Question 328: AC fundamentals

Question

An e.m.f. $E=E_0 \cos \omega t$ is applied to circuit containing L and R in series. If $\mathrm{X}_{\mathrm{L}}=2 \mathrm{R}$, then the power dissipated in the circuit is

Accepted Answer

$$\begin{aligned}
&	ext { Impedance in the circuit is given by, }\
&\begin{aligned}
& \mathrm{Z}^2=\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2=\mathrm{R}^2+(2 \mathrm{R})^2=5 \mathrm{R}^2 \
& \mathrm{P}=\frac{\mathrm{E}_0}{\sqrt{2}} 	imes \frac{\mathrm{I}_0}{\sqrt{2}} \cos 	heta \
& =\frac{\mathrm{E}_0}{\sqrt{2}} 	imes \frac{\mathrm{E}_0}{\sqrt{2} \mathrm{Z}} 	imes \frac{\mathrm{R}}{\mathrm{Z}} \
& =\frac{E_0^2 R}{2 Z^2} \
& =\frac{\mathrm{E}_0^2 \mathrm{R}}{10 \mathrm{R}^2} \
& \ldots .\left(\because Z^2=5 R^2ight) \
& =\frac{\mathrm{E}_0^2}{10 \mathrm{R}}
\end{aligned}
\end{aligned}$$

Answer

$\frac{\mathrm{E}_0{ }^2}{12 \mathrm{R}}$

Answer

$\frac{\mathrm{E}_0{ }^2}{10 \mathrm{R}}$

Answer

$\frac{\mathrm{E}_0{ }^2}{8 \mathrm{R}}$

Answer

$\frac{\mathrm{E}_0{ }^2}{6 \mathrm{R}}$

AC fundamentals

An e.m.f. E=E_0 \cos \omega t is applied to circuit containing L and R in series. If \mathrm{X}_{\mathrm{L}}=2 \mathrm{R}, then the power dissipated in the circuit is

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