The circuit in Figure 2- contains two resistors, R_1 = 2.00 k Ohm and R_2 = 3.00

Question

The circuit in Figure 2- contains two resistors, R_1 = 2.00 k Ohm and R_2 = 3.00 k Ohm, and two capacitors, C_1 = 2.00 mu F and C_2 = 3.00 mu F, connected to a battery with emf epsilon = 120 V. No charge is on either capacitor before switch S is closed. Determine the charges q_1 and q_2 on capacitors C_1 and C_2, respectively, after the switch is dosed. (Suggestion: First reconstruct the circuit so that it becomes a simple RC circuit containing a single resistor and single capacitor in series, connected to the battery, and then determine the total charge q stored in the equivalent circuit)

Explanation / Answer

Since capacitor are in parallel

net capacitanceCnet==C1+C2=5 uF

Net Resistance =Rnet=(R1*R2)/(R1+R2)=1.2 kohm

now charge across equivalent circuit

q=Qo(1-e^(-t/Rnet*Cnet)=Qo(1-e^(-t/0.006)

Qo=Cnet*V=600 uC

q=(600uC)(1-e^(-t/0.006)

charge in C1=(C1/(C1+C2))*q=(240uC)(1-e^(-t/0.006)

charge in C2=(C2/(C1+C2))*q=(360uC)(1-e^(-t/0.006)

after long time

charge on C1=240 uC

charge on C2=360 uC

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