DC circuits (emf, battery power, resistance)

  • Thread starter travh2007
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Homework Statement



Why is the following situation impossible? A battery has an emf of E = 9.2V and an internal resistance of r = 1.2 ohms. A resistance R is connected across the battery and extracts from it a power of P = 21.2W.

Homework Equations



My book gives these equations throughout the section:
[tex]\Delta[/tex]V = [tex]\epsilon[/tex] - Ir

[tex]\epsilon[/tex] = IR + Ir

I = [tex]\epsilon[/tex]/(R+r)

P=I[tex]\Delta[/tex]V


The Attempt at a Solution



I tried plugging in some of my values into some of these formulas, but I don't know the current, so therefore any formula with I in it, I couldn't do. Can someone steer me in the correct direction somehow please?
 

Answers and Replies

  • #2
gneill
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20,909
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Presumably they mean that the external resistor is to dissipate 21.2W .

Here are some additional relationships for power that you might find useful in your career:

P = I*V = I2R = V2/R

You should be able to write an expression for the power dissipated in the external resistor given its value, and determine the conditions for maximum possible dissipation.
 

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