Question

Two resistors of resistances R 1 = 100+/- 1 ohm and R2= 200 +/- 2 ohm are connected in parallel. the resistance of the parallel combination is what?

Answer 1

$\huge \underline \green{\sf{ Answer :- }}$

Here, R1(100±3)Ω;R2 = (200±4)Ω 

The equivalent resistance in parallel combination is

R11 = R11+R21,

Rp1 = 1001+2001 = 2003,

Rp = 3200 = 66.7Ω

The error in equivalent resistance is given by

Rp2ΔRp

=R12ΔR1+R22ΔR2;

ΔRp=ΔR1(R1Rp)2+ΔR2(R2Rp)2

= 3(10066.7)2+4(20066.7)2

= 1.8Ω 

Hence, the equivalent resistance along with error in parallel combination is (66.7 ± 1.8)Ω