Power dissipation in capacitor circuit

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In a capacitor circuit connected to a battery with internal resistance, the power dissipated as heat must consider both the voltage across the capacitor and the internal resistance of the battery. The heat dissipated is not solely determined by the capacitor's voltage and current; the internal resistance of the battery also contributes to energy loss. While some discussions suggest that only the capacitor's heat dissipation should be calculated, this overlooks the significant heat lost in the battery. The current does not flow through the capacitor itself but through the connecting wires and battery during charging. Therefore, a comprehensive analysis must include both components to accurately assess total power dissipation.
apkrishna1963
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Capacitor connected to a battery of certain internal resistance. If it is asked to calculate what is power dissipated as heat, Does -the voltage across capacitor X current through capacitor- gives heat dissipated or should we add heat dissipated in battery by considering its internal resistance.
 
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The capacitor will take some part of the energy released by the battery, only the fraction that is lost in the internal resistance gets converted to heat.
 
@mfb, I agree. But in the book the answer given for only heat dissipation in capacitor. They have not considered the dissipation in battery. I think they have neglected this fraction of heat lost. But strictly speaking one has to consider. Am I right?
mfb said:
The capacitor will take some part of the energy released by the battery, only the fraction that is lost in the internal resistance gets converted to heat.
 
apkrishna1963 said:
Capacitor connected to a battery of certain internal resistance. If it is asked to calculate what is power dissipated as heat, Does -the voltage across capacitor X current through capacitor- gives heat dissipated or should we add heat dissipated in battery by considering its internal resistance.
What current "through" capacitor? The current goes through the connecting wires and through the battery, during the charging process. But not through the capacitor.
Maybe you should write the full text of the problem.
 
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It may be shown from the equations of electromagnetism, by James Clerk Maxwell in the 1860’s, that the speed of light in the vacuum of free space is related to electric permittivity (ϵ) and magnetic permeability (μ) by the equation: c=1/√( μ ϵ ) . This value is a constant for the vacuum of free space and is independent of the motion of the observer. It was this fact, in part, that led Albert Einstein to Special Relativity.
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