Two capacitors C1 = 6.3 F, C2 = 13.7 F are charged individually to V1 = 14.1 V,
ID: 1531869 • Letter: T
Question
Two capacitors C1 = 6.3 F, C2 = 13.7 F are charged individually to V1 = 14.1 V, V2 = 5.8 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
Before capacitors are connected.
Q1 = C1*V1
= 6.3*14.1
= 88.83 micro C
Q2 = C2*V2
= 13.7*5.8
= 79.46 C
when They are connected,
Ceq = C1 + C2
= 6.3 + 13.7
= 20 micro F
Qeq = 88.83 + 79.46
= 168.29
potential diffrence between the plates, Veq = Qeq/Ceq
= 168.29/20
= 8.41 volts <<<<<<<<<<<<<<<<--------------------------Answer
amount of charge that flows from one capacitor to another, Q = C1*(V1 - Veq)
= 6.3*(14.1 - 8.41)
= 35.8 micro C <<<<<<<<<<<<<<<<--------------------------Answer
Ui = (1/2)*C1*V1^2 + (1/2)*C2*V2^2
= (1/2)*6.3*14.1^2 + (1/2)*13.7*5.8^2
= 857 micro J
Uf = (1/2)*Ceq*Veq^2
= (1/2)*20*8.41^2
= 707 micro J
Reduced energy stored, delta_U = Ui - Uf
= 857 - 707
= 150 micro J
= 1.5*10^-4 J <<<<<<<<<<<<<<<<--------------------------Answer
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