Question

An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance 'R' that must be put in series with the bulb so that the bulb delivers the same power is ________ $\Omega$.

Hint:

If the supply voltage is double the rated voltage, the series resistance required is always equal to the resistance of the device itself.

Answer 1

Correct Answer: 50

Step 1: Understanding the Concept:
For a bulb to deliver its rated power, it must have its rated voltage across its terminals. If the supply voltage is higher, a series resistor must be added to drop the excess voltage.
Step 2: Key Formula or Approach:
1. Resistance of the bulb: \(R_b = \frac{V_{rated}^2}{P}\).
2. Voltage Divider / Ohm's Law: \(V_{supply} = I(R + R_b)\).
Step 3: Detailed Explanation:
Bulb resistance \(R_b\):
\[ R_b = \frac{100 \times 100}{200} = 50 \text{ }\Omega \]
To deliver rated power, the voltage across the bulb must be 100 V.
The current required is:
\[ I = \frac{V_{rated}}{R_b} = \frac{100}{50} = 2 \text{ A} \]
Total resistance needed for the 200 V supply to provide 2 A:
\[ R_{total} = \frac{V_{supply}}{I} = \frac{200}{2} = 100 \text{ }\Omega \]
Series resistance \(R\):
\[ R = R_{total} - R_b = 100 - 50 = 50 \text{ }\Omega \]
Step 4: Final Answer:
The resistance \(R\) must be 50 \(\Omega\).