Two capacitors C 1 = 7.9 F, C 2 = 19.3 F are charged individually to V 1 = 15.9

Question

Two capacitors C1 = 7.9 F, C2 = 19.3 F are charged individually to V1 = 15.9 V, V2 = 8.0 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

A) charge on C1, Q1 = C1V1 = 7.9uF x 15.9 = 125.61 uC

charge on C2, Q2 = C2V = 19.3 uF x 8   = 154.4 uC

total charge on system = 125.61 + 154.4 = 280.01 uC

when capacitors are connected in parallel connection , then finally potential diff across them will
be same say V.

now using charge conservation, Q1' + Q2' = 280.01 uC ..........(i)

Q1' = C1V and Q2' = C2V

putting in eqtn,

C1V + C2V = 280.01 uC

V ( 7.9 + 19.3)uF = 280.01

V = 10.29 Volts

B) Q1' = 10.29 x 7.9 = 81.33 uC

Q2' = 10.29 x 19.3 = 198.60 uC

from C1 = 125.61 - 81.33 = 44.28 uC to C2.

c) initial stored energy = C1 V1^2 / 2 + C2 V2^2 /2

= ( 7.9 x 10^-6 x 15.9^2 / 2 ) + ( 19.3 x10^-6 x 8^2 / 2)

= 1.62 x 10^-3 J

final stored energy = (C1 + C2) V^2 /2 = (7.9 + 19.3)uF x 10.29^2 /2

= 1.44 x 10^-3 J

energy reduced = 1.62 - 1.44) x 10^-3 = 0.18 x 10^-3 J

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