10. An AC source operating at 60Hz with a maximum voltage of 172 V is connected
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Question
10. An AC source operating at 60Hz with a maximum voltage of 172 V is connected in series with a resistor (R= 1.8KQ) and a capacitor (C= 2.5F). (a) What is the maximum value of the current in the circuit? (b) What are the maximum values of the potential difference across the resistor and the capacitor? © When the current is zero, what are the magnitudes of the potential difference across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant? (d) When the current is at a maximum, what are the magnitudes of the potential differences across the resistor, the capacitor, and the AC source? How much charge is on the capacitor at this instant?Explanation / Answer
a)
Imax=Vmax/R = 172/1800 = 0.096 A
b)
V(R)max = 172.0 V
V(C)max = 172.0 V
c)
By KVL
Vmax – V(R) – V(c) = 0 ----------(1)
At t=0s,
V(c) = 0 V and Vmax = 172.0 V
Plug these values in (1)
V(R) = Vmax = 172.0V
Q(t) = Q(0)*[1-e^(-t/RC)]
Q(t) = CVmax*[1-e^(-t/RC)]
Plugging values,
e^-0 = 1
Q(t) = CVmax(1-1) = CVmax = 0 C
d)
At t=infinity s
I=0A => V(R) = 0 V, Vmax = 172.0 V
Plug these values in (1),
V(c) = Vmax = 172.0V
Q(t) = Q(0)*[1-e^(-t/RC)]
Q(t) = CVmax*[1-e^(-t/RC)]
Plugging values,
e^-infinity = 0
Q(t) = CVmax(1-0) = CVmax = 2.5*10^-6*172.0 = 0.43*10^-3 C = 0.43mC
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