Two electric bulbs whose resistances are in the ratio of `1:2` are connected inparallel to a constant voltage source. The powers dissipated in them have the ratio

Options

A. `1:2`
B. `1:1`
C. `1:4`
D. `2:1`

Answer

To solve the problem of finding the ratio of powers dissipated in two electric bulbs connected in parallel, we can follow these steps:

### Step 1: Define the resistances
Let the resistance of the first bulb be \( R_1 \) and the resistance of the second bulb be \( R_2 \). According to the problem, the ratio of their resistances is given as:
\[ \frac{R_1}{R_2} = \frac{1}{2} \]
This implies that we can express the resistances as:
\[ R_1 = R \quad \text{and} \quad R_2 = 2R \]
where \( R \) is a constant.

### Step 2: Use the power formula
The power dissipated in an electrical component is given by the formula:
\[ P = \frac{V^2}{R} \]
where \( P \) is the power, \( V \) is the voltage across the component, and \( R \) is the resistance.

### Step 3: Calculate the power for each bulb
Since the bulbs are connected in parallel, the voltage across both bulbs is the same. Let’s denote the voltage across the bulbs as \( V \).

- For the first bulb (with resistance \( R_1 \)):
\[ P_1 = \frac{V^2}{R_1} = \frac{V^2}{R} \]

- For the second bulb (with resistance \( R_2 \)):
\[ P_2 = \frac{V^2}{R_2} = \frac{V^2}{2R} \]

### Step 4: Find the ratio of powers
To find the ratio of the powers dissipated in the two bulbs, we can write:
\[ \frac{P_1}{P_2} = \frac{\frac{V^2}{R}}{\frac{V^2}{2R}} = \frac{V^2}{R} \cdot \frac{2R}{V^2} \]
The \( V^2 \) terms cancel out:
\[ \frac{P_1}{P_2} = \frac{2R}{R} = 2 \]

### Step 5: Express the ratio in standard form
Thus, we can express the ratio of powers as:
\[ \frac{P_1}{P_2} = 2:1 \]

### Conclusion
The ratio of the powers dissipated in the two bulbs is \( 2:1 \).

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