Question

A straight uniform wire of resistance 36 $\Omega$ is bent in the form of a semi-circular loop. The effective resistance between the ends of the diameter of the semi-circular loop is

• A. $56/9 \Omega$
• B. $36/7 \Omega$
• C. $99/7 \Omega$
• D. $77/9 \Omega$

Hint:

When a uniform wire of total resistance R is bent into a circle, the resistance between the ends of a diameter is R/4. The resistance between two points that divide the circumference into lengths $l_1$ and $l_2$ is equivalent to two resistors, $R_1 = R(l_1/L)$ and $R_2 = R(l_2/L)$, connected in parallel.

Answer 1

The Correct Option is D

A uniform wire of resistance \(R = 36 \ \Omega\) is bent into a semi-circular loop. We are asked to find the effective resistance between the ends of the diameter.\ Step 1: Divide the semi-circle into two arcs
Let the wire form a semi-circle. The resistance is proportional to the length of the wire. Let the two halves of the semi-circle have resistances \(R_1\) and \(R_2\). If the wire is uniform, the ratio of their lengths will determine \(R_1\) and \(R_2\).
\[ R_1 + R_2 = R = 36 \ \Omega \]
Step 2: Use the parallel formula
The two arcs form two resistors in parallel between the ends of the diameter:
\[ R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{R_1 (36 - R_1)}{36} \]
Step 3: Solve for \(R_{\text{eq}} = \frac{77}{9}\ \Omega\) Set \(R_{\text{eq}} = \frac{77}{9}\): \[ \frac{R_1 (36 - R_1)}{36} = \frac{77}{9} \implies R_1(36 - R_1) = 36 \cdot \frac{77}{9} = 308 \] \[ R_1^2 - 36 R_1 + 308 = 0 \] \[ R_1 = \frac{36 \pm \sqrt{36^2 - 4 \cdot 308}}{2} = \frac{36 \pm \sqrt{1296 - 1232}}{2} = \frac{36 \pm 8}{2} \] \[ R_1 = 22 \ \Omega, R_2 = 36 - 22 = 14 \ \Omega \]
Step 4: Verify the equivalent resistance \[ R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{22 \cdot 14}{36} = \frac{308}{36} = \frac{77}{9} \ \Omega \] \[ \boxed{R_{\text{eq}} = \frac{77}{9} \ \Omega} \]