In the circuit shown, C = 5.90 ?F, ? = 28.0 B, and the emf has negligible resist
ID: 1425624 • Letter: I
Question
In the circuit shown, C = 5.90 ?F, ? = 28.0 B, and the emf has negligible resistance. Initially the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2, so that the capacitor begins to charge.
(a) What will be the charge on the capacitor a long time after the switch is moved to position 2?
(b) After the switch has been in position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 ?C. What is the value of the resistance R?
Switch S in position 1 Switch S in position 2 74Explanation / Answer
a) we know that after a long time circuit approaches a steady state where the charge on the capacitor will be simply given by Q = C*E = 5.9*10^-6*28 = 1.652 *10^-4 C
b) Q=Qo[1-e^(-t/(R*C))] ======> R= -t/[(Cln(1-Q/Qo)]
R = -3*10^-3/[5.9*10^-6*ln(1 - 110/165.2)] = 463.85 ohm
c) We want to solve for t when q = 0.99*Q0
Now t = -RC*ln(1 - q/Q0) = -(463.85*5.9*10^-6)*ln[1 - 0.99Q0/Q0] = 0.0126 s