If voltage across a bulb rated 220 volt-100 watt drops by `2.5 %` of its value, the percentage of the rated value by which the power would decrease is

Options

A. `20%`
B. `2.5%`
C. `5%`
D. `10%`

Answer

To solve the problem, we need to determine how the power of a bulb changes when the voltage across it drops by 2.5%. Let's break this down step by step.

### Step 1: Identify the given values
- Rated voltage (V) = 220 volts
- Rated power (P) = 100 watts
- Percentage drop in voltage = 2.5%

### Step 2: Understand the relationship between power and voltage
The power (P) consumed by a resistive load (like a bulb) is given by the formula:
\[ P = \frac{V^2}{R} \]
where \( R \) is the resistance of the bulb.

### Step 3: Calculate the change in voltage
The change in voltage (\( dV \)) can be calculated as:
\[ dV = -2.5\% \text{ of } V = -0.025 \times 220 = -5.5 \text{ volts} \]

### Step 4: Calculate the new voltage
The new voltage (\( V' \)) after the drop is:
\[ V' = V + dV = 220 - 5.5 = 214.5 \text{ volts} \]

### Step 5: Calculate the new power
Using the power formula, we can express the new power (\( P' \)) as:
\[ P' = \frac{(V')^2}{R} \]

To find \( R \), we can use the rated power at the rated voltage:
\[ P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P} = \frac{220^2}{100} = 484 \text{ ohms} \]

Now substitute \( R \) into the equation for \( P' \):
\[ P' = \frac{(214.5)^2}{484} \]

### Step 6: Calculate \( P' \)
Calculating \( (214.5)^2 \):
\[ (214.5)^2 = 46006.25 \]

Now substitute this value:
\[ P' = \frac{46006.25}{484} \approx 95.0 \text{ watts} \]

### Step 7: Calculate the decrease in power
The decrease in power (\( \Delta P \)) is:
\[ \Delta P = P - P' = 100 - 95 = 5 \text{ watts} \]

### Step 8: Calculate the percentage decrease in power
The percentage decrease in power is given by:
\[ \text{Percentage decrease} = \left( \frac{\Delta P}{P} \right) \times 100 = \left( \frac{5}{100} \right) \times 100 = 5\% \]

### Final Answer
The percentage of the rated value by which the power would decrease is **5%**.

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