Question

2. In this ciruit, C-6.0 uF, emf-30 V with negligible resistance. Initially the capacior is uncharged and the switch S is in position 1. The switch is then moved to position 2, so that the capacitor begins to charge. (a) What will be the charge on the capacitor a long time after the switch is moved to position 2? (b) After the switch has been in position 2 for 3.0 ms, the charge on the capacitor is measured lo be 100 μο. what is the value of the resistance? (c) How long after the switch is moved to position 2 will the change on the capacitor be equal to 99% of the value found in part (a)?

Answer 1

(a)

C = Capacitance of the capacitor = 6 x 10^-6 F

E = emf of the battery = 30 Volts

After the switch has been in position 2 for long time, the potential difference across the capacitor is same as the emf of the battery. hence the charge stored by the battery is given as

Q = CE

Q = (6 x 10^-6) (30)

Q = 180 $\mu$C

b)

Q_t = Charge on the capacitor at time "t" = 100 $\mu$C

Q = maximum charge on the capacitor = 180 $\mu$C

T = Time constant = RC

t = time = 3 ms = 0.003 sec

During charging , charge at any time of the capacitor is given as

Q_t = Q (1 - e^-t/T)

100 = 180 (1 - e^-t/T)

e^-t/T = 0.444

e^-0.003/T = 0.444

T = 0.0037 sec

RC = 0.0037

R (6 x 10^-6) = 0.0037

R = 616.7 ohm

c)

Time constant is given as

T = 0.0037 sec

t = time

Q_t = Q (1 - e^-t/T)

Given that : Q_t = 0.99 Q

0.99 Q = Q (1 - e^-t/T)

0.99 = (1 - e^-t/T)

e^-t/T = 0.01

e^-t/0.0037 = 0.01

t = 0.017