Two capacitors C 1 = 7.2 F, C 2 = 18.3 F are charged individually to V 1 = 17.1

Question

Two capacitors C1 = 7.2 F, C2 = 18.3 F are charged individually to V1 = 17.1 V, V2 = 3.8 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?

Pick one capacitor, you know the charge on this capacitor before they were connected. Now that you know the potential difference after they are connected - remember the potential drop is the same for both of them - you can calculate the charge: Q=CV. You have to calculate the difference of Q before and after.

Explanation / Answer

Final potential difference:


Charge equals capacitance times voltage.

Q1 = C1 * V1 = 7.2 x 10^-6 * 17.1 = 12.3 x 10^-5

Q2 = C2 * V2 = 18.3 x 10^-6 * 3.8 = 6.95 10^-5

The total charge on the two caps is19.25 x 10^-5.

Since the combined parallel capacitance is 25.5 x 10^-6 or 2.55 x 10^-5, that corresponds to a voltage (once they are connected) of 19.2 x 10^-5 / 2.55 x 10^-5 or7.54volts.


Charge flow:

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

Q1 = C1 * V1 = 7.2 x 10^-6 * 7.54= 5.42 x 10^-5

Since its charge started out at 12.3x 10^-5, it has lost 6.88 x 10^-5 (which the larger cap has gained).

Q2 = C2 * V2 = 18.3 x 10^-6 * 7.54 = 13.79x 10^-5 which is 6.88x 10^-5 greater than its initial 6.91 x 10^-5 charge.


Total stored energy:

Energy is one half times the capacitance times times voltage squared.

J = (C * V^2) / 2

J1 = (7.2 x 10^-6 * 17.1^2) / 2 = 10.5 x 10^-4 J

J2 = (18.3 x 10^6 * 3.8^2) / 2 = 13.2 x 10^-4 J

Total energy separately = 13.2 x 10^-4 J


When paralleled:

J = (25.5 x 10^-6 * 7.54^2) / 2 = 7.24 x 10^-4 J

reduction in stored energy = 5.96 x 10^-4 J

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