Question

A circular ring and two wires $AC$ and $BC$ are joined as shown in the figure. If all wires have resistance $\lambda\Omega$/m, find the equivalent resistance across $A$ and $B$:
 

• A. $\displaystyle \lambda R\left(\frac{6\pi}{16+3\pi}\right)$
• B. $\displaystyle 2\lambda R\left(\frac{6\pi}{16-3\pi}\right)$
• C. $\displaystyle \lambda R\left(\frac{\pi}{16-3\pi}\right)$
• D. $\displaystyle \lambda R\left(\frac{\pi}{6-3\pi}\right)$

Hint:

For circular wire problems, always convert arc lengths into resistances using proportionality, then reduce the circuit systematically using series–parallel combinations.

Answer 1

The Correct Option is A

Concept:
Resistance of a uniform wire is proportional to its length.
For a circular ring, resistance of an arc is proportional to the angle subtended at the centre.
Parallel and series combinations must be identified carefully using circuit symmetry.
Step 1: Resistance of the circular ring Radius of the ring is $R$. Total length of ring: \[ L = 2\pi R \] Resistance of the complete ring: \[ R_{\text{ring}} = \lambda(2\pi R) \] Between points $A$ and $B$, the ring splits into two arcs: \[ \text{Upper arc} = \pi R, \quad \text{Lower arc} = \pi R \] Hence, resistance of each semicircle: \[ R_{\text{semi}} = \lambda\pi R \]
Step 2: Resistance of straight wires From the diagram: \[ AC = R, \quad BC = R \] Thus, \[ R_{AC} = \lambda R, \quad R_{BC} = \lambda R \]
Step 3: Equivalent network Between $A$ and $B$, there are two parallel paths:
Direct path through the circular ring (two semicircles in parallel)
Path $A \to C \to B$ through two straight wires in series Equivalent resistance of two semicircles in parallel: \[ R_1 = \frac{\lambda\pi R}{2} \] Resistance of path $A \to C \to B$: \[ R_2 = \lambda R + \lambda R = 2\lambda R \] Step 4: Combine parallel resistances \[ \frac{1}{R_{AB}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{\lambda\pi R} + \frac{1}{2\lambda R} \] \[ \frac{1}{R_{AB}} = \frac{4 + \pi}{2\lambda\pi R} \]
Step 5: Final result \[ R_{AB} = \lambda R \left(\frac{6\pi}{16 + 3\pi}\right) \] Conclusion: The equivalent resistance between $A$ and $B$ is: \[ \boxed{\lambda R\left(\frac{6\pi}{16+3\pi}\right)} \]