Question

A 6.40 μF capacitor that is initially uncharged is connected in series with a 4400 Ω resistor and a 502 V emf source with negligible internal resistance.

a. Just after the circuit is completed, what is the voltage drop across the capacitor?

b. Just after the circuit is completed, what is the voltage drop across the resistor?

c. Just after the circuit is completed, what is the charge on the capacitor?

d. Just after the circuit is completed, what is the current through the resistor?

e. A long time after the circuit is completed (after many time constants), what are the voltage drop across the capacitor and across the resistor?

f. A long time after the circuit is completed (after many time constants), what is the charge on the capacitor?

g. A long time after the circuit is completed (after many time constants), what is the current through the resistor?

Answer 1

a)

Just after the switch is close, there is no charge stored in the capacitor and hence it behaves as short circuit.

So V_c = voltage drop across capacitor = 0 volts

b)

E = emf source voltage = 502 Volts

V_r = voltage drop across resistor

voltage drop across resistor is given as

V_r = E

V_r = 502 volts

c)

Q_c = charge on the capacitor

Charge on the capacitor is given as

Q_c = C V_c

Q_c = (6.40 x 10^-6) (0)

Q_c = 0 C

d)

current through the resistor using ohm's law is given as

i = V_r /R = 502/4400 = 0.114 A