Question

The resistance of metal is given by V$=$IR. The voltage in the resistance is $V=(80.5\pm 0.1)V$ and current in the resistance is $I=(20.2\pm 0.2)A$, the value of resistance with its percentage error is :

• A. $\left(3.98\%\right)\Omega$
• B. $\left(2.34\%\right)\Omega$
• C. $\left(1.11\%\right)\Omega$
• D. $\left(2.4\%\right)\Omega$

Answer 1

The correct option is C $\left(1.11\%\right)\Omega$

$V=IR$

$V=80.5v$

$I=20.2A$

$R=\frac{V}{I}=3.985\Omega$

$\frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}$

%error $=(\frac{0.1}{80.5}+\frac{0.2}{20.2})\times 100=1.11$%