### JEE Main 12 Jan 2019 Morning Question 22

##### Question: 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 [JEE Main Online Paper Held On 12-Jan-2019 Morning]

#### Options:

A) $ P _1=9W,P _2=16W $

B) $ P _1=16,W,P _2=4W $

C) $ P _1=4,W,P _2=16W $

D) $ P _1=16,W,P _2=9W $

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

Resistance across 25 W bulb

$ R _1=\frac{V^{2}}{P _1}=\frac{220\times 220}{25}=1936\Omega $

Resistance across 100 W bulb $ R _2=\frac{V^{2}}{P _2}=\frac{220\times 220}{100}=484\Omega $

Total Resistance $ R _1+R _2=2420\Omega $ Total current $ I=\frac{V}{R}=\frac{220}{2420}=0.09A $

Power consumed by 25 W bulb $ I^{2}R _1=0.09\times 0.09\times 1936\cong 16W $

Power consumed by 100 W bulb $ I^{2}R _2=0.09\times 0.09\times 484\cong 4W $