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. 25W
B. <25W
C. >25W
D. None of these

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the given information
We are given:
- Power rating of the lamp (P) = 100 W
- Voltage rating of the lamp (V) = 220 V
- Voltage of the power supply (V') = 110 V

### Step 2: Calculate the resistance of the filament
Using the formula for power:
\[ P = \frac{V^2}{R} \]
we can rearrange this to find the resistance (R):
\[ R = \frac{V^2}{P} \]

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

### Step 3: Calculate the power dissipated when connected to a 110 V supply
Using the power formula again, we can find the power (P') when the lamp is connected to a 110 V supply:
\[ P' = \frac{(V')^2}{R} \]
Substituting the values:
\[ P' = \frac{110^2}{R} \]
\[ P' = \frac{110^2}{484} \]
\[ P' = \frac{12100}{484} \]
\[ P' \approx 25 \, W \]

### Step 4: Consider the increase in resistance with temperature
The problem states that the resistance of the filament increases with temperature. Therefore, the actual resistance when the lamp is connected to the 110 V supply will be greater than 484 Ω.

### Step 5: Determine the effect on power
Since power is inversely proportional to resistance, if the resistance increases, the power dissipated will be less than the calculated 25 W.

### Conclusion
Thus, the power dissipated in the lamp when connected to a 110 V supply will be less than 25 W.

### Final Answer
The power dissipated will be less than 25 W.

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