Question

In ohm's law experiment, potential drop across a resistance was measured as v = 5.0 volt and current was measured as i = 2.00 amp. Find the maximum permissible error in resistance.

Answer 1

$R=\frac{v}{i}=v\times i^{-1}{\left(\frac{\Delta R}{R}\right)}_{max}=\frac{\Delta v}{v}+\frac{\Delta i}{i}$
$v=5.0volt\to \Delta v=0.1volt$
$i=2.00amp\to \Delta i=0.01amp\%{\left(\frac{\Delta R}{R}\right)}_{max}=\left(\frac{0.1}{5.0}+\frac{0.01}{2.00}\right)\times 100\%=2.5\%$
value of R from the observation R = $\frac{v}{i}=\frac{5.0}{2.00}=2.5\Omega$
So we can write R = (2.5 $\pm$ 2.5 %) $\Omega$