Two C 1 = 3.7 μF, C 2 = 18.7 μF are charged individually to V 1 = 14.2 V, V 2

Question

Two C1 = 3.7 μF, C2 = 18.7 μF are charged individually to V1 = 14.2 V, V2 = 7.0 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

In parallel, they will naturally come to an equilibrium where voltage is the same. Mathematically, Q=VC and Q is conserved. So you have:
V1C1 + V2C2 = V(C1+C2) where V is the equilibrium voltage.

Use Q=VC to calculate the amount of charge before and after being wired together.

Stored energy, U = 1/2 C V^2. Just calculate it before and after...
the change would be 1/2 * (C1*V1*V1 + C2*V2*V2) - 1/2 * V^2 * (C1+C2), where I have chosen to write V1^2 as V1*V1 to avoid confusion.

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