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When you have learned to integrate find the resistance of the conductor of the previous problem and the voltage across it.

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SOLUTION :The resistance of the conductors is `R=underset(0)overset(1)int =(p DX)/(S)=(4P)/pi underset(0)overset(1)int (dx)/(y^2)`
Let us change the variables, nothing that y=a for x=0 and y=D for x=1. DIFFERENTIATING, we obtain `(dy)-dx=(D-a)/l, so dx=(l dy)/(D-a)`
Substituting into expression for the resistance, we obtain
`R=(4pl)/(pi (D-a)) underset(a)overset(D)int (dy)/(y^2)=(4pl)/(pi (D-a)) [-1/y]_(a)^(D) =(4pl)/(pi aD)`


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