Question

If current flowing through a resistance is, $I=2.00\pm 0.01\text{A}$, whereas the resistance $R$ is $R=10.00\pm 0.01\Omega$; Then the maximum error in the calculation of voltage is

• A. $0.21\text{V}$
• B. $0.12\text{V}$
• C. $0.02\text{V}$
• D. $0.2\text{V}$

Answer 1

The correct option is B $0.12\text{V}$

$I=2.00\pm 0.01\text{A}$
$R=10.00\pm 0.01\Omega$
We know,
$V=IR$ (Ohm's law)
$\frac{\Delta V}{V}=\frac{\Delta I}{I}+\frac{\Delta R}{R}$
$\frac{\Delta V}{V}=\frac{\left(\Delta I\right)R+I\left(\Delta R\right)}{IR}$
$\Delta V=\left(\Delta I\right)R+I\left(\Delta R\right)$
$=0.01\times 10+2.00\times 0.01$
$=0.1+0.02$
$\Delta V=0.12\text{V}$