Problem with discharging capacitors in a circuit SA 50 We study what happens whe
ID: 1874757 • Letter: P
Question
Problem with discharging capacitors in a circuit SA 50 We study what happens when the switch of the cir- cuit sketched in the Figure is closed and opened. We start by assuming that the switch in the circuit has been closed for a very long time (ideally t 40 v 30 15 10 HF a) What can you tell right away about the current I3? Give its numerical value and explain your answer. b) Find the currents I and I2 c) Find the voltage over the capacitor d) What is the charge on the capacitor? At t0 the switch is opened. e) At which time will the charge on the capacitor drop to 1/2 of the initial value you calculated in part d? Let's assume now that an insulator with dielectric constant K-5 is inserted between the plates of the capacitor, and that the switch is again closed for a very long time (ideally t-o). Finally, the switch is opened again f) Will now the capacitor discharge slower or faster? Calculate the new time constant that charac- terizes the time-dependence of the charge on the capacitorExplanation / Answer
a) when the switch is closed for a long time the capcitor is fully charged to the maximum and no more charging current passes through the capacitor
hence I3 =0
b) current passes through only 50 and 30 ohms which are in series
I1 = I2 = 40/(30+50) = 0.5 A
c)30-ohm and the capcitor are in paralel and the voltage across them is same
= 30*0.5 = 15V
d) charge on the capacitor Q=CV = 15*10e-6 = 150 e-6 C
e) when the switch is opened the capcitor start discharging through 30 + 15 ohms
time const RC = 45*10e-6 = 450e-6 s
voltage across the capacitor V(t) = Vo exp(-t/RC)
for t= 0.693RC the voltage is half of the intial value
= 0.693*450 = 311.85 us
f) with the dielctric the capcitance C increase by k
C = 5*10 = 50 uF
new time const RC = 45*50e-6 = 22.5e-4 s
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