Two capacitors C 1 = 3.9 F, C 2 = 17.2 F are charged individually to V 1 = 17.9
ID: 3162958 • Letter: T
Question
Two capacitors C1 = 3.9 F, C2 = 17.2 F are charged individually to V1 = 17.9 V, V2 = 4.9 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
C1 = 3.9 F= 3.9 x10 -6 F
C2 = 17.2 F= 17.2 x10 -6 F
V1 = 17.9 V
V2 = 4.9 V.
(a). final potential difference across the plates of the capacitors once they are connected is
V= [C1V1+C2V2]/[C1+C2]
= [(3.9 x10 -6)(17.9)+(17.2 x10 -6 )(4.9)]/[(3.9 x10 -6)+(17.2 x10 -6 )]
= 7.302 volt
(b). Charge in C1 capacitor before connecting Q1 = C1V1 = (3.9 x10 -6)(17.9)
Charge in C1 capacitor after connecting Q = C1V = (3.9 x10 -6)(7.302)
Required answer = Q1-Q = 41.32 x10 -6 coulomb
(c).Energy stored before connect U = (1/2)C1V1 2 +(1/2)C2V2 2
= (1/2)(3.9 x10 -6)(17.9) 2+(1/2)(17.2 x10 -6 )(4.9) 2
= 8.312x10 -4 J
Energy stored after connection U ' = (1/2)(C1+C2) V 2
= (1/2) [(3.9 x10 -6)+(17.2 x10 -6 )] (7.302) 2
= 5.625 x10 -4 J
Required answer = U - U '
= 2.686x10 -4 J
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