Question

What is the resistance of a $40\, W$ lamp which is lighted at full brilliance by a current of $ \frac{\text{1}}{\text{3}}\text{A} $ ?

• A. $ 450\,\Omega $
• B. $ 360\,\Omega $
• C. $ 120\,\Omega $
• D. $ 13.33\,\Omega $

Answer 1

Answer: (B)

Power of electric bulb is given by
$P=\frac{V^{2}}{R}$
But from Ohm's law,
$V = i R$
So, $P=\frac{(i R)^{2}}{R}=i^{2} R$
$\therefore R=\frac{p}{i^{2}}$
Here, $P=40\, W , i=\frac{1}{3} A$
Hence, $ R=\frac{40}{(1 / 3)^{2}}=40 \times 9=360\, \Omega$