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Questions  

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

a
66.7±1.8
b
300±7
c
150.8±2
d
92.3±3

detailed solution

Correct option is A

The equivalent resistance of parallel combination is R'=R1R2R1+R2=2003=66.7ΩFrom 1R'=1R1+1R2, we get ΔR'R'2=ΔR1R12+ΔR2R22ΔR'=R'2ΔR1R12+R'2ΔR2R22=66.710023+66.720024=1.8ΩHence, R'=(66.7±1.8)Ω

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