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. `26 Omega`
B. `13 Omega`
C. `230 Omega`
D. `66 Omega`

Explanation

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

### Step-by-Step Solution:

1. **Determine the resistance of the bulb:**
The power \( P \) and voltage \( V \) ratings of the bulb are given. We can use the formula for power:
\[ P = \frac{V^2}{R_b} \]
where \( R_b \) is the resistance of the bulb. Rearranging this gives:
\[ R_b = \frac{V^2}{P} \]
Substituting the values:
\[ R_b = \frac{100^2}{500} = \frac{10000}{500} = 20 \, \Omega \]

2. **Calculate the current flowing through the circuit:**
When the bulb operates at its rated power of \( 500 \, W \), we can find the current \( I \) using:
\[ P = I^2 R_b \]
Rearranging gives:
\[ I^2 = \frac{P}{R_b} \]
Substituting the values:
\[ I^2 = \frac{500}{20} = 25 \implies I = 5 \, A \]

3. **Apply Ohm's Law to find the total resistance in the circuit:**
The total voltage supplied is \( 230 \, V \). According to Ohm's Law:
\[ V = I \cdot R_{total} \]
where \( R_{total} = R + R_b \). Thus:
\[ 230 = 5 \cdot (R + 20) \]
Rearranging gives:
\[ R + 20 = \frac{230}{5} = 46 \]
Therefore:
\[ R = 46 - 20 = 26 \, \Omega \]

4. **Conclusion:**
The value of the resistance \( R \) that needs to be connected in series with the bulb is:
\[ R = 26 \, \Omega \]