An initially 465 mu F capacitor is connected in series with a 4.31 k Ohm resisto
ID: 1549145 • Letter: A
Question
An initially 465 mu F capacitor is connected in series with a 4.31 k Ohm resistor, a 250 V battery with negligible internal resistance, and a switch. (a) At an instant after the switch is closed, what is the voltage drop across the resistor and the current in the resistor? (b) What is the final voltage drop across the capacitor and final charge on the capacitor? (c) What is the voltage drop across the capacitor 2 s after the switch is closed? (d) What is the current in the circuit 2 s after the switch is closed?Explanation / Answer
(A) voltage drop acoss resistor = 250 Volt
I = V / R = 250 / (4.31 x 10^3)
= 0.058 A Or 58 mA
(B) Vf = 250 Volt
Q = C V = 465 x 10^-6 x 250
= 0.116 C
(C) Vc = 250 [ 1 - e^(-t/RC)]
RC = 4.31 x 10^3 x 465 x 10^-6 = 2 s
Vc = 250 [1 - e^-1] = 158 V
(D) V - Vc - VR = 0
Vr = 250 - 158 = 91.97 V
I = Vr / R = 0.0213 A Or 21.3 mA