Two resistors of resistances R1=100±3Ω and R2=200±4Ω are connected in parallel, then the equivalent resistance in parallel is (in ohm) (Use 1R'=1R1+1R2 and ΔR'ΔR'2=ΔR1R12+ΔR2R22)
66.7±1.8
300±7
150.8±2
92.3±3
The equivalent resistance of parallel combination is R'=R1R2R1+R2=2003=66.7 Ω
From, 1R'=1R1+1R2, we get
ΔR'R'2=ΔR1R12+ΔR2R22
ΔR'=(R'2)ΔR1R12+(R'2)ΔR2R22
=(66.7100)23+(66.7200)24=1.8 Ω
Hence, R'=(66.7±1.8)Ω