Verified answer

(a)how much energy does the capacitor store when it is fully charged?

E = 0.5CV^2

= (0.5) ( 3x10^-6)(24^2)

= 8.64 x 10^-4 J

(b)what is the capacitor's voltage when it has only half of its maximum energy? Is it 12V?

E = 0.5CV^2

solve for V when E= 8.64 x10^-4 /2

V = sqrt[ E / ((0.5(C)) ]

=sqrt [ 4.32 x 10^-4 / (( 3x10^-6)(0.5)))

=16.9 V

alternate solution 0.5 = V^2 / 24^2 since Energy is proportional to the square of the voltage

V = sqrt [ 0.5 x24^2]

V =16.9 volts

(c)what resistance is necessary to cause the capacitor to have only 50% of its energy left after 0.50s of discharge?

16.9 = 24 (e^(-t/RC))

.704 = (e^(-t/RC))

ln (.704) = - t /RC

0.350 = 0.5 / RC

R = 0.5 / ((.350)(3x10^-6)

R = 4.76 x 10^5 ohms

(d)what is the current in the resistor at this time?

I = V/R

I 16.9 /4.76x10^5

I = 3.55x10^-5 amps