MHT CET · Physics · Alternating Current
An e.m.f. \(\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}\) is applied to a circuit containing 'L' and 'R' in series. If \(X_{L}=R\), then the power dissipated in the circuit is
- A \(\frac{\mathrm{E}_{0}^{2}}{4 \mathrm{R}}\)
- B \(\frac{\mathrm{E}_{0}}{2 \mathrm{R}}\)
- C \(\frac{\mathrm{E}_{0}}{4 \mathrm{R}}\)
- D \(\frac{\mathrm{E}_{0}^{2}}{2 \mathrm{R}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{E}_{0}^{2}}{2 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\mathrm{E}_{0} \sin \omega \mathrm{t}\)
\(P=\frac{E_{0}}{\sqrt{2}} \times \frac{I_{0}}{\sqrt{2}} \cos \theta\)
\(=\frac{E_{0}}{\sqrt{2}} \times \frac{E_{0}}{\sqrt{2} Z} \times \frac{R}{Z}\)
\(=\frac{E_{0}^{2} R}{2 Z^{2}} \quad \quad Z^{2}=R^{2}+\omega L^{2}=2 R^{2}\)
\(=\frac{E_{0}^{2} R}{4 R^{2}}=\frac{E_{0}^{2}}{4 R}\)
\(P=\frac{E_{0}}{\sqrt{2}} \times \frac{I_{0}}{\sqrt{2}} \cos \theta\)
\(=\frac{E_{0}}{\sqrt{2}} \times \frac{E_{0}}{\sqrt{2} Z} \times \frac{R}{Z}\)
\(=\frac{E_{0}^{2} R}{2 Z^{2}} \quad \quad Z^{2}=R^{2}+\omega L^{2}=2 R^{2}\)
\(=\frac{E_{0}^{2} R}{4 R^{2}}=\frac{E_{0}^{2}}{4 R}\)
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