Two capacitors C 1 = 5.6 F, C 2 = 15.1 F are charged individually to V 1 = 18.0

Question

Two capacitors C1 = 5.6 F, C2 = 15.1 F are charged individually to V1 = 18.0 V, V2 = 5.7 V. The two capacitors are then connected together in parallel with the positive plates together and the negative plates together.
a) Calculate the final potential difference across the plates of the capacitors once they are connected.

b) Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together.

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

Explanation / Answer

The charges on each are
Q1 = CV = 5.6x 18V

= 100.8µC
Q2 = CV = 15.1µF x 5.7v

= 86.07µC

when they are connected together the total charge is the sum, 186.87µC
Total C is the sum, 20.7µF
V = Q/C = 186.87 / 20.7µF

= 9.027 volts

b)

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

Q1 = C1 * V1

= 5.6* 9.027

= 50.552 micro coloumb.

Charge flown will be

100.8-50.552

=50.248 micro coloumb

c)

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

J = (C * V2) / 2

J1 = (5.6 x 10-6 * 182) / 2

= 9.072 x 10-4 J

J2 = (15.1 x 10-6 * 5.72) / 2 = 2.45 x 10-4 J

Total energy separately = 11.52 x 10-4 J


When paralleled:

J = (20.7 x 10-6 * 9.0272) / 2 = 8.43 x 10-4 J

reduction in stored energy = 3.09x 10-4 J

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