In the figure below, a 17.0 V battery is connected across capacitors of capacita

Question

In the figure below, a 17.0 V battery is connected across capacitors of capacitances with the following values.

C1 = C6 = 2.80 ?F
C3 = C5 = 4.40 ?F
C2 = C4 = 2.2 ?F

I have a-b, but having trouble with the rest. Thanks.

In the figure below, a 17.0 V battery is connected across capacitors of capacitances with the following values C2 = C4 = 2.2 C5 C2 C4 (a) What is the equivalent capacitance Ceg? 3.03 IF (b) What is the charge stored by Ceq? 51.51e-6 (c) What is V1 of capacitor 1? 4.59 (d) What is q1 of capacitor 1? (e) What is V2 of capacitor 2? (f) What is q2 of capacitor 2? (g) What is V3 of capacitor 3? (h) What is q3 of capacitor 3?

Explanation / Answer

Here C2, C4 are in parallel and both are in parallel with series combination of C3 and C5.

The voltage across C2 = voltage across C4 = voltage across c3+voltage across c5.

Equivalent capacitance of c2, c4 and c3+c5:

Ceq_1= c2||c4||(c3 series c5) = 2.2+2.2+4.4||4.4 = 6.6 uF

Now c1 and c6 are in parallel and equivalent capacitance will be c1+c6,

Ceq_2 = 2.8+2.8 = 5.6 uF

a)

hence equivalent capacitance of the circuit = 3.03 uF

b) Charge stored by Ceq = Ceq*V = 17*3.03 = 51.5 uC

c) In series charges on each capacitor will be same,

hence Ceq_1 and Ceq_2 will have same charge and that is equal to 51.5 uC

hence voltage across C1 = voltage across Ceq_1 = Q/C = 51.5/5.6 = 9.2 V

d) charge in capacitor 1 = q1 = 9.2*2.8 = 25.75 uC

e) V2 on capacitor 2 = 51.5/6.6 = 7.8 V

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