Problem with discharging capacitors in a circuit SA 50 We study what happens whe

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-o) 40 v 30 15 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 h. c) Find the voltage over the capacitor d) What is the charge on the capacitor? At t-0 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-oo). 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 capacitor.

Explanation / Answer

a) I3 = 0

because, after a long time the capacitor acts as open ckt.

b) I1 = I2 = 40/(50 + 30)

= 0.5 A

c) voltage across the capacitor, Vc = voltage across 30 ohm resistor

= 30*0.5

= 15 V

d) charge on the capacitor, Q = C*Vc

= 10*15

= 150 micro C or 1.5*10^-4 C

e) when the switch is opened the capacitor is discharged through 15 and 30 ohm.

time constant, T = R*C

= (15 + 30)*10

= 450 micro C or 4.5*10^-4 s

so, use,

q = Qmax*e^(-t/T)

let at time t, q = Qmax/2

Qmax/2 = Qmax*e^(-t/T)

1/2 = e^(-t/T)

t = -T*ln(1/2)

= -4.5*10^-4*ln(1/2)

= 3.12*10^-4 s

f) the capacitor will be discharged slower.

new time constant, T' = R*C'

= R*5*C

= 5*R*C

= 5*T

= 5*4.5*10^-4

= 2.25*10^-3 s or 2.25 ms

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