The bulb of a motorcycle head-lamp is marked 40W, 6V. The resistance of the filament when it is switched on is
\(\frac{6^2}{40^2}\)ohms
\(\frac{40}{6^2}\)ohms
\(\frac{40}{6}\)ohms
6 x 40 ohms
\(\frac{6^2}{40}\)ohms
Correct answer is E
Power \(\frac{V^2}{R}\)
40 = \(\frac{6 \times 6}{r}\)
R = \(\frac{36}{40}\)\(\Omega\)