Question

If the electric current in a lamp decreases by $5\%$, then the power output decreases by:

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

Answer 1

Answer: (B)

Let the original current be $I$ and the resistance of the lamp be $R$

Then, power is $P=I^{2}R..........\left(i\right)$

Reduced current value is $95\%$ of original current value.

$I^{\prime }=0.95I$

New power is:

$P^{\prime }=I^{\prime 2}R$

$P^{\prime }={0.95}^2{I}^{2}R$

$P^{\prime }=0.9025I^{2}R$

$P^{\prime }\approx 0.9R$