A `100 W` bulb designed to operate on `100 V` is to be connected across a `500 V` source . Find the resistance to be put in series so that bulb consumes `100 W` only.

Options

A. `100 Omega`
B. `500 Omega`
C. `400 Omega`
D. `300 Omega`

Explanation

To solve the problem step-by-step, we need to find the resistance that should be connected in series with a 100 W bulb designed for 100 V, so that it consumes only 100 W when connected to a 500 V source.

### Step 1: Understand the given values
- Power of the bulb (P) = 100 W
- Voltage rating of the bulb (V_bulb) = 100 V
- Source voltage (V_source) = 500 V

### Step 2: Calculate the current through the bulb
Using the formula for power:
\[ P = V \times I \]
We can rearrange this to find the current (I):
\[ I = \frac{P}{V_bulb} \]
Substituting the values:
\[ I = \frac{100 \, \text{W}}{100 \, \text{V}} = 1 \, \text{A} \]

### Step 3: Determine the voltage drop across the resistor
Since the bulb is rated for 100 V and we are connecting it to a 500 V source, the voltage drop across the resistor (V_R) can be calculated as:
\[ V_R = V_{source} - V_{bulb} \]
Substituting the values:
\[ V_R = 500 \, \text{V} - 100 \, \text{V} = 400 \, \text{V} \]

### Step 4: Calculate the resistance needed
Using Ohm's Law, which states:
\[ V = I \times R \]
We can rearrange this to find the resistance (R):
\[ R = \frac{V_R}{I} \]
Substituting the values:
\[ R = \frac{400 \, \text{V}}{1 \, \text{A}} = 400 \, \Omega \]

### Conclusion
The resistance that should be connected in series with the bulb is:
\[ R = 400 \, \Omega \]