If electric bulbs having resistances in the ratio 2 : 3 are connected in parallel to a voltage sources of 220 V. The ratio of the power dissipated in them is

Options

A. 2: 3`
B. 3 : 2 `
C. 1 : 1 `
D. 2: 5 `

Answer

To solve the problem of finding the ratio of power dissipated in two electric bulbs with resistances in the ratio 2:3 when connected in parallel to a voltage source of 220 V, we can follow these steps:

### Step 1: Define the resistances
Let the resistances of the two bulbs be \( R_1 \) and \( R_2 \). According to the problem, we have:
\[ \frac{R_1}{R_2} = \frac{2}{3} \]
This implies that we can express \( R_1 \) and \( R_2 \) as:
\[ R_1 = 2k \quad \text{and} \quad R_2 = 3k \]
for some constant \( k \).

### Step 2: Use the power formula
The power dissipated in a resistor when a voltage \( V \) is applied across it is given by:
\[ P = \frac{V^2}{R} \]
For the two bulbs, the power dissipated can be expressed as:
\[ P_1 = \frac{V^2}{R_1} \quad \text{and} \quad P_2 = \frac{V^2}{R_2} \]

### Step 3: Substitute the values of resistances
Substituting the expressions for \( R_1 \) and \( R_2 \) into the power equations:
\[ P_1 = \frac{V^2}{2k} \quad \text{and} \quad P_2 = \frac{V^2}{3k} \]

### Step 4: Find the ratio of powers
To find the ratio of the powers dissipated in the two bulbs, we calculate:
\[ \frac{P_1}{P_2} = \frac{\frac{V^2}{2k}}{\frac{V^2}{3k}} = \frac{V^2 \cdot 3k}{V^2 \cdot 2k} \]
The \( V^2 \) and \( k \) terms cancel out:
\[ \frac{P_1}{P_2} = \frac{3}{2} \]

### Step 5: Conclusion
Thus, the ratio of the power dissipated in the two bulbs is:
\[ P_1 : P_2 = 3 : 2 \]

### Final Answer
The ratio of the power dissipated in the bulbs is \( 3 : 2 \).

---