Is Thermal Energy Equal to Work Done by Battery in Electric Circuits?

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SUMMARY

The discussion centers on the relationship between thermal energy and work done by a battery in electric circuits, specifically when a capacitor is involved. Participants assert that for non-ideal batteries, the work done includes both heat loss in the circuit and energy stored in the capacitor, while for ideal batteries, the work done does not result in thermal energy due to the absence of internal resistance. Key equations discussed include the work done by the battery as QV = Q(E - ir) and thermal energy across a resistor as i²Rt. The consensus is that the definitions of work and energy storage vary based on the ideality of the battery.

PREREQUISITES
  • Understanding of electric circuits and components, including capacitors and resistors.
  • Familiarity with the concepts of EMF (Electromotive Force) and internal resistance in batteries.
  • Knowledge of electrical power equations, specifically work done (QV) and thermal energy (i²Rt).
  • Basic principles of energy conservation in electrical systems.
NEXT STEPS
  • Study the impact of internal resistance on battery performance in real-world applications.
  • Learn about energy storage mechanisms in capacitors and their role in circuit behavior.
  • Explore the differences between ideal and non-ideal battery models in electrical engineering.
  • Investigate thermal management techniques in electric circuits to mitigate heat loss.
USEFUL FOR

Electrical engineers, physics students, and anyone interested in the dynamics of electric circuits, particularly in understanding the interplay between thermal energy and work done by batteries.

cupid.callin
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Homework Statement


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The Attempt at a Solution



1) Ummm ... well ...no idea :-p

2) I guess no. Because if a capacitor is connected with battery, work done by batery is QE (EMF) but thermal energy is just QE/2. Rest QE/2energy is stored in capacitor.

3) Same as (2)

4) i = E/(R+r)
--- r is internal resistance of battery
--- R is external resistance.
--- E is EMF
Work done by battery = QV = Q(E - ir) where r is internal resistance of battery
Thermal energy across in resistor = i2Rt = E2RT/(R+r)2

Now i don't know what to do.
But i still guess that for non-ideal its no and for ideal its yes ... don't know why

5) Well i thought of some explanation before i started typing but now i can't remember it :biggrin:

6) Yes, only for non ideal battery. At time of charging, V = E + ir
 

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In #1, the U = i*V*t formula does not tell the story of current dependence because the V across the resistor depends on current, V = i*R

I disagree with your answer for #2, especially when the resistor is connected. The work is converted to heat but we still say the work is "done" by the battery.

In #3 I would argue that the work done by the battery includes both heat lost in the circuit and electrical energy stored in the capacitor. Could be wrong, though.

#4 could be a matter of definition. To me, "the battery" includes its internal resistance. The work done by the battery is the i*V*t for the current coming out of the combination. I would answer YES to the first part and no change for the ideal battery.

No idea on #5! Good thinking on #6 - I missed it.
 
I think #5 is technically incorrect. Its same like using E.M.F. when we know that there is no such 'force'.
Heat is developed in a circuit (or resistor) due to the collision of moving electrons with the ions of the conductor and not due to the difference in temperature across the resistor.
 
Most welcome! We worked really well together.
 
Thank you guys for help!

But for #3, wouldn't energy stored in capacitor + heat developed be equal to work done by battery.

for#4, heat will also be developed in battery if its non-ideal due to its resistance, so shouldn't the heat developed in the resistor + battery be equal to battery's work?
and for ideal one, due to no resistance, no heat will develop in battery,,,,
 

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