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 step by step, we will follow the calculations as outlined in the video transcript.

### Step 1: Identify the Given Values
- Power (P) = 100 W
- Voltage (V) = 220 V

### Step 2: Calculate the Resistance of the Bulb
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 New Voltage after the Drop
The voltage drops by 2.5%. To find the new voltage (Vn):
\[ \text{Voltage drop} = 2.5\% \text{ of } 220 \]
\[ \text{Voltage drop} = \frac{2.5}{100} \times 220 = 5.5 \, V \]

Now, subtract this drop from the original voltage:
\[ V_n = 220 - 5.5 = 214.5 \, V \]

### Step 4: Calculate the New Power
Using the new voltage, we can calculate the new power (P_n):
\[ P_n = \frac{V_n^2}{R} \]

Substituting the values:
\[ P_n = \frac{(214.5)^2}{484} \]
\[ P_n = \frac{46056.25}{484} \]
\[ P_n \approx 95.06 \, W \]

### Step 5: Calculate the Percentage Decrease in Power
Now we find the percentage decrease in power:
\[ \text{Percentage decrease} = \frac{P - P_n}{P} \times 100 \]

Substituting the values:
\[ \text{Percentage decrease} = \frac{100 - 95.06}{100} \times 100 \]
\[ \text{Percentage decrease} = \frac{4.94}{100} \times 100 \]
\[ \text{Percentage decrease} = 4.94\% \]

### Conclusion
The percentage of the rated value by which the power would decrease is approximately **4.94%**, which can be rounded to **5%**.

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