Question

A 14.0 ??F capacitor is charged to a potential of 50.0 V and then discharged through a 180 ?? resistor. Part A How long does it take the capacitor to lose half of its charge? Part B How long does it take the capacitor to lose half of its stored energy?

Answer 1

(A) we know that the charge in the capacitor is given by
q= Q*e^-(t/RC)
where q is charge at any time t and Q is maximum charge
When q = Q/2
Q/2 = Q*e^-(t/RC)
(1/2) = e^-t/RC
tkaing log both side
ln(1/2) = -(t/RC)
-0.6931 = -(t/RC)
t = RC*(0.6931) = 180*14*10^-6*0.6931 = 1.747*10^-3 s
(B) Now for energy
U = U_o*(e^-2t/RC)
Now when U = U_o/2
U_o/2 = U_o*(e^-2t/RC)
(1/2) = e^-2t/RC
taking log both side
ln(1/2) = -(2t/RC)
-0.6913 = -(2t/RC)
t = RC*0.6913 /2
t = 180*14*10^-6*0.6913 / 2= 0.871*10^-3 s