A potential difference V = 4.760×102 V is applied to two capacitors connected in

Question

A potential difference V = 4.760×102 V is applied to two capacitors connected in series. One capacitor, C1, is 2.70 ?F and the other, C2, is 7.60 ?F. What is the charge on the positive plate of C1?What is the potential difference across C1?

What is the charge on the positive plate of C2?

What is the potential difference across C2?

What is the potential difference across C1?

What is the charge on the positive plate of C2?

What is the potential difference across C2?

Suppose the charged capacitors had been reconnected with plates of opposite sign together. What then would be the steady-state potential difference for C1 ? (Do not try this at home!)

Explanation / Answer

C1 and C2 are connected in series.

so there will be same charge flow through both capacitor.

suppose potential across C1 is V! and across C2 is V2.

and Q = CV

hence. C1 V1 = C2 V2

and V1 + V2= V

V1 + C1 V1 / C2 = V

V1 = C2 V / (C1 + C2) = 7.60 x 476 / (7.60 + 2.70)

V1 = 351.22 volt   .......Ans

V2 = V - V1 = 124.78 volt .......Ans

Q1 = Q2 = C1V1 = C2V2 = 948.30 uC .........Ans

--------------------------------------------

total charge will be conserved.

and they are connected in parallel hence PD across C1 and C2 will be same.

C1 V + C2 V = Q

V (C1 + C2) = 2 x 948.3

V = 184.14 Volt

V1 = V2 =184.14 volt .........Ans

Q1 = C1 V = 2.70 uF x 184.14 =497.17 uC   ......Ans

Q2 = C2 V = 7.60 x 184.14 = 1400 uC ..

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