Part A Two capacitors of capacitance 6.00 ? F and 8.00 ?F are connected in paral

Question

Part A

Two capacitors of capacitance 6.00 ?F and 8.00 ?F are connected in parallel. The combination is then connected in series with a 12.0-V voltage source and a 14.0-?F capacitor, as shown in the figure.

(a) What is the equivalent capacitance of this combination?

Part B

(b) What is the charge on the 6.00-?F capacitor?

Part C

(c) What is the potential difference across the 6.00-?F capacitor?

Part A Two capacitors of capacitance 6.00 ?F and 8.00 ?F are connected in parallel. The combination is then connected in series with a 12.0-V voltage source and a 14.0-?F capacitor, as shown in the figure. (a) What is the equivalent capacitance of this combination? Part B (b) What is the charge on the 6.00-?F capacitor? Part C (c) What is the potential difference across the 6.00-?F capacitor?

Explanation / Answer

Here,
In parallel for capacitor ,
Ceq = c1 + C2
and in series ,

1/ Ceq = 1/c1 + 1/c2 + ...

A)

For 6 and 8 uF in Parallel ,

Cp = 6 + 8

Cp = 14 uF

Now , Cp and 14 uF are in series

1/Ceq = 1/14 + 1/14

Ceq = 7 uF

the equivalent capacitance is 7 uF

B)

Now , Qeq = Ceq*V

Qeq = 7 * 12

Qeq = 84 uC

Now , charge on 6 uF

Q(6 uF) = 6 * 84/(6 + 8)

Q(6 uF) = 36 uC

the charge on 6 uF capacitor is 36 uC

C)

as Q = CV

for 6 uF capacitor

36 = V*6

V = 6 V

the potential difference across the capacitor is 6 V

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