A wire of uniform resistance $$\lambda$$Ω/m is bent into a circle of radius r and another piece of wire with length 2r is connected between points A and B (AOB) as shown in figure. The equivalent resistance between points A and B is ____Ω .

step 1: resistance of circular wire

Between A and B, the circle splits into two arcs:

• smaller arc (quarter circle): $$length=\frac{\pi r}{2}$$
$$→resis\tan ce=\frac{\pi r\lambda}{2}$$

• larger arc (remaining): $$length=\frac{3\pi r}{2}$$
$$→resis\tan ce=\frac{3\pi r\lambda}{2}$$

step 2: resistance of straight wire AB

Length = 2r

RAB=2rλ

step 3: all paths are in parallel

$$ \frac{1}{R_{eq}} = \frac{1}{\left( \frac{\pi r \lambda}{2} \right)} + \frac{1}{\left( \frac{3 \pi r \lambda}{2} \right)} + \frac{1}{2 \pi r \lambda} $$

solving this gives

$${R_{eq}}=r\lambda\cdot\frac{3\pi/4\times8}{3\pi+16}=r\lambda\cdot\frac{6\pi}{3\pi+16}$$