The resistance of the filament of a lamp increases with the increase in temperature. A lamp rated `100 W` and `220 V` is connected across `220 V` power supply. If the voltage drops by `10%`, then the power of the lamp will be

Options

A. `90 W`
B. `81 W`
C. between `90 W and 100 W`
D. between `81 W and 90 W`

Explanation

To solve the problem, we need to find the power consumed by the lamp when the voltage drops by 10%. Here’s a step-by-step solution:

### Step 1: Calculate the resistance of the lamp
The resistance \( R \) of the lamp can be calculated using the formula:
\[ R = \frac{V^2}{P} \]
where \( V \) is the voltage rating of the lamp and \( P \) is the power rating.

Given:
- \( V = 220 \, \text{V} \)
- \( P = 100 \, \text{W} \)

Substituting the values:
\[ R = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

### Step 2: Determine the new voltage after the drop
The voltage drops by 10%, so the new voltage \( V' \) is:
\[ V' = V - 0.1V = 0.9V = 0.9 \times 220 = 198 \, \text{V} \]

### Step 3: Calculate the power at the new voltage
Assuming the resistance remains constant (which is a simplification for this step), the power \( P' \) consumed by the lamp can be calculated using the formula:
\[ P' = \frac{(V')^2}{R} \]

Substituting the values:
\[ P' = \frac{(198)^2}{484} \]
Calculating \( (198)^2 \):
\[ 198^2 = 39204 \]
Now substituting back:
\[ P' = \frac{39204}{484} \approx 81 \, \text{W} \]

### Step 4: Consider the effect of temperature on resistance
Since the resistance of the filament increases with temperature, when the voltage drops, the power will not only depend on the calculated value but also on the fact that the resistance will decrease due to lower temperature. Hence, the actual power consumed will be slightly higher than 81 W.

### Step 5: Conclusion
Given the options, the power will be between 81 W and 90 W, but not exceeding 90 W since the original power was 100 W. Therefore, the final answer is:
\[ \text{The power of the lamp will be between } 81 \text{ W and } 90 \text{ W.} \]