A servo voltage stabiliser restricts the voltage output to `220 V +- 1%`. If an electric bulb rated at `220 V, 100 W` is connected to it, what will be the minimum and maximum power consumed by it?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the rated values of the bulb
The electric bulb is rated at:
- Voltage (V₀) = 220 V
- Power (P₀) = 100 W

### Step 2: Calculate the resistance of the bulb
Using the formula for resistance (R) based on the rated voltage and power:
\[ R = \frac{V^2}{P} \]
Substituting the values:
\[ R = \frac{220^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

### Step 3: Calculate the minimum and maximum voltage
The stabilizer restricts the output voltage to \(220 \, V \pm 1\%\).
- Minimum Voltage (V_min):
\[ V_{min} = 220 - 1\% \text{ of } 220 = 220 - 2.2 = 217.8 \, V \]
- Maximum Voltage (V_max):
\[ V_{max} = 220 + 1\% \text{ of } 220 = 220 + 2.2 = 222.2 \, V \]

### Step 4: Calculate the minimum and maximum current
Using Ohm's law, the current can be calculated as:
\[ I = \frac{V}{R} \]
- Minimum Current (I_min):
\[ I_{min} = \frac{V_{min}}{R} = \frac{217.8}{484} \approx 0.450 \, A \]
- Maximum Current (I_max):
\[ I_{max} = \frac{V_{max}}{R} = \frac{222.2}{484} \approx 0.459 \, A \]

### Step 5: Calculate the minimum and maximum power consumed
Using the formula for power (P) in terms of current and resistance:
\[ P = I^2 \cdot R \]
- Minimum Power (P_min):
\[ P_{min} = I_{min}^2 \cdot R = (0.450)^2 \cdot 484 \approx 98.01 \, W \]
- Maximum Power (P_max):
\[ P_{max} = I_{max}^2 \cdot R = (0.459)^2 \cdot 484 \approx 102.00 \, W \]

### Final Results
- Minimum Power Consumed: **98.01 W**
- Maximum Power Consumed: **102.00 W**

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