Two capacitors C1 = 3.9 F, C2 = 18.2 F are charged individually to V1 = 18.1 V,
ID: 581843 • Letter: T
Question
Two capacitors C1 = 3.9 F, C2 = 18.2 F are charged individually to V1 = 18.1 V, V2 = 6.3 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
C1:
W = (1/2)CV² = (1/2)(3.9e-6)(18.1)² = 0.0006388Joules
Q = CV = (3.9e-6)(18.1) = 0.0007059 Coulombs
C2:
W = (1/2)CV² = (1/2)(18.2e-6)(6.3)² = 0.0003611Joules
Q = CV = (19.6e-6)(6.3) = 0.00012348Coulombs
Before connecting them in parallel:
W = W1 + W2 = 0.0006388 Joules + 0.0006388Joules = 0.0012776 Joules
Q = Q1 + Q2 = 0.0007059 Coulombs + 0.00012348Coulombs = 0.00082938Coulombs
After connecting in parallel, the total charge will remain the same.
C1 + C2 =3.9 uF + 18.2 uF = 22.1 uF
Q is still equal to 0.00082938Coulombs.
Q = CV
V = Q/C = 0.00082938Coulombs/22.1 uF = 3.75 volts
B. You go ahead an calculate the charge in each capacitor with the new voltage. Compare this to the original charge in each cap and you can determine how much charge flowed which way.
C. Calculate the stored energy using the new voltage and cap total. W = (1/2)CV². This amount of energy will be less than the sum of the two original energies. Simply subtract for the answer.
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