Two resistors of resistance R1=100±3 Ω and R2=200±4 Ω are connected in parallel. The equivalent resistance of the parallel combination is:

A
$(66.7 \pm 1.8) Ω$
B
$(66.7 \pm 4.0) Ω$
C
$(66.7 \pm 3.0) Ω$
D
$(66.7 \pm 7.0) Ω$

Solution:

Here, $R_{1} = (100 \pm 3) Ω; R_{2} = (200 \pm 4) Ω$

The equivalent resistance in parallel combination is

$\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \Rightarrow \frac{1}{R_{p}} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200} \Rightarrow R_{p} = \frac{200}{3} = 66.7 Ω$

The error in equivalent resistance is given by:

$\frac{△ R_{p}}{R_{p}^{2}} = \frac{△ R_{1}}{R_{1}^{2}} + \frac{△ R_{2}}{R_{2}^{2}} △ R_{p} = △ R_{1} {(\frac{R_{p}}{R_{1}})}^{2} + △ R_{2} {(\frac{R_{p}}{R_{2}})}^{2} △ R_{p} = 3 {(\frac{66.7}{100})}^{2} + 4 {(\frac{66.7}{200})}^{2} = 1.8 Ω$