A capacitorcharging circuit has a time constant of 40 ms Whe

A capacitor-charging circuit has a time constant of 40 ms. When the switch is closed, the initial current to the 60 F capacitor is 45 mA .

What is the capacitor\'s voltage after 20 ms ? Assume the capacitor was completely uncharged when the switch closed.

Express your answer with the appropriate units.

*Please show all work and steps

Time constant = RC C=60 * 10^-6 F

and time constant = 40 ms

so R = 40 *10^-3/60*10^-6 =666.67

Now when Switch is closed capacitor would act as short circuit

So V=IR where V is battery voltage

V= 666.67 * 45 * 10^-3 =30 V

Now Vc= V(1-e^-t/RC) = 30 (1-e^-1/2) = 11.80 V