has been closed for a very lo

76. The switch in FIGURE CP31.76 has been closed for a very long time a. What is the charge on the capacitor? bThe switch is opened at t 0 s. At what time has the charge on the capacitor decreased to 10% of its initial value! 40 52 FIGURE CP11.76

Solution

a)

After a long time, the capacitor will act as an open circuit so, the 10 ohm resistor will virtually be out of the circuit

So, the net resistance of the circuit:

Rnet = 60 + 40 = 100 ohm

So, net current in the circuit = 100 /100 = 1 A

So, voltage across 40 ohm resistor = 40*1 = 40 V

So, voltage across capacitor = 40 V

So, charge on the capacitor = 2*10^-6*40 = 8*10^-5 C <-----answer

b)

Now, for the discharge circuit,

time constant of the circuit,

T = Rnet\'*C = (40+10)*2*10^-6 = 1*10^-4 s

Now, the charge on the capacitor is given by:

Q = Qo*e^(-t/T)

So, for Q = 0.1*Qo

0.1*Qo = Qo*e^(-t/T)

So, t = -ln(0.1)*T = -ln(0.1)*1*10^-4

So, t = 2.3*10^-4 s <------answer