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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?

User Dollyn
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1 Answer

23 votes
23 votes

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}

User Shrimpy
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