(a) Find the steady-state current in each resistor. I 1 = In Figure P28.67, supp

Question

(a) Find the steady-state current in each resistor.
I1 = In Figure P28.67, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged. ( e = 8.90 V, r1 = 10 k?, and r2 = 17 k?.) Find the steady-state current in each resistor. Find the charge Q on the capacitor. The switch is opened at t = 0. Write an equation for the current IR2 in R2 as a function of time. Find the time that it takes for the charge on the capacitor to fall to one-fifth its initial value.

Explanation / Answer

(a) For steady state condition there is no current through the resistor 3 k?. So I (3k?) = 0 ?A
The current passing through the remaining two resistors is same and is
I1 = I2 = V / (R1 + R2)
= 8.9 V / (27 x 103?)
= 329.629 uA

(b) The charge on the capacitor is
Q = C I2 R2
= 10 x 10-6x 329.629 x 10-6x 17 x 103
= 56.037x 10-6C
= 56.037uC
(c) time constant ? = (R2 + R3) C = 20 x 103x 10 x 10-6= 0.200
Therefore I = I2 e- t /t
= ( 286 e^-t/0.200 )

(d) Time taken t = - (R2 + R3) C ln(1 /5)
= (R2 + R3) C ln(5)
= 0.200 x ln(5)
= 321.8 ms

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