If a capacitor is charged the energy stored in it is, say E1.
E1 = 0.5*C1*V²
The charged capacitor is connected to another capacitor.
We have to find energy lost in the process.
In my textbook it is given that energy stored in second capacitor would be E2=0.5*C*V² where C is the resultant capacitance of combination.
Now why the hell the resultant capacitance is taken into account.
E2 must be 0.5*C2*V²
Where C2 is the capacitance of the second capacitor.
Please Explain in detail.
I am totally confused please tell me what actually is the energy stored in a capacitor.
Help! Capacitor problem.?
2015-07-13 12:02 pm
回答 (2)
2015-07-13 2:17 pm
Eo = 1/2 C1*V^2
If i put C2 in parallel to C1, then the equivalent circuit is given by C1 in series to C2 fed the voltage source V...so :
Ce = C1//C2 = C1C2/(C1+C2)
Q = Ce*V = V*(C1C2)/(C1+C2)
V1 = V*C2/(C1+C2)
V2 = V*C1/(C1+C2)
E = 1/2Ce*V^2
E1 = 1/2*C1*V1^2 = C1*V^2*C2^2/((C1^2+C2^2+2C1C2)*2)
E2 = 1/2*C2*V2^2 = C2*V^2*C1^1/((C1^2+C2^2+2C1C2)*2)
E1/E2 = (C1*V^2*C2^2/((C1^2+C2^2+2C1C2)*2))*(((C1^2+C2^2+2C1C2)*2)/(C2*V^2*C1^2) =
= C1*C2^2/(C2*C1^2) = C2/C1
in case C1 = C2 ...then
Ce = C1/2
E = Eo/2
E1 = E2 = E/2 = Eo/4
If i put C2 in parallel to C1, then the equivalent circuit is given by C1 in series to C2 fed the voltage source V...so :
Ce = C1//C2 = C1C2/(C1+C2)
Q = Ce*V = V*(C1C2)/(C1+C2)
V1 = V*C2/(C1+C2)
V2 = V*C1/(C1+C2)
E = 1/2Ce*V^2
E1 = 1/2*C1*V1^2 = C1*V^2*C2^2/((C1^2+C2^2+2C1C2)*2)
E2 = 1/2*C2*V2^2 = C2*V^2*C1^1/((C1^2+C2^2+2C1C2)*2)
E1/E2 = (C1*V^2*C2^2/((C1^2+C2^2+2C1C2)*2))*(((C1^2+C2^2+2C1C2)*2)/(C2*V^2*C1^2) =
= C1*C2^2/(C2*C1^2) = C2/C1
in case C1 = C2 ...then
Ce = C1/2
E = Eo/2
E1 = E2 = E/2 = Eo/4
2015-07-13 2:39 pm
During a transfer between capacitors, there is conservation of the charges, but not of the energy: the transfer does not occur without energetic exchanges with the environment.
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