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

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