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Two metal wires have the resistivity 1.62 10^-8 and 4.86 x 10^-8. Compare the resistance of the two wires if wire A is twice as long as wire B and half of its diameter. |
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Answer» riend,◆ ANSWER -R1/R2 = 1/6◆ EXPLAINATION-# Given-ρ1 = 1.62×10^-8 ohm-mρ2 = 4.86×10^-8 ohm-ml1/l2 = 2d1/d2 = 1/2# Solution-Resistance of wire A,R1 = ρ1l1 / A1Resistance of wire B,R2 = ρ2l2 / A2Ratio of resistances is -R1/R2 = (ρ1l1/A1) / (ρ2l2/A2)R1/R2 = (ρ1/ρ2) × (l1/l2) × (d2/d1)^2R1/R2 = (1.62×10^-8 / 4.86×10^-8) × (2) × (1/2)^2R1/R2 = (1/3) × (2) × (1/4)R1/R2 = 1/6 Therefore, resistance of wire B is SIX TIMES that of B.Hope it helps.. |
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