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. (script E = 9.50 V, r1 = 10 k, and r2 = 17 k.)

In Figure P28.67, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged (8 9.50 v ri 10 kn, and rz 17 kn.) o o W 10.0 AF 3.00 k2 Figure P28.67 (a) Find the steady-state current in each resistor. HA HA 13-kn HAA (b) Find the charge Q on the capacitor. (c) The switch is opened at t 0. Write an equation for the current IR2 n R2 as a function of time (299 HA)e-t/(0.200 s) O (352 uA)e-t/(0.200 s) (352 HA)et/(0.200 s) (299 HA)et/(0.200 s) (d) Find the time that it takes for the charge on the capacitor to fall to one-fifth its initial value. mS

Explanation / Answer

In steady state condition

Current in 3kohm is zero

current through r1 and r2 are same

I1 = I2 = e/(r1+r2) = 9.5/((10+17)*1000) = 3.52*10^-4 A = 352 uA

I3_kohm = 0 uA

B) charge is Q = C*dV = 10*10^-6*(9.5-(352*10^-6*10*1000)) = 59.8 uC

C) When switch is opened

current does not passes through r1

So maximum current passes through r2 and 3kohm is Imax = dV/R = (9.5-(352*10^-6*10*1000))/(17000+3000) = 299 uA

time constant is T = R*c = (17000+3000)*10*10^-6 = 0.2 sec

So equation is i = Imax*e^(-t/T)

i = 299 uA*e^(-t/0.2)

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d) q = Q*e^(-t/T)

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

1/5 = e^(-t/0.2)

t = 0.321 sec = 321 mSec

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