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Working with time constant RC

A capacitor is used in the electronic flash unit of a camera. A small battery with a constant voltage of 6 V is used to charge the capacitor with a constant current of 10 microamperes. How long does it take to charge the capacitor when C = 10 microfarads? My Answer:
V=6V
i=10mA
C=10microF
Wc=1/2CV^2*(t)
Wc=1/2(10^-5)6^2*(t)
Wc=18*10^-5*(t)
Wc=6V*.01A=.06J
Wc=.06J=18*10^-5*(t)
t=.06/(18*10^-5)=333.33
Final answer: 333.33 s

Solution Preview

We know that charge, Q = current * time = I * t

Also, for a capacitor, C = ...

Solution Summary

The answer with good explanation on why the student answered it wrongly.

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