Two capacitors C 1 = 5.9 ?F, C 2 = 16.2 ?F are charged individually to V 1 = 15.

Question

Two capacitors C1 = 5.9 ?F, C2 = 16.2 ?F are charged individually to V1 = 15.7 V, V2 = 4.9 V. The two capacitors are then connected together in parallel with the positive plates together and the negative plates together.
Calculate the final potential difference across the plates of the capacitors once they are connected.

Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together.

By how much (absolute value) is the total stored energy reduced when the two capacitors are connected?

Explanation / Answer

PART-A

The charges on each are

Q1 = CV = 5.9uF x 15.7v = 92.63uC

Q2 = CV = 16.2uF x 4.9v = 79.38uC

when they are connected together the total charge is the sum, 172.01uC

Total C is the sum, 5.9+16.2 =22.1uF

V = Q/C

V= 172.01uC / 22.1uF

V= 7.783 volts

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PART-B

Charge flow from 1st capacitor = C1 V1 - C1 V

Charge flow from 1st capacitor = 5.9uF x 15.7v -5.9 x7.783

Charge flow from 1st capacitor = 5.9*106-6( 15.7-7.783)

Charge flow from 1st capacitor=4.257*10^-5 C

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2nd capacitor = C2V2 - C2V

2nd capacitor = 16.2uF x 4.9v -16.2 x 7.783

2nd capacitor =16.2*10^-6(4.9-7.783)

Charge flow from 2nd capacitor=-4.67*10^-5 C

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PART-3

E1 = Q1^2 / 2C1

E1= (92.63uC)^2 /2*5.9 =7.2714*10^-4 J

E2 = Q2^2 / 2C2

E2 = (79.38uC)^2 /2*16.2 =1.9448*10^-4 J

E = Q^2 / 2C

E =(172.01uC)^2/2*22.1 =6.6939*10^-4 J

Energy change = E1 + E2 - E

Energy change = 7.2714*10^-4 +1.9448*10^-4- 6.6939*10^-4

Energy change =10^-4( 7.2714+1.9448-6.6939)

Energy change =2.5223*10^-4 J

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