Question

Initially, for the circuit shown, the switch S is open and the capacitor voltage is 150 V. The resistance and capacitance are 1.2 M? and 34 µF respectively. The switch S is closed at time t = 0. In Figure, the charge in µC on the capacitor, when the current in the circuit is 15 µA, is:

Answer 1

Current and Charge are given as the function of time by the below formula -
I(t) = I_o e^-t/τ
Q(t) = Q_o e^-t/τ

where:
C = 34*10^-6 Farads
R = 1.2*10^6 ohms
V = 150 v
Io = V/R = 150/(1.2 * 10^6) = 125 uA
Q_o = C* V = 34 * 10^-6 * 150 = 5100 uC
ε is the base of natural logarithms (-t/RC))


When current in the circuit is 15 µA -
I(t)  = I_o e^-t/τ
15 uA = 125 uA * e^-t/τ
e^-t/τ = 15/125
e^-t/τ = 0.12

Charge in the Capacitor =

Q(t) = Q_o e^-t/τ
Q(t) = 5100 uC _* 0.12
Q(t) = 612 uC

The charge in µC on the capacitor, Q(t) = 612 uC