The battery in FIGURE $EX28.17$is short-circuited by an ideal ammeter having zero resistance.

a. What is the battery’s internal resistance?

b. How much power is dissipated inside the battery?

We have given that the battery is short-circuited by an ideal ammeter having zero resistance.

We need to find that the battery's internal resistance.

The real batteries have an internal resistance $r$due to their components while the ideal batteries have zero internal resistance. When an ideal wire with zero resistance is connected to the two terminals of the battery, it will make a short circuit where the current is maximum. For a battery that is connected to a circuit, the current that flows through the circuit is given as

localid="1648896776181" $I=\frac{\in }{R+r}...\left(1\right)$

In a short circuit, the resistance $R=0$. So, equation (1) could be rearranged for $r$to be in the form

localid="1648896790084" $r=\frac{\in }{I}...\left(2\right)$

The ammeter reads current $I=2.3A$.Plug the value of $\in$and $I$into equation (2) to get $r$

$$\begin{gathered} r=\frac{\in }{I} \\ =\frac{1.5V}{2.3A} \\ =0.65\Omega \end{gathered}$$

We have given that the battery is short-circuited by an ideal ammeter having zero resistance.

We need to find that how much power is dissipated inside the battery.

The energy is dissipated when the current flows through a resistor. The rate of the dissipated energy is the power. It is given as

$P_{bat}=\frac{{\left(\in \right)}^2}{r}...\left(3\right)$

Putting the values for $r$and $\in$into equation (3) to get $P_{bat}$

$$\begin{gathered} P_{bat}=\frac{{\left(\in \right)}^2}{r} \\ =\frac{{\left(1.5V\right)}^2}{0.65\Omega } \\ =3.5W \end{gathered}$$