If the current in an electric bulb drops by `1%` the power decreases by

Options

A. 0.01
B. 0.02
C. 0.04
D. `0.5%`

Explanation

To solve the problem of how much the power decreases when the current in an electric bulb drops by 1%, we can follow these steps:

### Step 1: Understand the relationship between power, current, and resistance.
The power \( P \) dissipated in an electric bulb can be expressed using the formula:
\[ P = I^2 R \]
where \( I \) is the current flowing through the bulb and \( R \) is the resistance of the bulb.

### Step 2: Determine the new current after a 1% decrease.
If the current decreases by 1%, the new current \( I' \) can be calculated as:
\[ I' = I - 0.01I = 0.99I \]

### Step 3: Calculate the new power with the decreased current.
Using the new current \( I' \), we can find the new power \( P' \):
\[ P' = (I')^2 R = (0.99I)^2 R = 0.9801 I^2 R \]

### Step 4: Find the decrease in power.
The decrease in power \( \Delta P \) can be calculated as:
\[ \Delta P = P - P' = I^2 R - 0.9801 I^2 R \]
Factoring out \( I^2 R \):
\[ \Delta P = I^2 R (1 - 0.9801) = I^2 R (0.0199) \]

### Step 5: Conclusion.
Thus, the decrease in power when the current drops by 1% is approximately:
\[ \Delta P \approx 0.0199 I^2 R \]