- #1
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?