Calculating Energy in a 10μF Capacitor Charged to 24V

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Homework Help Overview

The discussion revolves around calculating the energy stored in a 10μF capacitor charged to 24V and exploring the difference between the energy stored in the capacitor and the energy supplied by the battery during charging. The subject area includes concepts of capacitors, energy storage, and energy conservation principles in electrical circuits.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the formula E = 1/2 CV^2 for calculating energy stored in the capacitor and question the discrepancy between the energy supplied by the battery and the energy stored. Some suggest considering energy dissipation due to resistance in the circuit and explore how different source impedances might affect the calculations.

Discussion Status

The discussion is active with multiple participants sharing their thoughts on the energy calculations and the reasons for the differences in energy values. There is an exploration of concepts related to energy conservation and dissipation, with participants questioning assumptions about the charging process and the role of resistance.

Contextual Notes

Participants are considering the effects of energy dissipation in the circuit, including heat loss in the wires and capacitors, as well as the implications of varying source resistance on the energy calculations. There is an acknowledgment of the complexity of the problem, with no consensus reached yet.

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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