Q. A bulb has specification of one kilowatt and $250\ volts,$ the resistance of bulb is
A) $125\ \Omega$
B) $62.5\ \Omega$
C) $0.25\ \Omega$
D) $625\ \Omega$
Explanation
(b) $R = \frac{V^{2}}{P} = \frac{(250)^{2}}{10^{3}} = 62.5\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.