Question

If voltage across a bulb rated $220 \,V$- $100\, W$ drops by $2.5\%$ of its rated value, the percentage of the rated value by which the power would decrease is

• A. $20\,\%$
• B. $2.5\,\%$
• C. $5\,\%$
• D. $10\,\%$

Answer 1

The Correct Option is C

To solve this problem, we need to understand how the power of a bulb changes with voltage. The bulb is rated at 220 \, \text{V} and 100 \, \text{W}. Let's break down the solution step-by-step:

1. The power (P) absorbed by an electrical device can be expressed as:

   P = \frac{V^2}{R}

   where V is the voltage across the device and R is its resistance.

2. We know the rated power (P_0 = 100 \, \text{W}) is when the voltage is 220 \, \text{V}. The resistance (R) can be found using:

   R = \frac{V_0^2}{P_0} = \frac{(220)^2}{100} = 484 \, \Omega

3. The voltage drops by 2.5\% of its rated value:

   V_{\text{new}} = 220 \, \text{V} - \frac{2.5}{100} \times 220 \, \text{V} = 220 \, \text{V} - 5.5 \, \text{V} = 214.5 \, \text{V}

4. Calculate the new power:

   P_{\text{new}} = \frac{V_{\text{new}}^2}{R} = \frac{(214.5)^2}{484} \approx 95.0625 \, \text{W}

5. Determine the decrease in power and express it as a percentage of the rated power:

   \Delta P = P_0 - P_{\text{new}} = 100 \, \text{W} - 95.0625 \, \text{W} = 4.9375 \, \text{W}

   \text{Percentage decrease} = \left(\frac{\Delta P}{P_0}\right) \times 100 = \left(\frac{4.9375}{100}\right) \times 100 = 4.9375\%

6. Therefore, the closest percentage given among the options is 5\%. Thus, the correct answer is:

   5%