ExamBro
ExamBro
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

  1. A \(\frac{\mathrm{E}_0^2}{\sqrt{2} \mathrm{R}}\)
  2. B \(\frac{E_0^2}{4 R}\)
  3. C \(\frac{\mathrm{E}_0^2}{2 \mathrm{R}}\)
  4. D \(\frac{\mathrm{E}_0^2}{8 \mathrm{R}}\)
Verified Solution

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 \(X_L=R\)
\(\therefore \quad \mathrm{Z}=\sqrt{\mathrm{R}^2+\mathrm{R}^2}\)
\(=\sqrt{2} \mathrm{R}\)
\(\therefore \quad P=\frac{E_0^2}{4 R}\)
Same subject
Explore more questions on app