Question

If two electric bulbs have $40\, W$ and $60\, W$ rating at $220\, V$, 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 $= \frac{W}{t}=\frac{V i t}{t}$
$(V=i \,R)$
$\Rightarrow P \frac{V^{2}}{R}$
Where $V$ is potential difference, $R$ is resistance.
Given, $P_{1}=40\, W$
$P_{2}=60\, W $
$\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.