An electric bulb rated for 500 watts at 100 volts is used in a circuit having a 200 volts supply. The resistance R that must be put in series with the bulb, so that the bulb delivers 500 watt is …………ohm.

Explanation

To solve the problem, we need to find the resistance \( R \) that must be put in series with a 500-watt bulb rated at 100 volts, when the supply voltage is 200 volts.

### Step-by-Step Solution:

1. **Determine the resistance of the bulb:**
The power rating \( P \) and voltage rating \( V \) of the bulb are given as:
\[ P = 500 \text{ watts}, \quad V = 100 \text{ volts} \]
We can use the formula for resistance \( R \) derived from the power formula:
\[ R = \frac{V^2}{P} \]
Substituting the values:
\[ R = \frac{100^2}{500} = \frac{10000}{500} = 20 \text{ ohms} \]
So, the resistance of the bulb is \( 20 \, \Omega \).

2. **Set up the circuit:**
We have a total supply voltage of \( 200 \) volts. The bulb has a resistance of \( 20 \, \Omega \) and we need to find the additional resistance \( R \) that must be connected in series.

3. **Calculate the current through the circuit:**
For the bulb to deliver \( 500 \) watts, it must have a voltage drop of \( 100 \) volts across it. Using Ohm's law:
\[ V = I \cdot R \]
where \( V = 100 \) volts and \( R = 20 \, \Omega \):
\[ 100 = I \cdot 20 \]
Solving for \( I \):
\[ I = \frac{100}{20} = 5 \text{ amperes} \]

4. **Apply Kirchhoff's Voltage Law:**
The total voltage in the circuit is \( 200 \) volts. The voltage drop across the bulb is \( 100 \) volts and the voltage drop across the resistance \( R \) is \( V_R \). Thus, we can write:
\[ 200 = 100 + V_R \]
Therefore:
\[ V_R = 200 - 100 = 100 \text{ volts} \]

5. **Calculate the resistance \( R \):**
Using Ohm's law again for the additional resistance \( R \):
\[ V_R = I \cdot R \]
Substituting the values:
\[ 100 = 5 \cdot R \]
Solving for \( R \):
\[ R = \frac{100}{5} = 20 \text{ ohms} \]

### Final Answer:
The resistance \( R \) that must be put in series with the bulb is \( 20 \, \Omega \).

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