How Does Battery EMF Affect Power Dissipation in a DC Circuit?

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Homework Help Overview

The discussion revolves around the effects of battery electromotive force (emf) on power dissipation in a direct current (DC) circuit involving two batteries with different emfs and internal resistances.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the emfs of the batteries and the power dissipated in the circuit, with one participant questioning their initial analysis and seeking clarification on the energy conversion process in the smaller battery.

Discussion Status

Some participants have provided insights into the application of Kirchhoff's voltage law and have derived an expression for the energy conversion in the smaller battery. There is acknowledgment of the reasoning presented, but no explicit consensus on the correctness of the final answer.

Contextual Notes

Participants are navigating the complexities of internal resistance and the specific configuration of the batteries, with some uncertainty about the implications of their calculations and assumptions regarding energy dissipation.

eprparadox
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Homework Statement


The positive terminals of two batteries with emf’s of E1 and E2, respectively, are connected together. Here E2 > E1. The circuit is completed by connecting the negative terminals. If each battery has an internal resistance r, the rate with which electrical energy is converted to chemical energy in the smaller battery is:

A. E1^2/r
B. E1^2/2r
C. (E2 − E1)*E1/r
D. (E2 − E1)*E1/2r
E. E2^2/2r


Homework Equations


P = V^2/R


The Attempt at a Solution


I'm not entirely sure how to analyze this. It seems to me that the power dissipated by the resistors if (E2 + E1)^2/2r. I know that's not what the question is asking for, but is that correct by itself? Whatever energy is created chemically will be dissipated in the internal resistance of the batteries.

Now, if I was subtract off the rate of chemical energy provided by the E2 (the larger) source, then I should get the remaining energy, which is that provided by the E1 (the smaller) source. This is probably all jibberish, but if anyone has any ideas pointing me in the right direction that'd be great.

Thanks a lot for the help.
 
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eprparadox said:
It seems to me that the power dissipated by the resistors if (E2 + E1)^2/2r. I know that's not what the question is asking for, but is that correct by itself?

You get E1+E2 when the positive terminal of one battery is connected to the negative terminal of the other. This is not the case here. Read the problem.
 
Alright, my fault. But I think I get it now.

So I ended up using kirchhoffs voltage law over the loop that would be if you were to draw this out and I came up with:

E2 - i*r - i*r - E1 = 0. I multiplied by i to both sides in order to get an energy (rate of energy) equation and I also brought the E1 to the other side because we're looking for the energy that E1 converts to chemical energy:

E2*i - 2*i^2*r = E2*i. Now, i is just equal to (E2 - E1) / 2*r. I substitute that in and get as an answer:

(E2 - E1)*E1 / *2r which is answer D. Does this reasoning sound ok?
 
Sounds OK.
 

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