The power dissipated by a light bulb with 4 ohm resistance when connected in parallel to 12V battery is

Options

A. 36 W
B. 3.6 W
C. 0.36 W
D. 0.036 W

Explanation

To find the power dissipated by a light bulb with a resistance of 4 ohms when connected in parallel to a 12V battery, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Resistance (R) = 4 ohms
- Voltage (V) = 12 volts

2. **Use the Power Formula:**
The power (P) dissipated in an electrical circuit can be calculated using the formula:
\[ P = V \times I \]
where \(I\) is the current flowing through the circuit.

3. **Apply Ohm's Law:**
According to Ohm's Law, the current (I) can be expressed as:
\[ I = \frac{V}{R} \]
Substituting the known values:
\[ I = \frac{12 \text{ V}}{4 \text{ ohms}} = 3 \text{ A} \]

4. **Substitute the Current Back into the Power Formula:**
Now we can substitute the value of current back into the power formula:
\[ P = V \times I = 12 \text{ V} \times 3 \text{ A} = 36 \text{ W} \]

5. **Conclusion:**
The power dissipated by the light bulb is:
\[ P = 36 \text{ watts} \]

### Final Answer:
The power dissipated by the light bulb is **36 watts**.

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