Question

A wire of resistance \( R \) is bent in the form of a circular loop. Two points on the circle separated by a quarter circumference are connected to a battery of emf \( E \) and negligible internal resistance. The heat generated in the wire per second is:

• A. \( \frac{E^2}{4R} \)
• B. \( \frac{16E^2}{3R} \)
• C. \( \frac{E^2}{R} \)
• D. \( \frac{2E^2}{3R} \)

Hint:

For circuits involving parallel resistances, first determine equivalent resistance before applying \( P = \frac{E^2}{R} \).

Answer 1

The Correct Option is A

Step 1: Resistance Calculation The wire is bent into a circular shape, meaning the total resistance remains \( R \), but the quarter circumference separation implies current flows through two parallel paths each of resistance \( R/4 \). The equivalent resistance of the parallel paths: \[ R_{\text{eq}} = \frac{\frac{R}{4} \times \frac{R}{4}}{\frac{R}{4} + \frac{R}{4}} \] \[ = \frac{R^2}{16} \times \frac{1}{R/2} = \frac{R}{8} \] Step 2: Heat Generated per Second Using Joule's Law: \[ P = \frac{E^2}{R_{\text{eq}}} \] \[ = \frac{E^2}{R/4} = \frac{E^2}{4R} \] Conclusion Thus, the correct answer is: \[ \frac{E^2}{4R} \]