Question

Two resistors $A$ and $B$ have values $(3.0\pm 0.1)k\Omega$ and $(9.0\pm 0.3)k\Omega$ respectively. If they are connected in parallel, the percentage error in the equivalent resistance is_________$\%$.

Answer 1

Answer: 3.33

$A=3.0k\Omega ,\Delta A=0.1k\Omega$
$B=9.0k\Omega ,\Delta B=0.3k\Omega$
Now, equivalent parallel resistance $R_P$ is given by
$\frac{1}{R_P}=\frac{1}{A}+\frac{1}{B}...(i)$
$\therefore R_P=\frac{AB}{A+B}=\frac{3\times 9}{3+9}$
$R_P=2.25k\Omega$
Differentiating equation (i), we get,
$\frac{-\Delta R_P}{R_P^2}=\frac{-\Delta A}{A^2}-\frac{\Delta B}{B^2}$
$\therefore \Delta R_P=\Delta A{\left(\frac{R_P}{A}\right)}^2+\Delta B{\left(\frac{R_P}{B}\right)}^2$
$=(0.1){\left(\frac{2.25}{3.0}\right)}^2+(0.3){\left(\frac{2.25}{9.0}\right)}^2$
$=0.05625+0.01875$
$\Delta R_P=0.075k\Omega$
$\therefore$ Percentage error in equivalent resistance is,
$\frac{\Delta R_P}{R_P}\times 100=\frac{0.075}{2.25}\times 100$
$=3.33\%$