MHT CET · Physics · Alternating Current
In a \(\mathrm{L}-\mathrm{R}\) circuit, the inductive reactance is equal to the resistance ' \(R\) ' in the circuit. An emf \(\mathrm{E}=\mathrm{E}_0 \cos \omega \mathrm{t}\) is applied to the circuit. The power consumed in the circuit is
- A \(\frac{\mathrm{E}_0^2}{\sqrt{2} \mathrm{R}}\)
- B \(\frac{E_0^2}{4 R}\)
- C \(\frac{\mathrm{E}_0^2}{2 \mathrm{R}}\)
- D \(\frac{\mathrm{E}_0^2}{8 \mathrm{R}}\)
Answer & Solution
Correct Answer
(B) \(\frac{E_0^2}{4 R}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{P}=\mathrm{E}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi\)
\(\cos \phi=\frac{R}{Z}\)
Also, \(I_{\text {rms }}=\frac{E_{\text {rms }}}{Z}=\frac{E_0}{Z \sqrt{2}}\)
\(\therefore \quad P=\frac{E_0}{\sqrt{2}} \times \frac{E_0}{Z \sqrt{2}} \times \frac{R}{Z}\)
\(=\frac{E_0^2 R}{2 Z^2}\)
Given \(X_L=R\)
\(\therefore \quad \mathrm{Z}=\sqrt{\mathrm{R}^2+\mathrm{R}^2}\)
\(=\sqrt{2} \mathrm{R}\)
\(\therefore \quad P=\frac{E_0^2}{4 R}\)
\(\cos \phi=\frac{R}{Z}\)
Also, \(I_{\text {rms }}=\frac{E_{\text {rms }}}{Z}=\frac{E_0}{Z \sqrt{2}}\)
\(\therefore \quad P=\frac{E_0}{\sqrt{2}} \times \frac{E_0}{Z \sqrt{2}} \times \frac{R}{Z}\)
\(=\frac{E_0^2 R}{2 Z^2}\)
Given \(X_L=R\)
\(\therefore \quad \mathrm{Z}=\sqrt{\mathrm{R}^2+\mathrm{R}^2}\)
\(=\sqrt{2} \mathrm{R}\)
\(\therefore \quad P=\frac{E_0^2}{4 R}\)
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