connected to a battery with emf The circuit in the figure below contains two res

Question

connected to a battery with emf The circuit in the figure below contains two resistors, R1 2.30 and R2 are no charges on the capacitors before switch S is closed. 3.20 kn, and two capacitors, C1-2.40 F and C2 3.10 105 V. There Ri C1 R2 C2 (a) Determine the charge on capacitor Ci as a function of time (in ms), after the switch is closed. (Use the following as necessary: t.) Q1-KNO Response) C (b) Determine the charge on capacitor C2 as a function of time (in ms), after the switch is closed.(Use the following as necessary: t.) 92-No Response) HC

Explanation / Answer

R12 = equivalent resistance = R1 R2 /(R1 + R2) = (2.3) (3.2)/(2.3 + 3.2) = 1.34 kohm

equivalent capacitance is given as

C = C1 + C2 = 2.4 + 3.1 = 5.5 uF

T = time constant = R12 C = (1.34 x 103) (5.5 x 10-6) = 0.0074

Voltage across the combination of capacitor at any time is given as

V = E (1 - e-t/T)

V = 105 (1 - e-t/0.0074)

V = 105 (1 - e-135.14t)

charge on C1 is given as

Q1 = C1 V = (2.4 x 10-6) (105) (1 - e-135.14t) = (252) (1 - e-135.14t)

b)

Q2 = C2 V = (3.1 x 10-6) (105) (1 - e-135.14t) = (326) (1 - e-135.14t)

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