Two capacitors (C_1 = 4.6 mu F, C_2 = 14.1 mu F) are charged individually to (V_

Question

Two capacitors (C_1 = 4.6 mu F, C_2 = 14.1 mu F) are charged individually to (V_1 = 19.5 V, V_2 = 3.5 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? Due in 19 hours, 20 minutes

Explanation / Answer

Charge equals capacitance times voltage.

Q1 = C1 * V1 = 4.6 x 10^-6 * 19.5= 8.97x 10^-5

Q2 = C2 * V2 = 14.1 x 10^-6 * 3.5 = 4.90 x 10^-5

The total charge on the two caps is 13.87 x 10^-5.

Since the combined parallel capacitance is 18.7 x 10^-6 or 1.87 x 10^-5, that corresponds to a voltage (once they are connected) of 13.87 x 10^-5 / 1.87 x 10^-5 or 7.42 volts.


Charge flow:

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

Q1 = C1 * V1 = 4.6 x 10^-6 * 7.42 = 3.41 x 10^-5

Since its charge started out at 8.4 x 10^-5, it has lost 5.56 x 10^-5 (which the larger cap has gained).

Q2 = C2 * V2 = 14.1 x 10^-6 * 7.42 = 1.04 x 10^-4 which is 5.56 x 10^-5 less than its initial 4.9 x 10^-5 charge.


Total stored energy:

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

J = (C * V^2) / 2

J1 = (4.6 x 10^-6 * 19.5^2) / 2 = 8.75 x 10^-4 J

J2 = (14.1 x 10^-6 * 19.5^2) / 2 = 2.68 x 10^-4 J

Total energy separately = 11.43 x 10^-4 J


When paralleled:

J = (18.7 x 10^-6 * 7.42^2) / 2 = 5.15x 10^-4 J

reduction in stored energy = 5.15x 10^-4 J

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