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 2.5 Ohm resistor. What is the potential difference between the terminals of the battery? Express your answer using two significant figures. What fraction of the battery's power is dissipated by the internal resistance? Express your answer using two significant figures.

Explanation / Answer

The internal resistance is effectively in series with the external load of 2.5 ohms.
So the combined resistance is 2.5 + 1 = 3.5 ohms.
Current = v/r = 1.5/3.5 = 0.429 amp
drop across internal resistance = I x R = 0.429 x 1 = 0.429 V
P.D = 1.5 - 0.429 = 1.071 V
Power dissipated = I ² x R = 0.429 ² x 1 = 0.184 watts
Power supplied = I ² x R(total) = 0.429² x 3.5 = 0.644 W
So 0.184 / 0.644 = 0.286 is the fraction of power dissipated in the internal resistance

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