An e.m.f. $E=E_0 \cos \omega t$ is applied to circuit containing $L$ and $R$ in series. If $X_L=2 R$, then…
An e.m.f. $E=E_0 \cos \omega t$ is applied to circuit containing $L$ and $R$ in series. If $X_L=2 R$, then the power dissipated in the circuit is
- $\frac{E_o{ }^2}{12 R}$
- $\frac{\mathrm{E}_0{ }^2}{10 \mathrm{R}}$
- $\frac{\mathrm{E}_0{ }^2}{8 \mathrm{R}}$
- $\frac{E_0{ }^2}{6 R}$
Solution
Impedance in the circuit is given by,
$\begin{aligned}
Z^2= & R^2+X_L^2=R^2+(2 R)^2=5 R^2 \\
P= & \frac{E_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \cos \theta \\
& =\frac{E_0}{\sqrt{2}} \times \frac{E_0}{\sqrt{2} Z} \times \frac{R}{Z} \\
& =\frac{E_0^2 R}{2 Z^2} \\
& =\frac{E_0^2 R}{10 R^2} \\
& =\frac{E_0^2}{10 R}
\end{aligned} \quad \ldots\left(\because Z^2=5 R^2\right)$
Asked in: MHT CET 2024 (09 May Shift 2)
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