If resistance of the filament increases with temperature, what will be power dissipated in a `220 V - 100 W` lamp when connected to `110 V` power supply

Options

A. `25 W`
B. `lt 25 W`
C. `gt 25 W`
D. None of these

Explanation

To solve the problem of finding the power dissipated in a `220 V - 100 W` lamp when connected to a `110 V` power supply, we will follow these steps:

### Step 1: Calculate the resistance of the lamp.
The power of the lamp at its rated voltage can be calculated using the formula:
\[ P = \frac{V^2}{R} \]
Rearranging this formula to find the resistance \( R \):
\[ R = \frac{V^2}{P} \]
Substituting the values \( V = 220 \, V \) and \( P = 100 \, W \):
\[ R = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

### Step 2: Calculate the power dissipated at the new voltage.
Now, we need to find the power dissipated when the lamp is connected to a `110 V` power supply. The power dissipated can be calculated using the formula:
\[ P' = \frac{V'^2}{R} \]
Where \( V' = 110 \, V \) and \( R = 484 \, \Omega \):
\[ P' = \frac{(110)^2}{484} = \frac{12100}{484} \approx 25 \, W \]

### Step 3: Consider the effect of temperature on resistance.
Since the resistance of the filament increases with temperature, the actual power dissipated will be greater than the calculated power of `25 W`. This is because as the filament heats up, its resistance increases, leading to a decrease in current and thus an increase in power dissipation.

### Conclusion:
Therefore, when the `220 V - 100 W` lamp is connected to a `110 V` power supply, the power dissipated will be more than `25 W`.