The values of the components in a simple series RC circuit containing a switch (
ID: 1520611 • Letter: T
Question
The values of the components in a simple series RC circuit containing a switch (see figure below) are C = 1.00 µF, R = 2.00 106 , and e m f = 10.0 V. At the instant 14.5 s after the switch is closed, calculate the following.
(a) the charge on the capacitor
(b) the current in the resistor
(c) the rate at which energy is being stored in the capacitor
(d) the rate at which energy is being delivered by the battery
a) To compute the charge on the capacitor at this instant in time, first we compute the time constant RC. We have
s.
Next we substitute this time constant and the time interval after the switch is closed into the following equation to find the charge on the capacitor at this instant.
× 106 C = µC
(b) We find the current in the resistor from the following.
= × 108 A
(c) Since the energy stored in the capacitor is
U = q2/2C,
= × 107 W.
(d) The battery power output is given by the following.
RC = × 106 × 106 F 1 V A · 1 A · s V · F =s.
Next we substitute this time constant and the time interval after the switch is closed into the following equation to find the charge on the capacitor at this instant.
q = C 1 et/RC = F V 1 e s / s =× 106 C = µC
(b) We find the current in the resistor from the following.
I = dq dt = d dt C 1 e(t/RC) = R e(t/RC) = V × 106 e s / s= × 108 A
(c) Since the energy stored in the capacitor is
U = q2/2C,
the rate of storing energy is dU dt = d dt 1 2 q2 C = q C dq dt = q C I = × 106 C × 106 C/V × 108 A= × 107 W.
(d) The battery power output is given by the following.
Pbattery = I = × 108 A V = × 107 WExplanation / Answer
a) Time constant, T = RC = (2 * 106) * (1 * 10-6) = 2 s
Charge on the capacitor, Q = C(1 - e-t/RC)
At t = 14.5 s, Q = (1 * 10-6) * 10 * (1 - e-14.5/2) = 9.99 * 10-6 C
b) Current in the circuit, I = dQ/dt = (/R)e-t/RC
At t = 14.5 s, I = [10 / (2 * 106)] * e-14.5/2 = 3.55 * 10-9 A
c) Energy stored in the capacitor, U = Q2/2C
=> dU/dt = (1/2C)[2Q(dQ/dt)] = QI/C
At t = 14.5 s, dU/dt = (9.99 * 10-6) * (3.55 * 10-9) / (1 * 10-6) = 3.55 * 10-8 W
d) Battery power output, Pbattery = I
At t = 14.5 s, Pbattery = 10 * (3.55 * 10-9) = 3.55 * 10-8 W
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