Question

The capacitor in an circuit is discharged with a time constant of 40.4 s. At what time, in microseconds, after the discharge begins is the charge on the capacitor reduced to half its initial value?

Answer 1

Apply:

`Q=Q_0e^{-(t)/(RC)}`

where:

RC= time constant = 40.4 s

Q0 = initial charge

Q = final charge

`(Q_0)/(2)=Q_0e^(-t/RC)`
`(Q_0)/(2Q_0)=e^(-t/RC)`
`(1)/(2)=e^(-t/RC)`
`ln((1)/(2))=-(t)/(RC)`
`ln((1)/(2))=-(t)/(40.4)`
`-0.693\text{ = -}(t)/(40.4)`
`t=\text{ 28 s}`