Question

Two capacitors C₁ = 5.1 ?F, C₂ = 19.8 ?F are charged individually to V₁ = 14.9 V, V₂ = 3.4 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.

C) By how much (absolute value) is the total stored energy reduced when the two capacitors are connected?

Part a is 5.76V

Part b is 4.66E-5 C

Please help me with part c! In J please

Answer 1

Final potential difference:


Charge equals capacitance times voltage.

Q1 = C1 * V1 = 5.1 x 10^-6 * 14.9= 7.6x 10^-5

Q2 = C2 * V2 = 19.8 x 10^-6 * 3.4 = 6.7 x 10^-5

The total charge on the two caps is 14.3 x 10^-5.

Since the combined parallel capacitance is 24.9x 10^-6 or 2.49 x 10^-5,, that corresponds to a voltage (once they are connected) of 14.3 x 10^-5 / 2.49 x 10^-5 or 5.76 volts.


Charge flow:

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

Q1 = C1 * V1 = 5.1 x 10^-6 * 5.76 = 2.94 x 10^-5

Since its charge started out at 7.6 x 10^-5, it has lost 4.7 x 10^-5 (which the larger cap has gained).

Q2 = C2 * V2 = 19.8 x 10^-6 * 5.76= 11.40 x 10^-5 which is 4.7 x 10^-5 greater than its initial 6.7 x 10^-5 charge.


Total stored energy:

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

J = (C * V^2) / 2

J1 = (5.1 x 10^-6 * 14.9^2) / 2 = 5.66x 10^-4 J

J2 = (19.8 x 10^-6 * 3.4^2) / 2 = 1.14 x 10^-4 J

Total energy separately = 6.8 x 10^-4 J


When paralleled:

J = (24.9 x 10^-6 * 5.76^2) / 2 = 4.13 x 10^-4 J

reduction in stored energy = 2.67 x 10^-4 J