A 3uF cap is charged by a 12V battery. It is disconnected from the battery and then connected to an uncharged 5uF cap. Determine the total stored energy (a) before the two capacitors are connected and (b) after they are connected. (c) What is the change in energy?
a) For this part I used U=(1/2)CV^2, and got the correct answer (2.2 x 10^-4 J)
b) I need some help here. What I did was use the formula for capacitance C=Q/V, to find the charge Q on the first cap - which I found to be 36uC. Then, when the first cap is connected to the second cap, the charge should be conserved, correct? So I calculated the total capacitance of the system:
For 2 caps in series: Total C = (C1*C2) / (C1 + C2) = 1.875uF
Then I used C=Q/V for the whole system, and plugged in the total capacitance (1.875uF) and the conserved charge Q (36uC) to find the voltage when the 2 caps are connected - and the answer I got is 19.2V (which doesn't make sense, does it?). So then plugging the total capacitance, total charge, and voltage of the system into U=(1/2)CV^2 I got:
3.456 x 10^(-4) J for the energy stored when the caps are connected.
Which again doesn't make sense because it should be smaller. The answer the book gives is 8.1 x 10^(-5) J.
Any help would be appreciated. Please explain the concepts involved with the answer because I'm having trouble understanding what's going on here.
Cheers.
a) For this part I used U=(1/2)CV^2, and got the correct answer (2.2 x 10^-4 J)
b) I need some help here. What I did was use the formula for capacitance C=Q/V, to find the charge Q on the first cap - which I found to be 36uC. Then, when the first cap is connected to the second cap, the charge should be conserved, correct? So I calculated the total capacitance of the system:
For 2 caps in series: Total C = (C1*C2) / (C1 + C2) = 1.875uF
Then I used C=Q/V for the whole system, and plugged in the total capacitance (1.875uF) and the conserved charge Q (36uC) to find the voltage when the 2 caps are connected - and the answer I got is 19.2V (which doesn't make sense, does it?). So then plugging the total capacitance, total charge, and voltage of the system into U=(1/2)CV^2 I got:
3.456 x 10^(-4) J for the energy stored when the caps are connected.
Which again doesn't make sense because it should be smaller. The answer the book gives is 8.1 x 10^(-5) J.
Any help would be appreciated. Please explain the concepts involved with the answer because I'm having trouble understanding what's going on here.
Cheers.
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You used the wrong formula for total capacitance.
When hooked together, they are in parallel, and (for capacitors, not resistors) Ctot = C1 + C2 = 8µF.
Try the rest of it that way.
When hooked together, they are in parallel, and (for capacitors, not resistors) Ctot = C1 + C2 = 8µF.
Try the rest of it that way.
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The following is derived from conservation of charge
and the fact that both capacitors have the same potential
difference after connection:
Ui = (1/2)*C*V^2 = 2.16*10^-4 J
Uf = Ui * [3/(3+5)] = 8.1*10^-5 J
Change in energy = Ui - Uf = 1.35*10^-4 J
If you're still having trouble, leave detail.
and the fact that both capacitors have the same potential
difference after connection:
Ui = (1/2)*C*V^2 = 2.16*10^-4 J
Uf = Ui * [3/(3+5)] = 8.1*10^-5 J
Change in energy = Ui - Uf = 1.35*10^-4 J
If you're still having trouble, leave detail.