Question

A wire of uniform cross-sectional area and resistivity has total resistance \(36\,\Omega\) and is bent into a circle as shown in the figure. The equivalent resistance between points \(A\) and \(B\) is:

• A. \(6\,\Omega\)
• B. \(5\,\Omega\)
• C. \(30\,\Omega\)
• D. \(42\,\Omega\)

Hint:

For a circular wire of total resistance \(R\): \[ R_{\text{arc}} = R\left(\frac{\theta}{360^\circ}\right) \] If two arcs connect the same terminals, they are in parallel. For an angle \(\theta\): \[ R_1=\frac{\theta}{360^\circ}R, \qquad R_2=\frac{360^\circ-\theta}{360^\circ}R \] and \[ R_{\text{eq}} = \frac{R_1R_2}{R_1+R_2} \]

Answer 1

The Correct Option is B

Step 1: Determine the resistances of the two arcs. The wire forms a complete circle of total resistance: \[ R=36\,\Omega \] The points \(A\) and \(B\) subtend an angle of \[ 60^\circ \] Therefore, resistance of the smaller arc is: \[ R_1 = 36\left(\frac{60}{360}\right) = 6\,\Omega \] The remaining arc subtends \[ 360^\circ-60^\circ=300^\circ \] Hence, \[ R_2 = 36\left(\frac{300}{360}\right) = 30\,\Omega \]

Step 2: Identify the combination. The two arcs connect the same points \(A\) and \(B\). Therefore, they are in parallel. \[ R_{AB} = \frac{R_1R_2}{R_1+R_2} \] \[ = \frac{(6)(30)}{6+30} \] \[ = \frac{180}{36} \] \[ = 5\,\Omega \]

Step 3: Identify the correct option. \[ \boxed{R_{AB}=5\,\Omega} \] Hence, the correct answer is: \[ \boxed{\text{(B)}} \]