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. 0.2
B. 0.025
C. 0.05
D. 0.1

Explanation

To solve the problem step by step, we will follow the process outlined in the video transcript.

### Step 1: Identify the given values
- Voltage (V) = 220 volts
- Power (P) = 100 watts

### 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 known values:
\[ R = \frac{220^2}{100} \]
\[ R = \frac{48400}{100} \]
\[ R = 484 \, \text{ohms} \]

### Step 3: Calculate the new voltage after a 2.5% drop
The drop in voltage is given as 2.5% of the original voltage:
\[ \text{Voltage drop} = 2.5\% \times 220 \]
\[ = \frac{2.5}{100} \times 220 \]
\[ = 5.5 \, \text{volts} \]

Now, we can find the new voltage \( V_N \):
\[ V_N = V - \text{Voltage drop} \]
\[ V_N = 220 - 5.5 \]
\[ V_N = 214.5 \, \text{volts} \]

### Step 4: Calculate the new power at the new voltage
Using the same power formula:
\[ P_N = \frac{V_N^2}{R} \]
Substituting the new voltage and resistance:
\[ P_N = \frac{(214.5)^2}{484} \]
Calculating \( (214.5)^2 \):
\[ (214.5)^2 = 46006.25 \]
Now substituting:
\[ P_N = \frac{46006.25}{484} \]
\[ P_N \approx 95.06 \, \text{watts} \]

### Step 5: Calculate the decrease in power
The decrease in power is:
\[ \text{Decrease in power} = P - P_N \]
\[ = 100 - 95.06 \]
\[ = 4.94 \, \text{watts} \]

### Step 6: Calculate the percentage decrease in power
The percentage decrease in power is given by:
\[ \text{Percentage decrease} = \left( \frac{\text{Decrease in power}}{P} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage decrease} = \left( \frac{4.94}{100} \right) \times 100 \]
\[ = 4.94\% \]

### Step 7: Round the percentage decrease
Approximately, the percentage decrease in power is:
\[ \approx 5\% \]

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

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