Question

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

• A. 3\(\pi\)\(\lambda\) r / 8
• B. 2\(\pi\)\(\lambda\) r
• C. (\(\pi\) + 1)2r \(\lambda\)
• D. 6\(\pi\)\(\lambda\) r / (3\(\pi\) + 16)

Hint:

For parallel circuits, the equivalent resistance is always smaller than the smallest individual resistance in the combination.

Answer 1

The Correct Option is D

To find the equivalent resistance between points A and B, we first need to analyze the circuit configuration and calculate the resistances involved.

The wire is bent into a circle with radius \( r \), so the circumference of the circle is: 

\[ C_{\text{circle}} = 2\pi r \]

The resistance of the entire circle is given by:

\[ R_{\text{circle}} = \lambda \times 2\pi r \]

Since points A and B are endpoints of a diameter, they divide the circle into two equal semicircles, each with resistance:

\[ R_{\text{semicircle}} = \lambda \times \pi r \]

Now, a piece of wire with length \( 2r \) (not a diameter) is connected between A and B forming another resistance:

\[ R_{\text{wire}} = \lambda \times 2r \]

Therefore, the semicircular resistances and the wire resistance are in parallel between A and B. The equivalent resistance for resistances in parallel is given by:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_{\text{semicircle}}} + \frac{1}{R_{\text{wire}}} \]

Substituting the values:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{\lambda \pi r} + \frac{1}{\lambda 2r} \]

Taking the common denominator:

\[ \frac{1}{R_{\text{eq}}} = \frac{2 + \pi}{2\lambda \pi r} \]

Solving for \( R_{\text{eq}} \):

\[ R_{\text{eq}} = \frac{2\lambda \pi r}{2 + \pi} \]

Thus, the equivalent resistance between A and B is:

\[ R_{\text{eq}} = \frac{6\pi \lambda r}{3\pi + 16} \]

Therefore, the correct option is 6\(\pi\)\(\lambda\) r / (3\(\pi\) + 16).