Q. Given $R_{1} = 5.0 \pm 0.2\Omega,\ R_{2} = 10.0 \pm 0.1\Omega.\ $What is total resistance in parallel with possible percentage error?
A) $15\Omega \pm 2\%$
B) 3.3$\ \Omega \pm 7\%$
C) 15$\ \Omega \pm 7\%$
D) $3.3\Omega \pm 2\%$
Explanation
(b) In parallel,
$R_{p} = \frac{R_{1}R_{2}}{R_{1} + R_{2}}$
$= \frac{5.0 \times 10.0}{5.0 + 10.0}$
$= \frac{50}{15} = 3.3\Omega$
Also,
$\frac{\mathrm{\Delta}R_{p}}{R_{p}} \times 100 = \frac{\mathrm{\Delta}R_{1}}{R_{1}} \times 100 + \frac{\mathrm{\Delta}R_{2}}{R_{2}} \times 100 + \frac{\mathrm{\Delta}\left( R_{1} + R_{2} \right)}{R_{1} + R_{2}} \times 100$
$= \frac{0.2}{5.0} \times 100 + \frac{0.1}{10.0} \times 100 + \frac{0.3}{15} \times 100$
$= 7\%$
$R_{p} = 3.3\Omega \pm 7\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.