the resistance that must be put in series with the bulb, we can use the equation for electrical power:
P = V^2 / R
where P is the power, V is the voltage, and R is the resistance.
We can rearrange this equation to solve for R:
R = V^2 / P
In this case, the power is 400 W and the voltage is 200 V, so the resistance is:
R = (200)^2 / 400
R = 200
Therefore, the resistance that must be put in series with the bulb is 200 ohms.
It's worth noting that the bulb is rated for 2 kW at 100 V, so using it in a circuit with a voltage of 200 V will likely cause it to operate at a higher power than it is rated for.