Question

A lightbulb is in series with a $2.0\Omega$resistor. The lightbulb dissipates $10\mathrm{W}$ when this series circuit is connected to a $9.0\mathrm{V}$ battery. What is the current through the lightbulb? There are two possible answers; give both of them.

Short answer

The current through the light bulb is$2\mathrm{A},2.5\mathrm{A}.$

Step-by-step solution

Given information

We have given,

The power dissipated by the bulb =$10\mathrm{W}$

voltage = $9\mathrm{V}$

Resistance = $2\Omega$

We have to find the current flow through bulb.

Simplify

Using the Kirchhoff voltage law in the circuit and let us consider R' as the resistance of the bulb

$9\mathrm{V}=(\mathrm{R}\text{'}\mathrm{\Omega })(\mathrm{I}\mathrm{A})+(2\mathrm{\Omega })(\mathrm{I}\mathrm{A})..........................\left(1\right)$

we know that power dissipation

$$\begin{gathered} \mathrm{P}={\mathrm{I}}^2\mathrm{R}=10\mathrm{W} \\ \mathrm{R}\text{'}=\frac{10\mathrm{W}}{{\mathrm{I}}^2} \\ 9=\mathrm{I}\left(\frac{10}{{\mathrm{I}}^2}+2\right) \\ 9\mathrm{I}=2{\mathrm{I}}^2+10..............\left(2\right) \end{gathered}$$

By solving this quadratic equation (1) and (2) we get the final result, which is

$\mathrm{I}=2\mathrm{A},2.5\mathrm{A}$