標題:
Electricity about resistance
Two copper wires, A and B, of the same length have the ratio of mass of 4 : 9. Assuming the cross-sectional are uniform, what is the ratio of the resistance of A to that of B ? The answer is 9 : 4. But why, how to calculate it ?? thx for answering. 更新: Steps are needed.
最佳解答:
Let Ma, Mb be the masses of wire A and B respectively Aa and Ab be their respective cross-sectional areas and L be their lengths Thus resistance of wire A, Ra = pL/Aa where p is the resistivity of copper resistance of wire B, Rb = pL/Ab Ra/Rb = [pL/Aa]/[pL/Ab] = Ab/Aa --------------------- (1) But Ma = density x volume = d.(Aa.L) (where d is the density of copper) and Mb = d.Ab.L thus Ma/Mb = Aa/Ab = 4/9 substitute into (1), Ra/Rb = 9/4
其他解答:
Electricity about resistance
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發問:Two copper wires, A and B, of the same length have the ratio of mass of 4 : 9. Assuming the cross-sectional are uniform, what is the ratio of the resistance of A to that of B ? The answer is 9 : 4. But why, how to calculate it ?? thx for answering. 更新: Steps are needed.
最佳解答:
Let Ma, Mb be the masses of wire A and B respectively Aa and Ab be their respective cross-sectional areas and L be their lengths Thus resistance of wire A, Ra = pL/Aa where p is the resistivity of copper resistance of wire B, Rb = pL/Ab Ra/Rb = [pL/Aa]/[pL/Ab] = Ab/Aa --------------------- (1) But Ma = density x volume = d.(Aa.L) (where d is the density of copper) and Mb = d.Ab.L thus Ma/Mb = Aa/Ab = 4/9 substitute into (1), Ra/Rb = 9/4
其他解答:
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