Question

If voltage across a bulb rated 220 Volts,100 Watt drops by 2.5% of its rated value the percentage of the rated value by which the power would decrease is :

• A. 5%
• B. 10%
• C. 20%
• D. 2.5%

Answer 1

The correct option is A 5%

Since, the bulb is rated 220v and 100 watt

and,

we know that,

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

where

$P\to$ power

$V\to$ voltage

$R\to$ resistance of bulb (in this case)

$\Rightarrow R=\frac{V^2}{P}=\frac{220\times 220}{100}$

$\Rightarrow R=484\Omega$

Now,

Voltage drops by 2.5% of its rated value

$\Rightarrow$ New voltage $\left(V_n\right)=220-\frac{2.5\times 220}{100}$

$=220-5.5$

$=214.5V$

$\Rightarrow$ New power $\left(P_n\right)=\frac{(V_n{)}^2}{R}$

$=\frac{214.5\times 214.5}{484}$

$\Rightarrow P_n=95.06W$

$\Rightarrow$ % decrease in power $\frac{(100-P_n)\times 100}{100}$

$=\frac{(100-95.06)}{4.94\%}$

$\simeq 5$ %