Two resistors of 200 \, and 400 \, are connected in series with a...

The resistance of the bulb is calculated using its power rating:

$R_b = \dfrac{V^2}{P} = \dfrac{200^2}{100} = 400 \, \Omega$

The bulb is connected in parallel with the $400 \, \Omega$ resistor. The equivalent resistance of this parallel combination is:

$R_p = \dfrac{400 \times 400}{400 + 400} = 200 \, \Omega$

This parallel combination is in series with the $200 \, \Omega$ resistor. The total equivalent resistance of the circuit is:

$R_{eq} = 200 + R_p = 200 + 200 = 400 \, \Omega$

The current drawn from the battery is:

$I = \dfrac{V}{R_{eq}} = \dfrac{100}{400} = 0.25 \text{ A}$

The potential drop across the bulb is the same as the potential drop across the parallel combination:

$V_p = I \times R_p = 0.25 \times 200 = 50 \text{ V}$

Answer: $50$