Two resistors of \(200 \, \Omega\) and \(400 \, \Omega\) are connected in series with a battery of \(100\) V. A bulb rated at \(200\) V, \(100\) W is connected across the \(400 \, \Omega\) resistance. The potential drop across the bulb is _______ V.

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: \(50\)