A capacitor (7.30 F ) is connected in a parallel arrangement with a second capac

Question

A capacitor (7.30 F ) is connected in a parallel arrangement with a second capacitor (2.50 F ) and in series with a 12-V battery. The battery is then removed, leaving the two capacitors isolated.

A).If the smaller capacitor's capacitance is now doubled, by how much does the charge on the larger capacitor change?

B).By how much does the charge on the smaller capacitor change?

C).By how much does the voltage across the larger capacitor change? How much does the voltage change for the smaller one?

Thank you

Explanation / Answer

let C1 = 7.3 micro F

C2 = 2.5 micro F

V = 12 volts

charge on the capacitors when the battery is connected,

Q = Cnet*V

= (C1+C2)*V

= (7.3 + 2.5)*12

= 117.6 micro C

Q1 = V*C1 = 12*7.3 = 87.6 micro C
Q2 = V*C2 = 12*2.5 = 30 micro C

A) when C2 is doubled,

Ceq' = C1 + 2*C2

= 7.3 + 2*2.5

= 12.3 micro F

V' = Q/Ceq'

= 117.6/12.3

= 9.56 volts

A) change in the charge on the larger capacitor change = C1*(V - V')

= 7.3*(12 - 9.56)

= 17.8 micro C

B) change in the charge on the smaller capacitor change = C2'*V' - C2*V

= 5*9.56 - 2.5*12

= 17.8 micro C

C) delta_V1 = delta_V2 = V - V'

= 12 - 9.56

= 2.44 volts

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