6. 4/6 points | Previous Answers SerPSE9 28.P075.scln. My Notes Ask Your Teache

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6. 4/6 points | Previous Answers SerPSE9 28.P075.scln. My Notes Ask Your Teache In the figure below, suppose the switch has been closed for a time interval sufficiently long for the capacitor to become fully charged. (Assume R1 = 11.0 kQ, R2 = 17.0 kQ, R3 = 8.00 kQ, and C = 12.0 R2 9.00 V (a) Find the steady-state current in each resistor. 11 = 321.429 12= 321.429 (b) Find the charge Qmax on the capacitor. 65.57 (c) The switch is now opened at t = 0, write an equation for the current in R2 as a function of time. (Use the following as necessary: t. Do not enter units in your answers. Assume the current is in microamperes, and t is in seconds.) 321.429ex 0.200 (d) Find the time interval required for the charge on the capacitor to fall to one-fifth its initial value. 442.595 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. ms

Explanation / Answer

given that the switch has been closed for a long time

hence the capacitors will now act as open switches

R1 = 11,000 ohm

R2 = 17,000 ohm

R3 = 8000 ohm

C = 12*10^-6 F

a. current in R1 = R2 = i

V = i(R1 + R2)

9/28,000 = i = 321.4285 micro A

current in R2 = 321.4285 micro A

curretn in R3 = 0 A

b. voltage drop across capacitor = V - iR1 = 9 - 321.4285*11,000*10^-6 = 5.464 V

hence

Qmax = C*Vc = 65.571 micro C

c. when the switch is opened, the battery gets out of circuit and the capacitor discharges

RC constant = (R2 + R3)C = 25000*12*10^-6 = 0.3

hence

current through R2 = i

i = (Qmax/RC)e^(-t/RC) = 0.21857*e^(-3.33t) milli A

d. Q = Qmax*e^(-3.33t)

Q/Qmax = 1/5 = e^(-3.33t)

ln(5) = 3.33t

t = 0.4833 s

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