In Figure P28.71, suppose the switch has been closed for a time interval suffici
ID: 1322976 • Letter: I
Question
In Figure P28.71, suppose the switch has been closed for a time interval sufficiently long for the capacitor to become fully charged. Find (a) the steady-state current in each resistor and (b) the charge Q on the capacitor. (c) The switch is now opened at t 5 0. Write an equation for the current in R2 as a function of time and (d) find the time interval required for the charge on the capacitor to fall to one-fifth its initial value.
In Figure P28.71, suppose the switch has been closed for a time interval sufficiently long for the capacitor to become fully charged. Find (a) the steady-state current in each resistor and (b) the charge Q on the capacitor. (c) The switch is now opened at t 5 0. Write an equation for the current in R2 as a function of time and (d) find the time interval required for the charge on the capacitor to fall to one-fifth its initial value.Explanation / Answer
after a long time of Switch been closed,
Total resisatnce R = 15 and 3 are in parallel
R = 15*3/18
R = 2.5 ohms
Rnet = 2.5 + 12 = 14.5 k ohms
so
current i = v/R
i = 9/14.5 k
i = 0.62 mA
current across 12K ohms is i = 9/12k = 0.75 mA
Current across 15 kohms = 9/15k = 0.6 mA
Current across 3k ohms = 9/3k = 3 mA
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Charge Qmax = CV
Qmax = 10 e-6 * 9
Qmax = 90 uC
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apply Q = Qo (1-e^-t/RC)
e^-t./Rc = 1-(1/5)
e^-t/RC = 0.8
-t./RC = ln (0.8)
-t/RC = -0.223
t = 0.223 * 14500* 10 e-6
t = 32 msecs
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