Two capacitors C1 = 7.6 F, C2 = 18.5 F are charged individually to V1 = 17.7 V,

Question

Two capacitors C1 = 7.6 F, C2 = 18.5 F are charged individually to V1 = 17.7 V, V2 = 5.0 V. The two capacitors are then connected together in parallel with the positive plates together and the negative plates together.

1) Calculate the final potential difference across the plates of the capacitors once they are connected.

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

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

Explanation / Answer

1) Initial charge on the capacitors,

Q1 = C1*V1 = 7.6*17.7 = 134 micro C

Q2 = C2*V2 = 18.5*5 = 92.5 micro C

net capaciatnce when they connected paralle,

Cnet = C1 + C2

= 7.6 + 18.5

= 26.1 micro F

Qnet = 134 + 92.5

= 226.5 micro C

final potential difference across the plates, Vnet = Qnet/Cnet

= 226.5/26.1

= 8.68 volts

2) charge on each capacitor after connected in parallel,

Q1' = C1*Vnet

= 7.6*8.68

= 66 micro C

amount of charge (absolute value) that flows from one capacitor to the other,
Q1 - Q1' = 134 - 66

= 68 micro C <<<<<<<<<-----------Answer

3) U - U' = 0.5*C1*V1^2 + 0.5*C2*V2^2 - 0.5*Cnet*Vnet^2

= 0.5*7.6*10^-6*17.7^2 - 0.5*18.5*10^-6*5^2 - 0.5*26.1*10^-6*8.68^2

= 2.4*10^-5 J <<<<<<<<<-----------Answer

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