Switch open i0 q0 R b SolutionA Q CV V 40 x 106 9 x 106

Switch open i=0 q=0 R b

A) Q = CV

V = (40 x 10^-6) / (9 x 10^-6) = 40/9 V ... .PD across capacitor

If current is 3A then

PD across capacitor, Vc = e - IR

40/9 = e - (15 x 3)

e = 49.44 volt

b) Q = Q0 [ 1 - e^(-t / RC) ]

CV = CV0 [ 1 - e^(-t / RC) ]

40/9 = 49.44 [ 1 - e^(-t/RC) ]

0.91 = e^(-t/RC)

- t / RC = ln(0.91) = - 0.0942

t = 0.0942 x 15 x 9 x 10^-6 = 1.27 x 10^-5 s

c) POwer supplied by battery = eI = 49.44 x 3 = 148.33 W

Power dissipated by resistor = I^2 R = 3^2 x 15 = 135 W

Rate of energy stored by capacitor = 148.33 - 135 = 13.33 W

d) POwer supplied by battery = eI = 49.44 x 3 = 148.33 W