a 2 microfarad and an electrostatic voltmeter are connected to a 500 Volts supply. after disconnection the voltmeter reading falls to 250 volts in 15 minutes. find the the leakage resistance?

Answer 1

initial Vi = 500
initial Qi = C Vi = 2*10^-6 *500 = 10^-3 coulombs
that leaks out with current i = -dQ/dt
V(t) = i R = Q /C
so
-dQ/dt = Q/RC
dQ/Q = -(1/RC) dt
Q = constant e^-t/RC = constant e^-t/RC
at t = 0 , Q = 10^-3 so
Q = 10^-3 e^-t/RC
in 15 min = 900 seconds
Q = 10^-3 e^-900/RC
but V = Q / C
so
250*2*10^-6 = 10^-3 e*-900/RC
5*10^-4 = e^-900/(2*10^-6 R)
ln 0.0005 = -900/(2*10^-6R)
-7.6 R =- 900*10^6 /2