Question

A wire of uniform cross-section \(A\), length \(l\) and resistance \(R\) is bent into a complete circle. The equivalent resistance between any two diametrically opposite points will be:

• A. \(\dfrac{R}{4}\)
• B. \(\dfrac{R}{8}\)
• C. \(4R\)
• D. \(\dfrac{R}{2}\)

Hint:

For a circular wire of total resistance \(R\): \[ \text{Diametrically opposite points} \Rightarrow \frac{R}{2} \parallel \frac{R}{2} \] Therefore, \[ R_{\text{eq}} = \frac{R}{4} \] This is a very common CUET/JEE/NEET formula-based question.

Answer 1

The Correct Option is A

Step 1: Divide the circular wire into two equal halves. The total resistance of the wire is: \[ R \] Since diametrically opposite points divide the circle into two equal semicircles, resistance of each semicircle is: \[ R_1=R_2=\frac{R}{2} \]

Step 2: Identify the combination. The two semicircular resistances connect the same pair of points. Hence they are in parallel. \[ R_{\text{eq}} = \frac{R_1R_2}{R_1+R_2} \]

Step 3: Substitute the values. \[ R_{\text{eq}} = \frac{\left(\frac{R}{2}\right)\left(\frac{R}{2}\right)} {\frac{R}{2}+\frac{R}{2}} \] \[ = \frac{\frac{R^2}{4}}{R} \] \[ = \frac{R}{4} \] Therefore, the equivalent resistance between diametrically opposite points is \[ \boxed{\frac{R}{4}} \]