The voltage across a bulb is decreased by 2%. Assuming that the resistance of the filament remains unchanged, the power of the bulb will

Options

A. decrease by 2%
B. increase by 2%
C. decrease by 4%
D. increase by 4%

Explanation

To solve the problem, we need to analyze how a decrease in voltage affects the power of the bulb, given that the resistance remains constant.

### Step-by-Step Solution:

1. **Understanding the Power Formula**:
The power \( P \) consumed by a bulb (or any resistive load) can be expressed using the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the bulb and \( R \) is the resistance.

2. **Initial Power Calculation**:
Let the initial voltage be \( V_0 \). The initial power \( P_0 \) can be expressed as:
\[ P_0 = \frac{V_0^2}{R} \]

3. **Calculating the New Voltage**:
The problem states that the voltage is decreased by 2%. Therefore, the new voltage \( V' \) can be calculated as:
\[ V' = V_0 - 0.02V_0 = 0.98V_0 \]

4. **New Power Calculation**:
Now, we can calculate the new power \( P' \) using the new voltage:
\[ P' = \frac{(V')^2}{R} = \frac{(0.98V_0)^2}{R} \]
Simplifying this, we get:
\[ P' = \frac{0.9604V_0^2}{R} = 0.9604 \cdot \frac{V_0^2}{R} = 0.9604P_0 \]

5. **Finding the Percentage Decrease in Power**:
To find the percentage decrease in power, we can use the formula:
\[ \text{Percentage Decrease} = \frac{P_0 - P'}{P_0} \times 100 \]
Substituting \( P' = 0.9604P_0 \):
\[ \text{Percentage Decrease} = \frac{P_0 - 0.9604P_0}{P_0} \times 100 = \frac{0.0396P_0}{P_0} \times 100 = 3.96\% \]
Rounding this, we can say the power decreases by approximately 4%.

6. **Conclusion**:
Therefore, the power of the bulb will decrease by 4%. The correct answer is option C.