4.8mF capacitor in series with 500 ohm resistor is connected, via a switch, to a 12V battery. What is the current through the resistor at t=1.0s after the switch is closed?

Question

I tried doing I^2R = 1/2 CV^2 and solving for I but the answer wasn't correct.

can some1 please help? thanks.
the correct answer is 16mA.

Answer 1

Current varies with time as the capacitor charges up. This is a transient circuit problem that needs to be solved with a differential equation. The solution is

I = (V/R)[1 - e^(-t/(RC))]

V/R = 24 mA
RC = 2.4 seconds
t/RC = 0.4167
e^(-0.4167) = 0.659
I = 24*.659 = 15.8 mA

They have rounded it off to two significant figures, to get 16 mA. That is what should be done, since 12V has only two sig. figures.

Answer 2

I don't know why you used the power in the reisistor is equal to the energy in the capacitor.

I=12/R * e^-t/RC

Answer 3

Bob's formula is correct. Forget the "1 - " in my answer

I = (V/R) * e^-t/RC
That is the one I used to do the calculation anyway