Four capacitors are arranged in the circuit shown in the figure. The capacitors

Question

Four capacitors are arranged in the circuit shown in the figure. The capacitors have the values C1 = 34.5 ?F, C2 = 13.5 ?F, C3 = 45.5 ?F, C4 = 85.5 ?F, and the power supply is set at voltage V = 30.5 V. What is the equivalent capacitance of the circuit?

Four capacitors are arranged in the circuit shown in the figure. The capacitors have the values C1-34.5 F, C2 = 13.5 lf, C3 = 45.5 F C4-85.5 , and the power supply is set at voltage V= 30.5 V What is the equivalent capacitance of the circuit? Number C. What is the charge on capacitor C3? Number 4 What is the potential difference across capacitor C4? Number What is the potential energy stored in capacitor C2? What is the potential energy stored in capacitor z? Number

Explanation / Answer

Here from the diagram C2,C3 and C4 are in sereis then the equivalent capaciatance is given by

1/Ceq =1/C2+1/C3+1/C4 =1/13.5 F+1/45.5 F+1/ 85.5 F =0.0740+0.0219+0.01169

Ceq =9.29*uF

Now Ceq is in parallel with C1 , then total equivalent capacitance is given by

Ceqtotal =C1+Ceq =34.5+9.29 =43.79uF

Now C1 and Ceq are in parallel in parallel combination voltage across the capacitors will be the same

Given that voltage is (V) =30.5V

Now C2 ,C3 and C4 are in series then charge remains the same but voltage across the each capacitor will vary

Now the total charge across the circuit is given by

Q =Ceqtotal*V =43.79uF*30.5 =1335.595uC =1.333*10-3C =1.33mC

Now the charge across the capacitance C1 is given by

Q1 =C1V =34.5*30.5 =1052.25uC =1.052mC

Now the total charge is Q =Q1+Q2

Now Q2 =Q-Q1 =1.33mC -1.052mC =0.277mC is the charge across the capacitor C3

The potential difference across C4 is gvien by

Q =CV

V4 =Q2/C4 =0.277*10-3/85.5*10-6 =3.24V

Now the potential energy stored across the capacitor C2 is given by

U =(1/2)Q2/C2 =(0.277*10-3)2/2*13.5*10-6 =2.84*10-3J =2.84mJ

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