The resistance of the bulb is calculated using its power rating:
Rb=PV2=1002002=400Ω
The bulb is connected in parallel with the 400Ω resistor. The equivalent resistance of this parallel combination is:
Rp=400+400400×400=200Ω
This parallel combination is in series with the 200Ω resistor. The total equivalent resistance of the circuit is:
Req=200+Rp=200+200=400Ω
The current drawn from the battery is:
I=ReqV=400100=0.25 A
The potential drop across the bulb is the same as the potential drop across the parallel combination:
Vp=I×Rp=0.25×200=50 V
Answer: 50