Question

If percentage change in current through resistor is $1\%$, then the change in power through it would be :

• A. 1%
• B. 2%
• C. 1.7%
• D. 0.5%

Answer 1

Answer: (B)

Maximum percentage error arises due to limit of accuracy of the measured value.
By Joule's law of heating, the power change due to current (i), through resistor $(R)$ is given by
$P=i^{2}R$
Taking log on both sides, we get
$\log P=2\log i+\log R$
Taking partial differentiation, we have
$\frac{\Delta P}{P}\times 100=2\frac{\Delta i}{i}\times 100+\frac{\Delta R}{R}\times 100$
Given, $\frac{\Delta i}{i}=1$ and $\frac{\Delta R}{R}=0$
$\therefore \frac{\Delta P}{P}\times 100=2\times 1\%$
$=2\%$