Two resistors R1 = (100 ± 3) Ω and R2 = (200 ± 4) Ω are connected in parallel. Choose the correct resistance of this parallel combination.

Here,R1=(100±3)Ω,R2=(200±4)ΩThe equivalent resistance in parallel combination is1Rp=1R1+1R2⋅1Rp=1100+1200=3200,Rp=2003=66.7ΩThe error in equivalent resistance is given byΔRpRρ2=ΔR1R12+ΔR2R22;ΔRp=ΔR1RpR12+ΔR2RpR22=366.71002+466.72002=1.8ΩHence, the equivalent resistance along with error in parallel combination is(66.7±1.8)Ω