Question

1 A wire of diameter d, length l and resistivity ρ forms a circular loop. A current enters
and leaves the loop at points P and Q as shown in the Figure. Show that the resistance R of
the wire is given by the expression

Answer 1

Explanations & Calculations

• The resistances of the wire segments $\small x$ and $\small (l-x)$ create a situation similar to two resistors connected in parallel between a couple of nodes.
• It's to be known that the equivalent resistance is given by $\small R_{EQ} = \large\frac{R_1R_2}{R_1+R_2}$ : which is the simplified form (only) for 2 resistors.
• Therefore,

Resistance of the $\small x$ long wire segment = $\large \frac{\rho x}{A}$

Resistance of the $\small (l-x)$ long segment = $\large \frac{\rho (l-x)}{A}$

$\qquad\qquad \begin{aligned} \small R &= \small \frac{\frac{\rho x}{A}\times\frac{\rho (l-x)}{A}}{\frac{\rho x}{A}+\frac{\rho (l-x)}{A}}\\ &= \small \frac{\rho \times x(l-x)}{Al}\\ &= \small \frac{\rho \times 4x(l-x)}{\pi d^2l} \,\,\,\,\,\,\,\,\,\,\,\, \because A = \frac{\pi d^2}{4}\\ &= \small \bold{ \frac{4\rho x(l-x)}{\pi d^2l}} \end{aligned}$