Q. Two bulbs of 100 W and 200 W working at 220 V are joined in series with 220 V supply. Total power consumed will be
A) 65 W
B) 33 W
C) 300 W
D) 100 W
Explanation
(a) $R_{1} = \frac{(220)^{2}}{100} = 484\ \Omega$
And $R_{2} = \frac{(220)^{2}}{200} = 242\ \Omega$
So, $i = \frac{220}{484 + 242} = 0.3\ A$
Hence, total power consumed will be
$P = i^{2}R = (0.3)^{2} \times (484 + 242) = 65.34 \approx 65\ W$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.