If the current in a electric bulb drops by 2% then the power decreases by

Options

A. 0.01
B. 0.02
C. 0.04
D. 0.16

Answer

To solve the problem of how much the power decreases when the current in an electric bulb drops by 2%, we can follow these steps:

### Step 1: Understand the relationship between power and current
The power (P) in an electrical circuit can be expressed as:
\[ P = I^2 R \]
where \( I \) is the current and \( R \) is the resistance.

### Step 2: Define the initial current
Let’s assume the initial current \( I_1 \) is 100 A (this is an arbitrary choice for calculation purposes).

### Step 3: Calculate the final current after a 2% drop
If the current drops by 2%, the final current \( I_2 \) can be calculated as:
\[ I_2 = I_1 - (0.02 \times I_1) \]
\[ I_2 = 100 - (0.02 \times 100) = 100 - 2 = 98 \text{ A} \]

### Step 4: Calculate the initial power
The initial power \( P_1 \) can be calculated using the initial current:
\[ P_1 = I_1^2 R = (100)^2 R = 10000 R \]

### Step 5: Calculate the final power
The final power \( P_2 \) can be calculated using the final current:
\[ P_2 = I_2^2 R = (98)^2 R = 9604 R \]

### Step 6: Calculate the decrease in power
The decrease in power \( \Delta P \) is given by:
\[ \Delta P = P_1 - P_2 = 10000 R - 9604 R = 396 R \]

### Step 7: Calculate the percentage decrease in power
To find the percentage decrease in power, we use the formula:
\[ \text{Percentage decrease} = \frac{\Delta P}{P_1} \times 100 \]
Substituting the values we have:
\[ \text{Percentage decrease} = \frac{396 R}{10000 R} \times 100 = \frac{396}{100} = 3.96\% \]

### Conclusion
Thus, if the current in an electric bulb drops by 2%, the power decreases by approximately **3.96%**, which can be rounded to **4%**.

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