Two capacitors C1 = 4.7 F, C2 = 15.1 F are charged individually to V1 = 15.6 V,

Question

Two capacitors C1 = 4.7 F, C2 = 15.1 F are charged individually to V1 = 15.6 V, V2 = 8.0 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

Final potential difference:

The charges on each are

Q1 = CV = 4.7µF x 15.6v = 7.33µC

Q2 = CV = 15.1µF x 8.0v =120.8µC

when they are connected together the total charge is the sum, 194.12µC

Total C is the sum, 19.8*10^-5

V = Q/C = 194.12*10^-5 / 19.8*10^-5 = 9.80 volts

Charge flow:

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

Q1 = C1 * V1 = 4.7 x 10^-6 * 7.6 = 3.431 x 10^-5

Since its charge started out at 3.43x 10^-5, it has lost 3.89 (which the larger cap has gained).

Q2 = C2 * V2 = 15.6 x 10^-6 * 7.6= 11.38 x 10^-5 which is .7 x 10^-5 greater than its initial 15.6 x 10^-5 charge.


Total stored energy:

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

J = (C * V^2) / 2

J1 = (4.7 x 10^-6 * 15.61^2) / 2 = 5.7 x 10^-4 J

J2 = (15.1 x 10^6 *8^2) / 2 = 4.8 x 10^-4 J

Total energy separately = 10.5 x 10^-4 J


When paralleled:

J = (24.4 x 10^-6 * 9.67^2) / 2 = 11.4 x 10^-4 J

reduction in stored energy = 1.7 x 10^-4 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.