Given data

• Bulb rating: $200\,\text{W}$ at $100\,\text{V}$
• Supply voltage available: $200\,\text{V}$
• We need a series resistance $R$ so that the bulb still works at its rated power.

Find the bulb resistance at rated condition Since the bulb is rated $200\,\text{W}$ at $100\,\text{V}$, $P = \frac{V^2}{R_b}$ where $R_b$ is the bulb resistance.

So, $R_b = \frac{V^2}{P} = \frac{100^2}{200} = \frac{10000}{200} = 50\,\Omega$

Find the current through the bulb at rated condition $I = \frac{P}{V} = \frac{200}{100} = 2\,\text{A}$

For the bulb to deliver the same power, it must still have:

• voltage across it $= 100\,\text{V}$
• current through it $= 2\,\text{A}$

Voltage to be dropped across the series resistor Total supply is $200\,\text{V}$, while bulb needs only $100\,\text{V}$.

Therefore series resistor must drop: $V_R = 200 - 100 = 100\,\text{V}$

Compute the required series resistance The same current $2\,\text{A}$ flows through the series resistor.

Using Ohm's law, $R = \frac{V_R}{I} = \frac{100}{2} = 50\,\Omega$

Final answer $\boxed{50\,\Omega}$

Comparison with stored correct answer Stored correct answer = $50$

My derived answer also = $50$.

Hence, the answer agrees with the stored correct answer.