Two similar headlight lamps are connected in parallel to each other. Together , they consume `48 W` from a `6 V` battery . What is the resistance of each filament ?

Options

A. `6 Omega`
B. ` 4 Omega`
C. `3.0 Omega`
D. `1.5 Omega`

Explanation

To find the resistance of each filament in the two similar headlight lamps connected in parallel, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Total power (P) = 48 W
- Voltage of the battery (V) = 6 V

2. **Understand the Configuration:**
- The two lamps (filaments) are connected in parallel.
- Let the resistance of each filament be R.

3. **Calculate Equivalent Resistance:**
- For two resistors in parallel, the equivalent resistance (R_eq) is given by:
\[ R_{eq} = \frac{R \cdot R}{R + R} = \frac{R^2}{2R} = \frac{R}{2} \]

4. **Use the Power Formula:**
- The power consumed in the circuit can also be expressed using the equivalent resistance:
\[ P = \frac{V^2}{R_{eq}} \]
- Substituting the values we have:
\[ 48 = \frac{6^2}{R_{eq}} \]

5. **Substitute for R_eq:**
- Replace \( R_{eq} \) with \( \frac{R}{2} \):
\[ 48 = \frac{36}{\frac{R}{2}} \]

6. **Rearranging the Equation:**
- This simplifies to:
\[ 48 = \frac{36 \cdot 2}{R} \]
- Which leads to:
\[ 48R = 72 \]

7. **Solve for R:**
- Dividing both sides by 48 gives:
\[ R = \frac{72}{48} = \frac{3}{2} = 1.5 \, \Omega \]

### Final Answer:
The resistance of each filament is **1.5 ohms**.