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. S 12.0 kn 10.0µF 9.00 V R=15.0 kn 3.00 kN Select one: O a. Q = 12µC(1-e/0.15) O b. Q= 90µCe t/0.15 O c. Q= 10µCe t/0.15 O d. Q = 50µCe/0.18 O e. Q= 90µCe-t/0.03 O f. Q= 15µC(1 - e 4/0.18) Q = 90µC(1 - e /0.15) O h. Q= 10µC(1 – e 0.03)

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. S 12.0 kn 10.0µF 9.00 V R=15.0 kn 3.00 kN Select one: O a. Q = 12µC(1-e/0.15) O b. Q= 90µCe t/0.15 O c. Q= 10µCe t/0.15 O d. Q = 50µCe/0.18 O e. Q= 90µCe-t/0.03 O f. Q= 15µC(1 - e 4/0.18) Q = 90µC(1 - e /0.15) O h. Q= 10µC(1 – e 0.03)

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

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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