Three capacitors having capacitances of 9.0 F , 8.2 F , and 4.9 F are connected
ID: 1333031 • Letter: T
Question
Three capacitors having capacitances of 9.0 F , 8.2 F , and 4.9 F are connected in series across a 33-V potential difference.
(A)What is the charge on the 4.9 F capacitor?
(B) What is the total energy stored in all three capacitors? U=?
(C)The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? V=?
(D) What is the total energy now stored in the capacitors?
Explanation / Answer
Let C1 = 9 micro F
C2 = 8.2 micro F
C3 = 4.9 micro F
A) 1/Cnet = 1/9 + 1/8.2 + 1/4.9
Cnet = 2.29 micro F
the charge on each capaciotr, Q = Cnet*V
= 2.29*33
= 7.55*10^-5 C or 75.5 micro C
B)
U_total = Q^2/(2*Cnet)
= (7.55*10^-5)^2/(2*2.29*10^-6)
= 1.24*10^-3 J
C) Net charge on 3 capacitrs, Qnet = 3*Q
= 3*7.55*10^-5
= 2.265*10^-4 C
Cnet = C1 + C2 + C3
= 9 + 8.2 + 4.9
= 22.1 micro F
Vnet = Qnet/Cnet
= 2.264*10^-4/(22.1*10^-6)
= 10.24 Volts
D) U_total = 0.5*Cnet*Vnet^2
= 0.5*22.1*10^-6*(10.24)^2
= 1.16*10^-3 J
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