Capacitor: charge and energy calculations

AI Thread Summary
A capacitor with a capacitance of 12 µF is charged by a 20V DC supply. The charge stored in the capacitor is calculated as 2.4e-4 C, while the energy delivered by the supply is 4.8e-3 J and the energy stored in the capacitor is 2.4e-3 J. The discrepancy between the energy delivered and stored can be attributed to energy lost to resistance in the circuit. As the capacitor charges, the voltage across it is lower than the supply voltage, leading to this loss. Understanding these energy dynamics is crucial for analyzing capacitor behavior in circuits.
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Homework Statement


A capacitor of capacitance 12*10^6 is fully charged from a 20V d.c. supply.
i Calculate the charge stored by the capacitor.
ii calculate the energy delivered by the dc supply
iii calculate the energy stored in the capacitor.
iv account for the difference between your answers for ii and iii.

Homework Equations


i charge is Q=CV
ii I think this is W=QV
iii I think this is W=1/2 CV^2[/B]

The Attempt at a Solution


iv I am not so sure it may be the fact that the dc current is equal to the voltage across the circuit or potential difference or something. I don't really get that.
i 2.4e-4
ii4.8e-3
iii2.4e-3
 

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Apparently this capacitor is actually ##10^{-6}## farads, not ##10^6## as written. With that understanding, your answers for i, ii and iii are correct.

For part iv, consider what is happening as the capacitor is halfway charged. The dc supply (ideally) is putting current out across a 20 V potential difference between its two leads. The capacitor (ideally) has a 10 V potential difference between its two leads. How can one account for the other 10 V of potential difference?
 
LOST TO RESISTANCE ?
 
Yes. You could elaborate on that answer by offering possibilities for where the resistance could exist. Thinking about conservation of energy, you could think about where the energy that is "lost to resistance" goes.
 
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