Question

Two resistances are measured in ohm and are given as:

$R_{1}=3 \, \Omega \pm1\%$

$R_{2}=6 \, \Omega \pm2\%$

When they are connected in parallel, the percentage error in equivalent resistance is

• A. $3\%$
• B. $4.5\%$
• C. $0.67\%$
• D. $1.33\%$

Answer 1

Answer: (D)

$\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$
$\frac{d R_{e q}}{R^{2}}=\frac{d R_{1}}{R_{1}^{2}}+\frac{d R_{2}}{R_{2}^{2}}\Rightarrow \frac{d R_{e q}}{R_{e q}}=\left(\frac{d R_{1}}{R_{1}^{2}} + \frac{d R_{2}}{R_{2}^{2}}\right)\times R_{e q}$
$\Rightarrow \left(\frac{1}{3} + \frac{2}{6}\right)\times 2$
$\Rightarrow \therefore \%$ error $=\frac{4}{3}\%$