A filament bulb `(500 W, 100 V)` is to be used in a `230 V` main supply. When a resistance `R` is connected in series, it works perfectly and the bulb consumers `500 W`. The value of `R` is

Options

A. `230 Omega`
B. `46 Omega`
C. `26 Omega`
D. `13 Omega`

Explanation

To solve the problem step by step, we need to find the value of the resistance \( R \) that must be connected in series with a filament bulb rated at \( 500 \, W \) and \( 100 \, V \) to operate it correctly on a \( 230 \, V \) supply.

### Step 1: Identify the given values
- Power of the bulb, \( P = 500 \, W \)
- Voltage rating of the bulb, \( V_b = 100 \, V \)
- Supply voltage, \( V_s = 230 \, V \)

### Step 2: Calculate the current through the bulb
Using the formula for power:
\[ P = V_b \times I \]
We can rearrange this to find the current \( I \):
\[ I = \frac{P}{V_b} = \frac{500 \, W}{100 \, V} = 5 \, A \]

### Step 3: Calculate the resistance of the bulb
Using the formula for resistance based on power:
\[ R_b = \frac{V_b^2}{P} \]
Substituting the values:
\[ R_b = \frac{(100 \, V)^2}{500 \, W} = \frac{10000}{500} = 20 \, \Omega \]

### Step 4: Apply Ohm's Law to find the total resistance in the circuit
The total voltage across the circuit is the supply voltage \( V_s \):
\[ V_s = I \times (R_b + R) \]
Rearranging this gives:
\[ R + R_b = \frac{V_s}{I} \]
Substituting the known values:
\[ R + 20 \, \Omega = \frac{230 \, V}{5 \, A} \]
Calculating the right side:
\[ R + 20 \, \Omega = 46 \, \Omega \]

### Step 5: Solve for the resistance \( R \)
Now, we can solve for \( R \):
\[ R = 46 \, \Omega - 20 \, \Omega = 26 \, \Omega \]

### Conclusion
The value of the resistance \( R \) that needs to be connected in series with the bulb is \( 26 \, \Omega \).