Power in a simple electric circuit

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Homework Help Overview

The discussion revolves around a circuit consisting of a battery with internal resistance and an external resistor. Participants are exploring the conditions under which maximum power is delivered to the load and the relationship between the resistances and power output.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the derivation of maximum power conditions, questioning the relationship between internal and external resistances. There is an exploration of the power equations and how they relate to the total power delivered by the battery.

Discussion Status

Some participants have provided insights into the definitions of power in the context of the circuit, while others are questioning the assumptions made regarding the power delivered to the load versus the total power from the battery. Multiple interpretations of the problem are being explored without a clear consensus.

Contextual Notes

There is a focus on the distinction between the power delivered to the external resistor and the total power from the battery, with some participants expressing uncertainty about the implications of these definitions in general circuit analysis.

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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