An RC circuit, hooked up to a battery as shown in the figure, starts with an unc

Question

An RC circuit, hooked up to a battery as shown in the figure, starts with an uncharged capacitor. The resistance in the circuit is R = 498.0 ? the capacitor has capacitance of C = 82.0 ?F and the battery maintains the emf of ? = 50.0 V. The switch is closed at time t = 0.0 s and the capacitor begins to charge .the time constant for this circuit = 4.084?10-2 s What is the charge on the capacitor after the switch has been closed for t = 4.33?10-2 s? What is the current through the circuit after the switch has been closed for t = 4.33?10-2 s? What is the voltage across the capacitor after the switch has been closed for t = 4.33?10-2 s?

Explanation / Answer

given t=4.33*10-2S

=4.084*10-2s

at t=o =>Qo=CV =82*10-6*50 =4.1*10-3C

Io=V/R =50/498 =0.1004A

a)Q=Qo(1-e-t/) =4.1*10-3(1-e-(4.33*10^-2/4.084*10^-2))=2.68*10-3C =2.68mC

b)I=Io(1-e-t/) =0.1004(1-e-4.33*10^-2/4.084*10^-2) =0.0656A =65.62mA

C)V=V0(1-e-t/) =50(1-e-4.33*10^-2/4.084*10^-2) =32.68V 32.7V

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