Power in a simple electric circuit

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SUMMARY

The discussion centers on the maximum power transfer theorem in electric circuits, specifically demonstrating that maximum power occurs when the load resistance (R) equals the internal resistance (r) of the battery. The derived formula for maximum power is confirmed as Pmax = ε2 / 4r, where ε represents the electromotive force (EMF) of the battery. The calculations show that the total power delivered by the battery is based on the current and the EMF, emphasizing the distinction between terminal voltage and EMF in power calculations.

PREREQUISITES
  • Understanding of Ohm's Law and circuit analysis
  • Familiarity with the concept of internal resistance in batteries
  • Knowledge of power calculations in electrical circuits
  • Basic grasp of the maximum power transfer theorem
NEXT STEPS
  • Study the maximum power transfer theorem in detail
  • Learn about the implications of internal resistance in battery performance
  • Explore advanced circuit analysis techniques using Kirchhoff's laws
  • Investigate the relationship between voltage, current, and power in AC circuits
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Electrical engineering students, circuit designers, and anyone interested in optimizing power delivery in electrical systems.

Karol
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Homework Statement


A circuit consists of a battery with internal resistance r and a resistor with resistance R.
It has been found that the maximum power generated is when R=r.
Prove that in those conditions ##P_{max}=\frac{ε^2}{4r}##

Homework Equations


Power invested in a resistor: P=i2R
Power invested between 2 points in a circuit: P=Vi

The Attempt at a Solution


The current: ##i=\frac{V}{2r}##
Voltage drop on the internal resistance: ##V=ir=\frac{V}{2}##
Power between the terminals of the battery: ##P=iV=\frac{V}{2r} \frac{V}{2}=\frac{V^2}{4r}##
But the power loss on one resistor (the internal or the external):
$$P=i^2r=\frac{V^2}{4r^2}\cdot r=\frac{V^2}{4r}$$
It's identical to the total power, and there are 2 resistors so they consume double the power, how come?
 
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Karol said:
It's identical to the total power, and there are 2 resistors so they consume double the power, how come?
The total power delivered by the battery is the current times the battery EMF (which is V), not the terminal voltage (V/2).
 
So:
$$P_{max}=\varepsilon\cdot i=\varepsilon\cdot \frac{\varepsilon}{2r}=\frac{\varepsilon^2}{2r}$$
The book asked about the maximum power of the battery and i guess this is the answer. the answer of the book is:
$$P_{max}=\frac{\varepsilon^2}{4r}$$
 
Karol said:
So:
$$P_{max}=\varepsilon\cdot i=\varepsilon\cdot \frac{\varepsilon}{2r}=\frac{\varepsilon^2}{2r}$$
The book asked about the maximum power of the battery and i guess this is the answer. the answer of the book is:
$$P_{max}=\frac{\varepsilon^2}{4r}$$

The maximum power the question is asking about is the maximum power delivered to the load, that's the external resistor only.
 
Thanks, i guess you are right. but is there, in general, a meaning to the multiplication of the terminals of the battery voltage and the current P=V⋅i? not in this specific problem but in general. i guess it's the power delivered to the rest of the circuit, right?
 
Last edited:
Karol said:
Thanks, i guess you are right. but is there, in general, a meaning to the multiplication of the terminals of the battery voltage and the current P=V⋅i? not in this specific problem but in general. i guess it's the power delivered to the rest of the circuit, right?
Yes. That's the power that the battery is delivering [to the external circuit]. In this problem, and in general.
 
Je Suis Charlie also, thanks
 

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