Calculating Energy in a 10μF Capacitor Charged to 24V

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A 10μF capacitor charged to 24V stores energy calculated using the formula E = 1/2 CV^2, resulting in a stored energy of 2.88 mJ. The energy required from the battery to charge the capacitor differs due to energy dissipation in the form of heat in the circuit's resistance. This discrepancy arises because not all energy from the battery is converted to stored energy in the capacitor; some is lost during the charging process. To calculate the energy dissipated, one must consider the source impedance and the resulting heat loss. Understanding these principles highlights the importance of energy conservation in capacitor charging scenarios.
dan greig
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A 10 micro farad capacitor is fully charged by connecting a 24v battery.

How much energy is stored in the capacitor?

How much energy would be required from the battery to charge the capacitor to the 24v level?

Explain why the answers are different.
 
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What you have got so far? Any ideas?
 
I think I have found the answer to the first question using,

E = 1/2 CV^2 or E = 1/2 QV

and I think the reason the two answers wouldn't be the same is because the rate of charge decreases as the charge increases therefore the capacitor can never fully charge.

I don't see how I can get a different answer in the second question?
 
dan greig said:
I think I have found the answer to the first question using,

E = 1/2 CV^2 or E = 1/2 QV

and I think the reason the two answers wouldn't be the same is because the rate of charge decreases as the charge increases therefore the capacitor can never fully charge.

I don't see how I can get a different answer in the second question?
This is a trick question that comes up often in basic capacitor problems. Assume some source impedance for the battery, and calculate the energy that is burned in that resistance as the battery charges the cap. Then try a different source resistance and re-calculate...do you see a pattern?
 
well, the second answer is different coz there might be energy dissipation in the form of heat in the capacitors and the wires connecting them...Thats all I can think of.Otherwise the energy given out by the battery should be equal to the energy gained by the capacitor-energy conservation. What do u say?
 
rammstein said:
well, the second answer is different coz there might be energy dissipation in the form of heat in the capacitors and the wires connecting them...Thats all I can think of.Otherwise the energy given out by the battery should be equal to the energy gained by the capacitor-energy conservation. What do u say?




This is the route i think i have to go down, how would i calculate the amount of energy dissipated?

Is it half the amount of energy procuced by the battery is converted into charge in the capacitor? If so is there a formula to calculate this?
 

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