Three capacitors having capacitances of 8.6 F , 9.0 F , and 4.2 F are connected

Question

Three capacitors having capacitances of 8.6 F , 9.0 F , and 4.2 F are connected in series across a 39-V potential difference.

What is the charge on the 4.2 F capacitor?

What is the total energy stored in all three capacitors?

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?

What is the total energy now stored in the capacitors?

Explanation / Answer

if connected in series then equivalent capacitance

1/Ceq = 1/C1 + 1/C2 + 1/C3

Ceq = 2.148 x 10^-6 F

Q = Ceq*V

Q = 83.78 x 10^-6 C

in series charge remain same for each capacitor

part a )

Q at . 2 uF capacitor = 83.78 x 10^-6 C

par b )

U = QV/2 = 1.634 x 10^-3 J

part c )

capacitor are disconnected , charge is remain cosntant

Ceq = C1 + C2 + C3

Ceq = 21.8 x 10^-6 F

V = Q/C

V = 3.84 V

it will be same for every capacitor

U = CV^2/2

U = 1.61 x 10^-4 J

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