A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V

Question

A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V emf in series with a resistor known as the "internal resistance", as shown in the figure . A typical battery has 1.0 Ohm internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.5 V, the value of the emf. Suppose the terminals of this battery are connected to a 2.4 Ohm resistor.

What fraction of the battery's power is dissipated by the internal resistance? change in P/P=%

I found the potential difference between terminals to be 1.1 V.
I can't seem to get the correct percentage used though. it would seem like it would be P=I^2*R but I get 0.19 and I said 19% and it's wrong.

Explanation / Answer

If is the emf of the battery, r is the internalresistor and R is the exernal resistor to which the battery isconnected is V= R / ( R+r) = 1.5(2.4)/ [2.4 +1 ] = 1.058volts The power dissipated throough internal resisitanceis P = I^2 r Current in the circuit is I = / (R+r) So the power is P = [ / (R+r ]^2 r = [ 1.5 / 3.4 ]^2 (1 ohm) = 0.194 W

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