SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

${\mathrm{v}}_{\mathit{s}}=12\cos \left(1000\mathit{t}+15°\right)\mathrm{V}$
$\mathrm{\omega }=1000\mathrm{rad}∕\sec ,\mathrm{R}=30\mathrm{\Omega },$ $\mathrm{L}=65\mathrm{mH},\mathrm{C}=40\mathrm{\mu }\mathrm{F}$
Form equations(1) & (2) we can write
${\mathrm{X}}_{\mathrm{L}}=\mathrm{\omega }\mathrm{L}=1000\times 65\times {10}^{-3}=65\mathrm{\Omega }$
${\mathit{X}}_{\mathrm{C}}=\frac{1}{\mathrm{\omega }\mathrm{C}}=\frac{1}{1000\times 40\times {10}^{-6}}$
${\mathrm{X}}_{\mathrm{C}}=25\mathrm{\Omega }$
From equations (3) & (4) we can write
Impedance magnitude $\left|\mathrm{Z}\right|=\sqrt{{30}^2+{\left(65-25\right)}^2}=50\mathrm{\Omega }$
Impedance angle $\mathrm{\theta }={\tan }^{-1}\left(\frac{65-25}{30}\right)$ $={\tan }^{-1}\left(\frac{4}{3}\right)$ degree
Then the current through the circuit can be given from equation(5) as,
$\mathit{i}\left(\mathit{t}\right)=\frac{12}{50}\cos \left(1000\mathit{t}+15°-\mathrm{\theta }\right)\mathrm{A}$
$\mathit{i}\left(\mathit{t}\right)=0.24\cos \left(1000\mathit{t}+15°-{\tan }^{-1}\frac{4}{3}\right)\mathrm{A}$