# Working with time constant RC

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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

https://brainmass.com/engineering/electronic-engineering/working-with-time-constant-rc-37632

#### 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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