If `R_(1) and R_(2)` are filament resistances of a 200 W and a 100 W bulb respectively, designed to operate on the same voltage, then :

Options

A. `R_(1)=R_(2)`
B. `R_(2)=2R_(1)`
C. `R_(2)=4R_(1)`
D. `R_(1)=4R_(2)`

Answer

To solve the problem, we will use the relationship between power, voltage, and resistance. The power \( P \) of an electrical device can be expressed in terms of voltage \( V \) and resistance \( R \) using the formula:

\[ P = \frac{V^2}{R} \]

### Step 1: Write down the power formulas for both bulbs.
For the 200 W bulb (let's denote its resistance as \( R_1 \)):
\[ P_1 = \frac{V^2}{R_1} \quad \text{(1)} \]
For the 100 W bulb (let's denote its resistance as \( R_2 \)):
\[ P_2 = \frac{V^2}{R_2} \quad \text{(2)} \]

### Step 2: Rearrange the formulas to express resistance in terms of power.
From equation (1), we can express \( R_1 \):
\[ R_1 = \frac{V^2}{P_1} \quad \text{(3)} \]
From equation (2), we can express \( R_2 \):
\[ R_2 = \frac{V^2}{P_2} \quad \text{(4)} \]

### Step 3: Substitute the power values into the resistance equations.
Substituting \( P_1 = 200 \) W and \( P_2 = 100 \) W into equations (3) and (4):
\[ R_1 = \frac{V^2}{200} \quad \text{(5)} \]
\[ R_2 = \frac{V^2}{100} \quad \text{(6)} \]

### Step 4: Find the relationship between \( R_1 \) and \( R_2 \).
Now, we can find the ratio of \( R_2 \) to \( R_1 \):
\[ \frac{R_2}{R_1} = \frac{\frac{V^2}{100}}{\frac{V^2}{200}} = \frac{200}{100} = 2 \]
This implies:
\[ R_2 = 2R_1 \quad \text{(7)} \]

### Conclusion
Thus, the relationship between the resistances is:
\[ R_2 = 2R_1 \]

### Final Answer
The correct option is that \( R_2 \) is equal to \( 2R_1 \).

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