Consider a capacitor that has a capacitance C1. It is charged by connecting it t

Question

Consider a capacitor that has a capacitance C1. It is charged by connecting it to a battery of V volts. After it is completely charged, it is disconnected from the battery and connected to another capacitor of capacitance C2. Before the connection, the second capacitor is completely discharged.

(a) What is the charge on the first capacitor when it is charged by the battery?

(b) How much charge remains on the first capacitor after it is connected to the second capacitor?

(c) What is the potential difference across the second capacitor after it is connected to the first capacitor?

(d) What is the energy stored in the first capacitor when it is completely charged by the battery?

(e) When the two capacitors are connected to each other, how much enegy is in each one of the capacitors?

Explanation / Answer

a)

Charge on 1st battery, Q = C1*V

b)

Let the charge on the 1st battery after it is conncted to 2nd battery be Q'

and by charge conservation, the charge on the second capacitor = Q - Q'

Now, Voltage across each is same .

So, Q'/C1 = (Q-Q')/C2

So, Q'(1/C1+1/C2) = Q/C2

So, Q' = (C1*C2/((C1+C2)*C2))*Q = (C1/(C1+C2))*Q <--------answer

c)

Potential difference across second capacitor, V = (Q-Q')/C2 = Q/(C1+C2)

d)

energy stored in first capacitor = 0.5*C1*V^2

e)

Energy in 1st capacitor, E1 = Q'^2/(2*C1) =0.5*(C1^3*V^2/(C1+C2)^2)

Energy in 2nd capacitor , E2 = (Q-Q')^2/(2*C2) = 0.5*(C1^2*C2/(C1+C2)^2)*V^2

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