# DC circuits (emf, battery power, resistance)

## 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:
$$\Delta$$V = $$\epsilon$$ - Ir

$$\epsilon$$ = IR + Ir

I = $$\epsilon$$/(R+r)

P=I$$\Delta$$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?

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gneill
Mentor
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.