JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given, power rating of bulb, PB=500W
Voltage across bulb, VB=100V
Supply voltage, VS=200V
If a resistance R is attached in series with the bulb, then the voltage across resistance will be 100 V.
Now, current flowing in circuit when bulb delivers power of 500 W is given as
PB=VBI
⇒500=100×I
⇒I=5A
Same amount of current will flow from the resistance as it is connected in series.
Using Ohm’s law,
V = IR
⇒ 100 = 5 × R
⇒ R = 20Ω
Thus, the resistance connected in series is 20Ω.