The circuit in the figure has a capacitor connected to a battery, two switches,
ID: 1895167 • Letter: T
Question
The circuit in the figure has a capacitor connected to a battery, two switches, and three resistors. Initially, the capacitor is uncharged and both of the switches are open.
(a) Switch S1 is closed. What is the current flowing out of the battery immediately after switch S1 is closed?
(b) After about 10 min, switch S2 is closed. What is the current flowing out of the battery immediately after switch S2 is closed?
(c) What is the current flowing out of the battery about 10 min after switch S2 has been closed?
(d) After another 10 min, switch S1 is opened. How long will it take until the current in the 200-? resistor is below 1 mA?
Explanation / Answer
a) Assuming S2 is still open.
Current flows through R1 and 3 only, so I = 6 / (300+100) = 15 mA
b) Assuming S1 is still closed.
Immediately after S2 is closed, you can regard C as a short. So you have R2 and 3 in parallel for a total of 100*200/300 = 66.67 ohm
That is in series with R1 for a total of 366.67. Total I = 6 / 366.67
= 16.3 mA
c) Assuming S1 is still closed.
after ten minutes, you can assume C is fully charged, so the only current is due to R1/3, see (a)
d) Assuming S2 is still closed.
voltage on a cap, discharging
v = ve^(–t/RC)
With S1 open, the cap discharges through R3 and R2.
But first we need the initial voltage on C, which is the voltage across R3, which is
V3 = (100)*6 / (300+100) = 1.5 volts
When current in R1 is 1 mA, the voltage across R3 and R2 is 300 mV.
So the cap starts out with 1.5 volts and we want the time for it to discharge to 0.30 volts.
RC = 300 x 4 mF = 1.2 seconds
v = ve^(–t/RC)
0.30 = 1.5e^(–t/1.2)
e^(–t/1.2) = 0.2
–t/1.2 = -1.609
t = 1.93 seconds
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