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A resistor and an inductor are connected in series to an AC source of voltage $150 \sin (100 \pi t+\pi)$ volt. If the current in the circuit is $5 \sin \left(100 \pi t+\frac{2 \pi}{3}\right)$ ampere, then the average power dissipated and the resistance of the resistor are respectively
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The correct answer is:
$187.5 \mathrm{~W}, 15 \Omega$
Given, peak voltage, $V_0=150 \mathrm{~V}$
Peak current,
$$
I_0=5
$$
Phase difference between voltage and current,
$$
\phi=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}=60^{\circ}
$$
Now, average power in $R-L$ circuit,
$$
\begin{aligned}
P_{\mathrm{av}}=V_{\mathrm{rms}} & \cdot I_{\mathrm{rms}} \cdot \cos \phi \\
& =\frac{V_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cos 60^{\circ}=\frac{150}{\sqrt{2}} \times \frac{5}{\sqrt{2}} \times \frac{1}{2} \\
P_{\mathrm{av}}= & 187.5 \mathrm{~W}
\end{aligned}
$$
$\begin{gathered}\text { Impendence, } Z=\frac{V_0}{I_0}=\frac{150}{5} \\ Z=30 \Omega \\ \frac{R}{Z}=\cos \phi \\ \frac{R}{30}=\cos 60^{\circ} \Rightarrow R=30 \times \frac{1}{2} \\ R=15 \Omega\end{gathered}$
Peak current,
$$
I_0=5
$$
Phase difference between voltage and current,
$$
\phi=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}=60^{\circ}
$$
Now, average power in $R-L$ circuit,
$$
\begin{aligned}
P_{\mathrm{av}}=V_{\mathrm{rms}} & \cdot I_{\mathrm{rms}} \cdot \cos \phi \\
& =\frac{V_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cos 60^{\circ}=\frac{150}{\sqrt{2}} \times \frac{5}{\sqrt{2}} \times \frac{1}{2} \\
P_{\mathrm{av}}= & 187.5 \mathrm{~W}
\end{aligned}
$$
$\begin{gathered}\text { Impendence, } Z=\frac{V_0}{I_0}=\frac{150}{5} \\ Z=30 \Omega \\ \frac{R}{Z}=\cos \phi \\ \frac{R}{30}=\cos 60^{\circ} \Rightarrow R=30 \times \frac{1}{2} \\ R=15 \Omega\end{gathered}$
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