Question

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

• A. 0.01
• B. 0.02
• C. 0.017
• D. 0.005

Answer 1

Answer: (B)

0.02

Answer 2

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 $\frac{\Delta R}{R}=0$
$\therefore \frac{\Delta P}{P} \times 100=2 \times 1 \%$
$=2 \%$