A 125 MUF capacitor is initially charged to a potential of 5

A 12.5 MUF capacitor is initially charged to a potential of 50.0 V and then discharged through a 75.0 Ohm resistor. How long after discharge begins does it take for the capacitor to lose 90% of its initial charge?

Solution

initial charge Q= CV= 12.5*10^-6 * 50=625*10^-6 C

Capacitor discharge (charge decay): Q = Q_oe^-(t/RC)

90 percent of initial charge is lost so remaining is 10 percent of initial

so

625*10^-6 *10/100 = 625*10^-6e^-(t/RC)

^{putting the value} of R =75 ohm and C= 12.5*10^-6 F

exp to the power(1066 t)=10

taking ln

1066tln(e)=ln(10)

because ln(e)=1

1066t=2.3026

t=2.1*10^-3 sec