Question

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

• A. 5%
• B. 20%
• C. 10%
• D. 2.5%

Answer 1

Answer: (A)

5%

Answer 2

The power consumed by a bulb can be calculated using the formula:
$P = \frac{V^2}{R}$
Given that the rated voltage (V) is 220 V and the rated power (P) is 100 W, we can calculate the resistance (R) using the above formula:
$100 = \frac{{220^2}}{{R}}$

$R = \frac{{220^2}}{{100}}$
R = 484 Ω
Now, if the voltage across the bulb drops by 2.5% of the rated value, the new voltage (V') can be calculated as:
$V' = V - \frac{2.5}{100} \times V$

$V' = 220 - \frac{2.5}{100} \times 220$

$V' = 220 - 5.5$
$V' = 214.5 V$
To find the new power (P'), we can use the formula:
$P' = \frac{{V'^2}}{{R}}$

$P' = \frac{{214.5^2}}{{484}}$

$P' = \frac{{46084.25}}{{484}}$
$P' ≈ 95.15 W$
Now, let's calculate the percentage decrease in power compared to the rated value:
Percentage decrease in power = $\left( \frac{{P - P'}}{{P}} \right) \times 100$

Percentage decrease in power =$\left( \frac{{100 - 95.15}}{{100}} \right) \times 100$
Percentage decrease in power $\approx 4.85\% \approx 5\%$
Therefore, options (A) is correct.