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(Figure 1) . A typical battery has 1.0 Ω 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 2.3 Ω resistor.

What is the potential difference between the terminals of the battery?

What fraction of the battery's power is dissipated by the internal resistance?

Answer 1

in the circuit, 1 ohm resistance and 2.3 ohm resistance are connected in series.

then net resistance in the circuit=1+2.3=3.3 ohms

current in the circuit=1.5/3.3=0.4545 A

then potential difference between the terminal=1.5-1*0.4545=1.0455 volts

part b:

battery's power=emf*current=1.5*0.4545=0.68175 W

power dissipated in internal resistance=current^2*resistance=0.4545^2*1=0.20657 W

hence fraction of power dissipated through internal resistance=0.20657/0.68175=0.303=30.3%