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(FigureJ.). A typical battery has 1.0 Ohms 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 2.0 ohms resistor.

Explanation / Answer

(a)With 2 ohms connected, the current is 1.5V/3 ohms = 0.5 amps. the voltave drop across the internal resistor is then IR = 0.5 amps x 1 ohm = 0.5 volts
So the terminal voltage is 1.5 -1 = 1 volt.
(B) Internal power dissipated = R x I^2 the internal power lost is 1.0 ohms x ( 0.5 amps)^2 = 0.25 watts. The total power drain on the battery is 3 ohms x (0.5 amps)^2 = 0.75 watts.
Hence, fraction dissipated internally = 0.25 W / 0.75 W = 1/3 W

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