An e.m.f. $E=E_0 \cos \omega t$ is applied to the L-R circuit. The inductive reactance is equal to the…
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
$\frac{\mathrm{E}_0^2}{\sqrt{2} \mathrm{R}}$
$\frac{\mathrm{E}_0^2}{2 \mathrm{R}}$
$\frac{\mathrm{E}_0^2}{4 \mathrm{R}}$
$\frac{\mathrm{E}_0^2}{\mathrm{R}}$
Solution
$\begin{aligned}
& P=E_{r m s} I_{r m s} \cos \phi \\
& \cos \phi=\frac{R}{Z} \\
& \text { Also, } I_{r m s}=\frac{E_{r m s}}{Z}=\frac{E_0}{Z \sqrt{2}} \\
& \therefore \quad P=\frac{E_0}{\sqrt{2}} \times \frac{E_0}{Z \sqrt{2}} \times \frac{R}{Z} \\
&=\frac{E_0^2 R}{2 Z^2}...(i)
\end{aligned}$
Given $\mathrm{X}_{\mathrm{L}}=\mathrm{R}$
$\begin{array}{ll}
\therefore & Z=\sqrt{R^2+R^2}=\sqrt{2} R \\
\therefore & P=\frac{E_0^2}{4 R}...[From(i)]
\end{array}$