In the circuit shown in the figure the switch has been closed for a long time so that the capacitor is fully charged. At t=0 the switch is opened. Write an expression for the charge on the capacitor as a function of time. 12.0 kN 10.0 μF 9.00 V Ro = 15.0 kN 3.00 kN

Physics for Scientists and Engineers, Technology Update (No access codes included)
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Author:Raymond A. Serway, John W. Jewett
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Chapter26: Capacitance And Dielectrics
Section: Chapter Questions
Problem 26.24P: Consider the circuit shown in Figure P26.24, where C1, = 6.00 F, C2 = 3.00 F. and V = 20.0 V....
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In the circuit shown in the figure the switch has been closed for a long time so that the capacitor is fully charged. At t=0 the switch is opened. Write an expression for the
charge on the capacitor as a function of time.
12.0 kN
10.0 µF
9.00 V
Ro = 15.0 kN
3.00 kN
Select one:
О а. Q3 15дС(1 — е 40.186)
O b. Q = 10µC(1 – e t/0.03% )
Q = 90µC(1 – e t/0.15s)
O d. Q = 50µCe t/0.03s
Q = 12µC(1 – e t/0.15s)
e.
Of.
Q = 90µCe t/0.15s
Q = 10µCe t/0.15s
O h. Q = 50µCe t/0.18s
a O o o O O O
Transcribed Image Text:In the circuit shown in the figure the switch has been closed for a long time so that the capacitor is fully charged. At t=0 the switch is opened. Write an expression for the charge on the capacitor as a function of time. 12.0 kN 10.0 µF 9.00 V Ro = 15.0 kN 3.00 kN Select one: О а. Q3 15дС(1 — е 40.186) O b. Q = 10µC(1 – e t/0.03% ) Q = 90µC(1 – e t/0.15s) O d. Q = 50µCe t/0.03s Q = 12µC(1 – e t/0.15s) e. Of. Q = 90µCe t/0.15s Q = 10µCe t/0.15s O h. Q = 50µCe t/0.18s a O o o O O O
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