Three capacitors having capacitances of 8.1 ? F , 8.2 ? F , and 4.6 ? F are conn

Question

Three capacitors having capacitances of 8.1?F , 8.2?F , and 4.6?F are connected in series across a 40-V potential difference.

1.) What is the charge on the 4.6?F capacitor?

2.) What is the total energy stored in all three capacitors?

3.) 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?

4.) What is the total energy now stored in the capacitors?

ANSWER ALL FOR BEST ANSWER

Explanation / Answer

Cnet in series = 1/Cnet = 1/C1 +1/C2 +1/C3

so

a. 1/Cnet = (1/8.1) +(1/8.2) +(1/4.6)

Cnet -= 2.16 uF

in series same charge exists

so

Qnet = CV = 40* 2.16 = 86.43 uC
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U    = 0.5 cv^2

U = 0.5 * 2.16e-6* 40*40

U = 6.914 mJ
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C) when they are connected in prallel they will have potentail difference across them

Qnet = 3*Qs

= 3*86.43*10^-6

= 259.29 micro C

Cnet = 8.1+8.2+4.6

= 20.9 uF

V = Qnet/Cnet

= 259.29/20.96

= 12.405 volts
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D) 0.5*Cnet*V^2

= 0.5*20.9*10^-6*12.405*12.406

U = 1.608 mJ

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