The hot resistance of a flashlight bulb is $\mathbf{2}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{\Omega }$, and it is run by a $\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{‐}\mathbf{V}$alkaline cell having a$\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{‐}\mathbf{\Omega }$ internal resistance. (a) What current flows? (b) Calculate the power supplied to the bulb using ${\mathbf{I}}^{\mathbf{2}}{\mathbf{R}}_{\mathbf{bulb}}$. (c) Is this power the same as calculated using$\frac{{\mathbf{V}}^{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{bulb}}}$?

Alkaline cell is use as the power source for various proposes. The emf of alkaline cell is given by equation,

$\mathbf{E}\mathbf{=}\mathbf{I}\left(r+R_{\mathrm{bulb}}\right)$

Here, r is the internal resistance of the battery, I is the current through the battery, and ${\mathbf{R}}_{\mathbf{bulb}}$is the resistance of the bulb.

The flashlight, whose bulb has a hot resistance ${\mathrm{R}}_{\mathrm{bulb}}=2.3\mathrm{\Omega }$and is run by a$\mathrm{E}=1.58\mathrm{V}$ alkaline cell with internal resistance $\mathrm{r}=0.1\mathrm{\Omega }$.

(a)

The current that flows through the circuit is obtained using the loop rule as

$$\begin{gathered} \mathrm{E}‐\mathrm{Ir}‐{\mathrm{IR}}_{\mathrm{bulb}}=0 \\ \mathrm{I}=\frac{\mathrm{E}}{\mathrm{r}+{\mathrm{R}}_{\mathrm{bulb}}} \\ \mathrm{I}=\frac{1.58\mathrm{V}}{\left(0.1+2.3\right)\mathrm{\Omega }} \\ \mathrm{I}=0.658\mathrm{A} \end{gathered}$$

Therefore, the current flows 0.658 A .

(b)

The power supplied to the bulb is

$$\begin{gathered} \mathrm{P}={\mathrm{I}}^2{\mathrm{R}}_{\mathrm{bulb}} \\ ={\left(0.658\mathrm{A}\right)}^2\times \left(2.3\mathrm{\Omega }\right) \\ =0.997\mathrm{W} \end{gathered}$$

Therefore, the power supplied is 0.997 W .

(c)

Calculation of the power supplied to the bulb using the formula $\mathrm{P}={\mathrm{V}}^2/{\mathrm{R}}_{\mathrm{bulb}}$if for V can be use the terminal voltage of the cell, because this is the same as the voltage of the light-bulb$\left({\mathrm{V}}_{\mathrm{bulb}}={\mathrm{IR}}_{\mathrm{bulb}}\right)$ . The terminal voltage is

$$\begin{gathered} \mathrm{V}=\mathrm{E}‐\mathrm{Ir} \\ =1.58\mathrm{V}-\left(0.658\mathrm{A}\right)\times \left(2.3\mathrm{\Omega }\right) \\ =1.514\mathrm{V} \end{gathered}$$

The power is then

$$\begin{gathered} \mathrm{P}=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_{\mathrm{bulb}}} \\ =\frac{{\left(1.514\mathrm{V}\right)}^2}{2.3\mathrm{\Omega }} \\ =0.997\mathrm{W} \end{gathered}$$

Therefore, the result is the same.