Two capacitors C 1 = 4.6 F, C 2 = 16.8 F are charged individually to V 1 = 18.0

Question

Two capacitors C1 = 4.6 F, C2 = 16.8 F are charged individually to V1 = 18.0 V, V2 = 6.8 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

Expression for charge:

Q = C. V

Find charge on first capacitor

Q1 = C1 * V1 = 4.6 x 10^-6 * 18 = 8.28 x 10^-5

Find charge on second capacitor

Q2 = C2 * V2 = 16.8 x 10^-6 * 6.8 = 1.1424 x 10^-4

Find total charge: Q = Q1 + Q2 = 1.97x 10^-4

As capacitors are in parallel, total capacitance is

C = C1 + C2 = 4.6 F + 16.8 F = 2.14 x 10^-5 F

Therefore, final potential difference across the plates

V = Q / C = ( 1.97x 10^-4 ) / ( 2.14 x 10^-5 F ) = 9.207 V

In three significant digits, V = 9.21 V

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The charge on the smaller capacitor is its capacitance times the final voltage:

Q1 = C1 * V1 = 4.6 x 10^-6 * 9.207 V = 4.235 x 10^-5 C

As its charge started out at 8.28 x 10^-5, it has lost 4.04 x 10^-5 C , it is equal to gain of larger capacitance

The charge after connected

Q2 = C2 * V2 = 16.8 x 10^-6 * 9.207 = 1.54 x 10^-4 which is 4.04 x 10^-5 greater than its initial 16.8 x 10^-6 charge.

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Find energy stored in each capacitor by using the following formula

J = (C * V^2) / 2

J1 = (4.6 x 10^-6 * 18^2) / 2 = 7.45 x 10^-4 J

J2 = (16.8 x 10-^6 * 6.8^2) / 2 = 3.88 x 10^-4 J

Total energy = J1 + J2 = = 1.165 x 10^-3 J

When paralleled:

J = (21.4 x 10^-6 * 9.207^2) / 2 = 8.94 x 10^-4 J

Therefore, the reduction in stored energy = 2.71 x 10^-4 J

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