Two capacitors C1 = 3.2 µF, C2 = 17.4 µF are charged individually to V1 = 14.8 V

Question

Two capacitors C1 = 3.2 µF, C2 = 17.4 µF are charged individually to V1 = 14.8 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

C1 = 3.2 µF, C2 = 17.4 µF V1 = 14.8 V, V2 = 4.9 V Q1 = C1V1 = 47.6µC Q2 = C2V2 = 85.26µC The two capacitors are then connected together in parallel with the positive plates together and the negative plates together. so their charge would be distributed equally on both of them so charge on each one, Q = (47.6+85.26)/2 = 66.43µC the final potential difference across the plates of the capacitors once they are connected. V1' = Q/C1 = 66.43/3.2 = 20.75volt V2' = Q/C2 = 66.43/17.4 = 3.81volts amount of charge flown = Q -Q1 = 66.43 - 47.6 = 18.83µC

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