Question

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?

Short answer

(a) The battery's internal resistance is$r=0.65\Omega$.

(b) The power dissipated inside the battery islocalid="1648896222422" $3.5W$.

Step-by-step solution

Part (a) step 1: Given information

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.

Part (a) step 2: Simplify

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 $r$

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

Part (b) step 1: Given information

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.

Part (b) step 2: Simplify

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 $P_{bat}$

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