A material B has twice the specific resistance of A. A circular wire made of B has twice the diameter of a wire made of A. Then for the two wires to have the same resistance, the ratio of lB $$/$$ lA of their respective lengths must be

Question

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Let the specific resistances, lengths, radii and areas of wires A and B be ρA,lA,rA,AA and ρB,lB,rB,AB respectively.Resistance of A $$=$$  $$RA=$$ρAlA$$π$$$$rA2Resistance$$ of $$B=$$ $$RB=$$ρBlB$$π$$$$rB2According$$ to given question, ρ$$B=2$$ρA,$$rB=2rA$$ and $$RA=RB$$ ThereforeρAlA$$π$$$$rA2=$$ρBlB$$π$$$$rB2Further$$, ρAlA$$π$$$$rA2=2$$ρAlB$$π$$$$2rA2$$∴ $$lBlA=21=2$$

A material B has twice the specific resistance of A. A circular wire made of B has twice the diameter of a wire made of A. Then for the two wires to have the same resistance, the ratio of l_B / l_A of their respective lengths must be

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A material B has twice the specific resistance of A. A circular wire made of B has twice the diameter of a wire made of A. Then for the two wires to have the same resistance, the ratio of lB / lA of their respective lengths must be

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A material B has twice the specific resistance of A. A circular wire made of B has twice the diameter of a wire made of A. Then for the two wires to have the same resistance, the ratio of l_B / l_A of their respective lengths must be