Question

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference. What is the resistance of (a) a 100-W bulb and (b) a 60-W bulb? (c) How much current does each bulb draw in normal use?

Short answer

1. The resistance of 100-W bulb is$144\mathrm{\Omega }$.
2. The resistance of 60-W bulb is $240\mathrm{\Omega }$.
3. The current drawn by each bulb is 0.833 A and 0.500 A.

Step-by-step solution

Determination of the resistance of 100-W and 60-W bulbs.

The connection of the bulb is across a 120-V potential difference.

The expression of power is,

$P=\frac{V^2}{R}$

Solve for R and 100 W bulb,

role="math" localid="1655719758468" $$\begin{gathered} R=\frac{V^2}{P} \\ =\frac{{(120\mathrm{V})}^2}{(100\mathrm{W})} \\ =144\mathrm{\Omega } \end{gathered}$$

Repeat the calculation for 60 W bulb,

$$\begin{gathered} R=\frac{V^2}{P} \\ =\frac{{(120\mathrm{V})}^2}{(60\mathrm{W})} \\ =240\mathrm{\Omega } \\ \mathrm{Thus},\mathrm{the}\mathrm{resistances}\mathrm{for}100-\mathrm{W}\mathrm{bulb}\mathrm{is}\mathrm{and}\mathrm{for}60\mathrm{W}\mathrm{bulb}\mathrm{is}144\mathrm{\Omega }\mathrm{and}\mathrm{for}60\mathrm{W}\mathrm{bulb}\mathrm{is}240\mathrm{\Omega } \end{gathered}$$

(b) Determination of the current drawn by each bulb is

Use the Ohm’s Law

$$\begin{gathered} \mathrm{V}=\mathrm{lR} \\ \mathrm{l}=\frac{\mathrm{V}}{\mathrm{R}} \\ \mathrm{For}\mathrm{the}100\mathrm{W}\mathrm{bulb}: \\ \mathrm{l}=\frac{(120\mathrm{V})}{(144\mathrm{\Omega })} \\ =0.833\mathrm{A} \\ \mathrm{For}\mathrm{the}60\mathrm{W}\mathrm{bulb}: \\ \mathrm{l}=\frac{(120\mathrm{V})}{(240\mathrm{\Omega })} \\ =0.500\mathrm{A} \\ \mathrm{Thus},\mathrm{the}\mathrm{current}\mathrm{drawn}\mathrm{by}100-\mathrm{W}\mathrm{bulb}\mathrm{and}\mathrm{the}60-\mathrm{W}\mathrm{bulb}\mathrm{is}0.833\mathrm{A}\mathrm{and}0.500\mathrm{A} \end{gathered}$$