If `R_(1)` and `R_(2)` are the resistances of the filaments of 200 W and 100 W electric bulbs operated at 220 V, then `((R_1)/(R_2))` is

Options

A. (a)2
B. (b)0.5
C. (c)4
D. (d)0.25

Answer

To solve the problem, we need to find the ratio of the resistances \( R_1 \) and \( R_2 \) of the filaments of two electric bulbs (200 W and 100 W) when operated at a voltage of 220 V.

### Step-by-Step Solution:

1. **Understand the relationship between power and resistance**:
The power \( P \) consumed by an electrical device is given by the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the device and \( R \) is the resistance.

2. **Write the power equations for both bulbs**:
- For the 200 W bulb:
\[ P_1 = 200 \, \text{W} = \frac{220^2}{R_1} \]
- For the 100 W bulb:
\[ P_2 = 100 \, \text{W} = \frac{220^2}{R_2} \]

3. **Rearranging the equations to find resistances**:
From the power equations, we can express the resistances in terms of power:
- Rearranging the equation for \( R_1 \):
\[ R_1 = \frac{220^2}{200} \]
- Rearranging the equation for \( R_2 \):
\[ R_2 = \frac{220^2}{100} \]

4. **Finding the ratio \( \frac{R_1}{R_2} \)**:
Now, we can find the ratio of the resistances:
\[ \frac{R_1}{R_2} = \frac{\frac{220^2}{200}}{\frac{220^2}{100}} = \frac{100}{200} = \frac{1}{2} \]

5. **Final result**:
Therefore, the ratio \( \frac{R_1}{R_2} \) is:
\[ \frac{R_1}{R_2} = 0.5 \]

### Conclusion:
The ratio \( \frac{R_1}{R_2} \) is \( 0.5 \).