The capacitor below is initially uncharged, and the switch is closed at t = 0. R

Question

The capacitor below is initially uncharged, and the switch is closed at t = 0. R1 Is 20 ohms. R2 Is 30 ohms, and R3 is 40 ohms. The battery is a 12V ideal battery, and the capacitor is 37.5 nF. Which of the circuits below is equivalent to the above circuit after the switch has been closed a long time? Explain. After the switch has been closed a very long time: What is the charge on the capacitor? How much energy is stored on the capacitor? The switch is opened at some new t=0s (after it has been closed a very long time). At what time has the charge on the capacitor decreased to 15% of its original value?

Explanation / Answer

a) figure 2

because after a long time capacitor acts as open ckt. there will not any effect of R2 in the ckt.

b)

Potentail across capaciotor, Vc = potentail across R3

= V*R3/(R1+R3)

= 12*40/(20 + 40)

= 8 volts

so, charge on capcator, Q = Vc*C

= 8*37.5*10^-9

= 3*10^-7 F or 300 nF or 0.3 micro F

c) UC = 0.5*C*Vc^2

= 0.5*37.5*10^-9*8^2

= 1.2*10^-6 J

d) Time constant, T = (R2+R3)*C

= (30+40)*37.5*10^-9

= 2.625*10^-6 s

now Apply,

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

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

0.15 = e^(-t/T)

e^(t/T) = 1/0.15

t = T*ln(1/0.15)

= 2.625*10^-6*ln(1/0.15)

= 4.98*10^-6 s

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