A typical small flashlight contains two batteries, each having an emf of$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathit{V}$, connected in series with a bulb having resistance$\mathbf{17}\mathbf{}\mathit{\Omega }$. (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{h}$what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

According to Ohm’s law, the current flowing through the conductor is directly proportional to the voltage across the two points.

$V=IR$

Where,$I$is current in ampere $\left(A\right),R$is resistance in ohms$\left(\Omega \right)$and $V$ is the potential difference volt $\left(V\right)$

If real source of emf $\epsilon$has internal energy $r,$then its terminal potential difference $V_{ab}$depends upon current.

So, formula for the terminal potential difference is:

$V_{\mathrm{ab}}=\epsilon -Ir$

Given that,

$R=17\Omega$

Also given that, two batteries connected in series with same voltage.

Since, the terminal voltage in the circuit is equal to the voltage of the two batteries

Therefore,$V=3.0V$

The power dissipated in the circuit is:

$$\begin{gathered} P=\frac{V^2}{R} \\ =\frac{3^2}{17} \\ =0.530\mathrm{W} \end{gathered}$$

Hence, if the internal resistance of the batteries is negligible, $0.530\mathrm{W}$power is delivered to the bulb

Given that,$t=5.0\mathrm{h}$

The energy consumed in given time is:

$$\begin{gathered} U=Pt \\ =\left(0.530\right)\left(5\times 3600\right) \\ =9540\mathrm{J} \end{gathered}$$

Hence, if the batteries last for $0.5\mathrm{h},9540\mathrm{J}$total energy delivered to the bulb.

The internal resistance of battery is:

$r=\frac{\epsilon -V}{1}$

Substitute$V=IR$and $I=\sqrt{\frac{P}{R}}$in internal resistance formula

$$\begin{gathered} r=\frac{\epsilon -IR}{\sqrt{{\displaystyle \frac{P}{R}}}} \\ =\frac{3.0-\left(0.125\right)\left(17\right)}{\sqrt{{\displaystyle \frac{\left(0.530/2\right)}{17}}}} \\ =\frac{3-2.125}{0.125} \\ =7\Omega \end{gathered}$$

Hence, the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value is $7\Omega$.