KCET · Physics · Alternating Current
An inductor of inductance \( L \) and resistor \( R \) are joined together in series and connected by a source of frequency \( \omega \). The power dissipated in the circuit is
- A \( \frac{V}{R^{2}+\omega^{2} L^{2}} \)
- B \( \frac{R^{2}+\omega^{2} L^{2}}{V} \)
- C \( \frac{V^{2} R}{\sqrt{R^{2}+\omega^{2} L^{2}}} \)
- D \( \frac{V^{2} R}{R^{2}+\omega^{2} L^{2}} \)
Answer & Solution
Correct Answer
(D) \( \frac{V^{2} R}{R^{2}+\omega^{2} L^{2}} \)
Step-by-step Solution
Detailed explanation
\(I=\frac{v}{\sqrt{R^{2}+\omega^{2} L^{2}}} \\ P=I^{2} R=\frac{v^{2} R}{R^{2}+\omega^{2} L^{2}} \)
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