Question

A conducting medium is shaped in a form of a quarter annulus of radius b and a (b > a) and thickness t. Resistivity of material is $\rho$.
[CD = AB = EF = GH = t]

• A. Resistance between faces ABCD and EFGH is $\frac{\pi \rho }{2tln\left(\frac{b}{a}\right)}$
• B. Resistance between faces BCGF and ADHE is $\frac{4\rho t}{\pi (b^2-a^2)}$
• C. Resistance between faces ABFE and CDHG is $\frac{2\rho }{\pi t}ln\left(\frac{b}{a}\right)$
• D. Resistance between faces ABCD and EFGH is $\frac{\rho \pi (a+b)}{\mu (b-a)t}$

Answer 1

Answer: (A), (B), (C)

Resistance between faces ABFE and CDHG is $\frac{2\rho }{\pi t}ln\left(\frac{b}{a}\right)$

For finding the resistance between faces ABCD and EFGH, let's consider an element of thickness 'dr' as shown in figure.
So, the resistance of the element will be:
$dR=\frac{\rho \pi r}{2t\left(dr\right)}$
$\frac{1}{R}=\frac{1}{\int dR}={\int }_a^b\frac{\pi tdr}{\rho \pi r}\Rightarrow \frac{1}{R}=\frac{2t}{\rho \pi }ln\left(\frac{b}{a}\right)$
$\therefore R=\frac{\pi \rho }{2tln\left(\frac{b}{a}\right)}$

Similarly, if we consider elements between the faces BCGF and ADHE, we will get $R=\frac{\rho t\times 4}{\pi (b^2-a^2)}$


Now, for faces ABFE and CDHG, the resistance of the element considered between them will be, $dR=\frac{\rho dr}{\frac{\pi r}{2}\times t}$
$\therefore R=\int dR=\frac{2\rho }{\pi t}{\int }_a^b\frac{dr}{r}=\frac{2\rho }{\pi t}ln\left(\frac{b}{a}\right)$