Twocapacitors C 1 = 4.0 µF and C 2 = 12µF are connected in series across 24 V ba

Question

Twocapacitors C1 = 4.0 µF and C2 = 12µF are connected in series across 24 V battery. Theyare carefully disconnected so that they are not discharged and arereconnected to each other with the positive plate to positive plateand negative plate to negative plate.

a. What is the potential difference across each capacitorafter they are reconnected?

b. How much charge is stored in each capacitor after they arereconnected?

c. Find the total initial and final energy stored in thecapacitors. Is energy conserved

     in this process?

I know to use these equations just not surewhere

Qe=VCe

and Qtotal=2Qe =>Ce=C1+C2 => V=Q/C1+C2

Explanation / Answer

Let Ct is the equivalent capacitor of the series. 1/Ct=1/C1+1/C2 so Ct=3F Q=U*C so Q=3C*24=7,2e-5(C) after they reconnect , Q'=2Q and now C1 is parallel to C2 Ct'=C1+C2=16F U=Q/C so U'=Q'/Ct' = 9(V) b) Q=U*C so the charge on C1 : Q1=UC1=3,6e-5 Q2=UC2=1,08e-4(C) c)The total energy W=U^2C/2 so W1=24^2*3F/2=8,64e-4(J) W2=9^2*16F/2=6,48e-4(J) The energy is lost.

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