In Figure P28.67, suppose the switch has been closed for a length of time suffic

Question

In Figure P28.67, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged. (r = 8.80 v, 12 kQ, and r2 = 16 kQ.) 10.0 F 3.00 Figure P28.67 (a) Find the steady-state current in each resistor (b) Find the charge Q on the capacitor. (c) The switch is opened at t = 0, write an equation for the current la, in R2 as a function of time. O (314 A)e-ti(0.190 s) O (265 )--0(0.190 s) (265 A)en(0.190 s) (314 A)et/(0.190s) (d) Find the time that it takes for the charge on the capacitor to fall to one-fifth its initial value ms

Explanation / Answer

a) After a long time the capacitor acts as open circuit.

so, I1 = I2 = V/(r1+r2)

= 8.8/(12000 + 16000)

= 314 micro A

I3 = 0

b) voltage across capacitor = voltage across r2

= V2

= I2*r2

= 314*10^-6*16000

= 5.02 V

charge on capacitor, Q = C*V2

= 10*10^-6*5.02

= 50.2 micro C

C) when switch is open current exponentially decreases.

Time constant of the ckt, T = (r2 + 3000)*C

= (16000 + 3000)*10*10^-6

= 0.19 s

so, I = Imax*e^(-t/T)

= 314 micro A)*e^(-t/0.19)

so 1st option is the correct answer.

d)

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

Qmax/5 = Qmax*e^(-t/T)

1/5 = e^(-t/T)

ln(1/5) = -t/T

==> t = -T*ln(1/5)

= -0.19*ln(1/5)

= 0.306 s

= 306 ms

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