# Power in a simple electric circuit

1. Feb 13, 2015

### Karol

1. The problem statement, all variables and given/known data
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 $P_{max}=\frac{ε^2}{4r}$

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

3. 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?

2. Feb 13, 2015

### Staff: Mentor

The total power delivered by the battery is the current times the battery EMF (which is V), not the terminal voltage (V/2).

3. Feb 13, 2015

### Karol

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:
$$P_{max}=\frac{\varepsilon^2}{4r}$$

4. Feb 13, 2015

### Dick

The maximum power the question is asking about is the maximum power delivered to the load, that's the external resistor only.

5. Feb 13, 2015

### Karol

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: Feb 13, 2015
6. Feb 13, 2015

### Staff: Mentor

Yes. That's the power that the battery is delivering [to the external circuit]. In this problem, and in general.

7. Feb 14, 2015

### Karol

Je Suis Charlie also, thanks