AP EAMCET · PHYSICS · Current Electricity
A wire of resistance ' \(R\) ' is bent in the form of a circular loop. Two points on the circle seperated 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 \(\frac{E^2}{4 R}\)
- B \(\frac{16 \mathrm{E}^2}{3 \mathrm{R}}\)
- C \(\frac{E^2}{R}\)
- D \(\frac{2 \mathrm{E}^2}{3 \mathrm{R}}\)
Answer & Solution
Correct Answer
(B) \(\frac{16 \mathrm{E}^2}{3 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
\( R_1 = \frac{R}{4}, R_2 = \frac{3R}{4} \) \( R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(\frac{R}{4})(\frac{3R}{4})}{\frac{R}{4} + \frac{3R}{4}} = \frac{\frac{3R^2}{16}}{R} = \frac{3R}{16} \) \( P = \frac{E^2}{R_{eq}} = \frac{E^2}{\frac{3R}{16}} = \frac{16E^2}{3R} \)
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