Question

Internal resistance of a $12{\text{V}}$ car battery is $0.02\Omega $. What is the theoretical maximum power which it can deliver?
$1)3600{\text{W}}$
$2)2400{\text{W}}$
$3)100{\text{W}}$
$4)1200{\text{W}}$

Answer 1

Hint: We know that the battery which is used as the source for power consists of some internal resistance to the flow of the current through it which is due to the electrolyte in the battery. A small voltage drop occurs inside the battery and a potential drop across the battery is less than the electromotive force of the battery.

Complete step by step solution:
The internal resistance is the resistance that is present within the battery that resists the current flow when connected to a circuit. Therefore it results in a voltage drop inside the battery when current flows through it. The internal resistance of the battery is due to the electrolyte and the electrodes present in the batteries.
Now to obtain maximum external power from a source battery with a finite internal resistance, the external resistance of the load in the circuit must be equal to the internal resistance of the source battery. Hence the total resistance of the circuit will be twice the internal resistance.
$I = \dfrac{V}{{{R_{IN}} + {R_{EXT}}}} = \dfrac{{12}}{{0.04}}$
Maximum power is given as $P = {I^2}{R_{}}$
Substituting the values we get
$P = {\left( {\dfrac{{12}}{{0.04}}} \right)^2} \times 0.04 = 3600{\text{W}}$
Hence option 1) is the correct option.

Note: The electromotive force is the potential difference of a source when no current is flowing. Terminal voltage is the voltage output of a device measured across its terminals. The terminal voltage of a battery is less than the electromotive force when it is discharging because of the voltage drop across the internal resistance of the device. Electromotive force is the cause and the terminal voltage or the potential drop is the effect.