I have everything correct except the power being stored on the capacitor at time
ID: 2280137 • Letter: I
Question
I have everything correct except the power being stored on the capacitor at time 1.9 s I've tried finding the potential energy by U=1/2QV^2 and by taking that answer and diving by time but I'm not coming up with the right answer... What am I doing wrong?
The values of the components in a simple series RC circuit containing a switch and an initially uncharged capacitor (see figure below) are C = 1.40
The values of the components in a simple series RC circuit containing a switch and an initially uncharged capacitor (see figure below) are C = 1.40 mu F, R = 2.70 M?, and epsilon = 10.0 V.Explanation / Answer
Pcapacitor:
U = 0.5 q^2/C
P = dU/dt = 0.5 (2 q q')/C = q' q/C
q = q0 (1 - e^(-t/RC)) = C V (1 - e^(-t/RC)) = 1.4e-6*10 * (1 - e^(-1.9/(1.4e-6*2.7e6))) = 5.531e-6 C
q' = q0 ((1/RC) e^(-t/RC)) = 1.4e-6*10 * (1/(1.4e-6*2.7e6)) * (e^(-1.9/(1.4e-6*2.7e6))) = 2.24047e-6 C/s
P = q' q/C = 2.24047e-6 * 5.531e-6/1.4e-6 = 8.85 uW