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.