Question

Figure shows a conductor of length $l$ having a circular cross-section. The radius of cross-section variance linearly from $a$ to $b$. The resistivity of the material is $\rho$. Find the resistance of the conductor.

Answer 1

dR, due to the small strip dx at a distance x d =

R = $\frac{fdx}{\pi Y^2}$ ..........( 1 )

$tan\theta =\frac{Y-a}{X}$ = $\frac{b-a}{L}$

$\frac{Y-a}{X}=\frac{b-a}{L}$

L ( Y - a ) = X ( b - a )

LY - La = Xb - Xa

$L\frac{dY}{dX}-0=b-a$ ( diff. w.r.t. x )

$L\frac{dY}{dX}=b-a$

dX = $\frac{LdY}{b-a}$ ...................( 2 )

Putting the value of dx in equation ( 1 )

dR = $\frac{fLdY}{\pi Y^2(b-a)}$

dR = $\frac{fL}{\pi (b-a)}$ $\frac{dY}{Y^2}$

${\int }_0^{R}dR=\frac{fL}{\pi (b-a)}$ ${\int }_a^b\frac{dY}{Y^2}$

R = $\frac{fL}{\pi (b-a)}$ $\frac{(b-a)}{ab}$

$=\frac{fL}{\pi ab}.$