Help PLease Currents??

Question

capacitance is 257nF with a charge of 1.5nC

connected to voltmeter 2.3Mohm's resistance

current through voltmeter is 2.5nA

when is the current equal to 20% of its original value??

ans. is .95s but how was that reached???

Answer 1

Try this:

I=Ii*e^(-t/RC)

Ii/I=e^(t/RC)

t/RC=ln(Ii/I)

t/RC=ln(20/100)

t=RC*ln(.2)

Ii=initial current

R=resistance

C=capacitance

e=2.718...