Question

If two electric bulbs have $40W$ and $60W$ rating at $220V$, then the ratio of their resistances will be :

• A. $3:2$
• B. $3:8$
• C. $4:3$
• D. $9:4$

Answer 1

Answer: (A)

The rate at which electrical energy is dissipated into other forms of energy is called electric power $P$.
Electrical power $=\frac{W}{t}=\frac{Vit}{t}$
$(V=iR)$
$\Rightarrow P\frac{V^2}{R}$
Where $V$ is potential difference, $R$ is resistance.
Given, $P_1=40W$
$P_2=60W$
$\therefore \frac{P_1}{P_2}=\frac{R_2}{R_1}$
$\Rightarrow \frac{R_1}{R_2}=\frac{P_2}{P_1}$
$=\frac{60}{40}=\frac{3}{2}$
Note : Higher the wattage of a bulb smaller is the resistance of its filament that is thicker is the filament.