MHT CET · Physics · Alternating Current
An e.m.f. \(E=E_0 \cos \omega t\) is applied to the L-R circuit. The inductive reactance is equal to the resistance ' \(R\) ' of the 'circuit. The power consumed in the circuit is
- A \(\frac{\mathrm{E}_0^2}{\sqrt{2} \mathrm{R}}\)
- B \(\frac{\mathrm{E}_0^2}{2 \mathrm{R}}\)
- C \(\frac{\mathrm{E}_0^2}{4 \mathrm{R}}\)
- D \(\frac{\mathrm{E}_0^2}{\mathrm{R}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{E}_0^2}{4 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& P=E_{r m s} I_{r m s} \cos \phi \\
& \cos \phi=\frac{R}{Z} \\
& \text { Also, } I_{r m s}=\frac{E_{r m s}}{Z}=\frac{E_0}{Z \sqrt{2}} \\
& \therefore \quad P=\frac{E_0}{\sqrt{2}} \times \frac{E_0}{Z \sqrt{2}} \times \frac{R}{Z} \\
&=\frac{E_0^2 R}{2 Z^2}...(i)
\end{aligned}\)
Given \(\mathrm{X}_{\mathrm{L}}=\mathrm{R}\)
\(\begin{array}{ll}
\therefore & Z=\sqrt{R^2+R^2}=\sqrt{2} R \\
\therefore & P=\frac{E_0^2}{4 R}...[From(i)]
\end{array}\)
& P=E_{r m s} I_{r m s} \cos \phi \\
& \cos \phi=\frac{R}{Z} \\
& \text { Also, } I_{r m s}=\frac{E_{r m s}}{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}...(i)
\end{aligned}\)
Given \(\mathrm{X}_{\mathrm{L}}=\mathrm{R}\)
\(\begin{array}{ll}
\therefore & Z=\sqrt{R^2+R^2}=\sqrt{2} R \\
\therefore & P=\frac{E_0^2}{4 R}...[From(i)]
\end{array}\)
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