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 -w- 10.0 μΕ 9.00 V R- 15.0 kN 3.00 kn Select one: O a. Q= 90µC(1 – e/0.15a) O b. Q= 10µC(1-e/0.0) c. Q= 12µC(1-e/0.15%) %3D d. Q = 10µCe t/0.15a O e. Q= 90µCet/0.15 f. Q= 50µCe /0.03 Q = 50µCe /0.18 15 uC(1 et/0.18)
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 -w- 10.0 μΕ 9.00 V R- 15.0 kN 3.00 kn Select one: O a. Q= 90µC(1 – e/0.15a) O b. Q= 10µC(1-e/0.0) c. Q= 12µC(1-e/0.15%) %3D d. Q = 10µCe t/0.15a O e. Q= 90µCet/0.15 f. Q= 50µCe /0.03 Q = 50µCe /0.18 15 uC(1 et/0.18)
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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 pF
9.00 V
Ro = 15.0 kn
3.00 kn
Select one:
a. Q= 90µC(1–e/0.15«)
O b. Q= 10µC(1 – e /0.03 )
O c. Q= 12µC(1 –e /0.15)
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O d. Q = 10µCe-t/0.15s
O e. Q= 90µCe t/0.15s
f. Q= 50µCe /0.03s
Q = 50µCe /0.18s
O h. Q= 15µC(1 – e /0.18 )
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