Calculate equivalent resistance of two resistors R1 and R2 in parallel where, R1=(6±0.2)ohm and R2=(3±0.1)ohm.

In parallel,1R=1R1+1R2  …….. (i)R=R1R2R1+R2=(6)(3)6+3=2ohmDifferentiating Eq. (i), we have-dRR2=-dR1R12-dR2R22Therefore, maximum permissible error in equivalent resistance may beΔR=ΔR1R12+ΔR2R22R2Substituting the values we get,ΔR=0.2(6)2+0.1(3)2(2)2=0.07ohmR=(2±0.07)ohm