(1) Three 5 mF capacitors (small) and three 10 mF capacitors (large) are linked
ID: 1782167 • Letter: #
Question
(1) Three 5 mF capacitors (small) and three 10 mF capacitors (large) are linked to form the network shown below. (a) (i) Compute the effective capacitance of the upper branch of this network. (i) Compute the effective capacitance of the lower branch of this network. (ii) Compute the effective capacitance of the entire network. (b) Suppose that a 3V battery is connected to this system and sufficient time has elapsed for electrostatic equilibrium to be established. (i) Determine the total amount of charge held on the network. (ii) Determine the amount of charge held on the upper branch of the network (ii) Determine the amount of charge held on the lower branch of the network. (e) Suppose that a 3V battery is connected to this system and sufficient time has elapsed for electrostatic equilibrium to be established. (G) Determine the total amount of energy stored in the network. (ii) Determine the amount of energy stored in the upper branch of the network. (ii) Determine the amount of energy stored in the lower branch of the network.Explanation / Answer
(A) (i) 3 capacitor are in series.
1 / Ceq = 1/C1 + 1/C2 + 1/C3
1/C1 = 1/5 + 1/5 + 1/10
C1 =2 mF
(ii) 1/C2 = 1/10 +1 /10 + 1/5
C2 = 2.5 mF
(iii) now C1 and C2 are in parallel,
Ceq = C1 + C2 = 4.5 mF
(B) (i) Q = Ceq V = 13.5 mC
(ii) Q1 = Q C1 / (C1 + C2) = 6 mC
(iii) Q2 = Q - Q1 = 7.5 mC
(C) (i) U = C V^2 /2 = (4.5 x 10^-3) (3^2)/2
= 0.0202 J
(ii) U1 = (6 x 10^-3)^2 / (2 x 2 x 10^-3)
= 0.009 J
(iii) U2 = U - U1 = 0.0112 J
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