The circuit in the figure below contains two resistors, R1 = 1.90 k ohm and R2 =

Question

The circuit in the figure below contains two resistors, R1 = 1.90 k ohm and R2 = 3.20 kohm, and two capacitors, C1 = 2.10 muF and C2 = 3.50 muF, connected to a battery with emf epsilon = 135 V. There are no charges on the capacitors before switch S is closed. Determine the charge on capacitor C1 as a function of time (in ms), after the switch is closed. (Use the following as necessary: Determine the charge on capacitor C2 as a function of time (in ms), after the switch is closed.(Use the following as necessary: t.)

Explanation / Answer

The equivalent resistance of R1 and R2 is R = (R1.R2)/(R1+R2) = 6.08/5.1 = 1.192 kilo ohms

The charge on the capacitor as a function of time is Q = C.V[1-e(-t/RC)]

Hence, charge on C1 is Q1 = 2.1*10-6*135[1-e(-t/1.192*10^3*2.1*10^-6)] = 283.5*10-6[1-e(-t/.0025)]

Taking t = 1ms, Q1 = 93.46 micro coulombs

Hence, charge on C2 is Q2 = 3.5*10-6*135[1-e(-t/1.192*10^3*3.5*10^-6)] = 472.5*10-6[1-e(-t/.0042)]

Taking t = 1ms, Q2 = 100.1 micro coulombs

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