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Charging capacitor by another capacitor

by bgq
Tags: capacitor, charging
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May7-13, 04:15 PM
P: 144

Consider a charged capacitor connected to a an empty capacitor. The charged capacitor starts to discharge while the empty capacitor starts charging. The whole process finishes when the two capacitors maintain the same potential difference. The problem is that when calculating the energies at the beginning and at the end, we find that they are different. Some books says that the lost energy is dissipated in the connecting wires, yet others say that the energy is transformed into electromagnetic waves.

How can we interpret the loss of the electric energy in this case?
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Philip Wood
May7-13, 05:16 PM
PF Gold
P: 963
Both types of energy transfer will take place, but which one dominates will depend on the details of the system such as the resistance and inductance of the wires used to join the capacitors.
May7-13, 05:30 PM
Sci Advisor
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tiny-tim's Avatar
P: 26,148
hi bgq!

the energy loss equals I2R, so it's certainly the loss through the resistor

from the pf library on capacitor

Inverse exponential rate of charging:

A capacitor does not charge or discharge instantly.

When a steady voltage [itex]V_1[/itex] is first applied, through a circuit of resistance [itex]R[/itex], to a capacitor across which there is already a voltage [itex]V_0[/itex], both the charging current [itex]I[/itex] in the circuit and the voltage difference [itex]V_1\,-\,V[/itex] change exponentially, with a parameter [itex]-1/CR[/itex]:

[tex]I(t) = \frac{V_1\,-\,V_0}{R}\,e^{-\frac{1}{CR}\,t}[/tex]

[tex]V_1\ -\ V(t) = (V_1\,-\,V_0)\,e^{-\frac{1}{CR}\,t}[/tex]

So the current becomes effectively zero, and the voltage across the capacitor becomes effectively [itex]V_1[/itex], after a time proportional to [itex]CR[/itex].

Energy loss:

Energy lost (to heat in the resistor):

[tex]\int\,I^2(t)\,R\,dt\ =\ \frac{1}{2}\,C (V_1\,-\,V_0)^2[/tex]

May9-13, 11:33 AM
P: 144
Charging capacitor by another capacitor

Thank you very much

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