In the circuit shown in the figure , switch S is closed at time t = 0 with no ch
ID: 1272063 • Letter: I
Question
In the circuit shown in the figure , switch S is closed at time t = 0 with no charge initially on the capacitor. The battery voltage is 74.0 V, R = 60.0 ?, C = 77.0 nF, and L = 19.0 mH,
(a) Find the reading of each ammeter just after S is closed.
ammeter 1: __ A
ammeter 2: __ A
ammeter 3: __ A
(b) Find the reading of each voltmeter just after S is closed.
voltmeter 1: __ V
voltmeter 2: __ V
voltmeter 3: __ V
voltmeter 4: __ V
(c) At what rate is the current increasing in the inductor just after S is closed?
__ A/s
(d) At what rate is the amount of charge separated by the capacitor increasing just after S is closed?
__ C/s
(e) Find the reading of each ammeter after a long time has elapsed.
ammeter 1: __ A
ammeter 2: __ A
ammeter 3: __ A
(f) Find the reading of each voltmeter after a long time has elapsed.
voltmeter 1: __ V
voltmeter 2: __ V
voltmeter 3: __ V
voltmeter 4: __ V
(g) Find the maximum charge separated by the capacitor. Give your answer in microcoulombs
Explanation / Answer
a) initially C acts as short ckt and L acts open ckt
Rnet = R+2*R = 3*R = 3*60 = 180 ohmns
I1 = V/Rnet = 74/180 = 0.41 A
I2 = 0
I3 = I1 = 0.41 A
b)
V1 = I*R = 0.41*60 = 24.67 volts
V2 = V4 = I3*2*R = 0.41*2*60 = 49.33 volts (beacuse 2*R and L are in parallel)
V3 = 0
V4 = I3*2*R = 49.33 volts
c)
Q = Qmax*(1- e^(-t/T))
here, T = C*3*R = 77*10^-9*3*60 = 13.86*10^-6 s
Qmax = 74*77*10^-9 = 5.698*10^-6 C
dQ/dt = -Qmax*(0 - e^-t/T)*(-1/T)
= 5.698*10^-6*e^(-t/T)/13.86*10^-6
at t = 0
dQ/dt = 0.41 C/s
d)
emf = L*dI/dt
V4 = L*dI/dt
==> dI/dt = v4/L
= 49.33/19*10^-3
= 2.596*10^3 A/s
e) after a long time L acts as short ckt
capacitor acts open ckt
I1 = 0 A
I2 = 0
I3 = 0
f) V1 = 0 volts
V2 = 0
V3 = 74 volts
V4 = 0
g) Qmax = C*v
= 77*10^-9*60
= 5.698 micro C