JEE PYQ: Current Electricity Question 73
Question 73 - 2019 (12 Jan 2019 Shift 2)
Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers $P_1$ and $P_2$ respectively, then:
(1) $P_1 = 16$ W, $P_2 = 4$ W
(2) $P_1 = 16$ W, $P_2 = 9$ W
(3) $P_1 = 9$ W, $P_2 = 16$ W
(4) $P_1 = 4$ W, $P_2 = 16$ W
Show Answer
Answer: (1)
Solution
$R_1 = \frac{220^2}{25}$, $R_2 = \frac{220^2}{100}$. Current $i = \frac{220}{R_1 + R_2}$. $P_1 = i^2 R_1 = \frac{220^2}{(R_1+R_2)^2} \cdot R_1 = \frac{220^2 \cdot \frac{220^2}{25}}{(\frac{220^2}{25} + \frac{220^2}{100})^2} = 16$ W. $P_2 = i^2 R_2 = 4$ W.
BETA
