Physics - EMI/AC Question with Solution | TestHub

as given $Z=\sqrt{{\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)}^2+R^2}$

$X_{\mathrm{L}}={\omega }_{\mathrm{L}}=100\times 100\times {10}^{-3}=10 \mathrm{\Omega }$

$X_{\mathrm{C}}=\frac{1}{{\omega }_{\mathrm{C}}}=\frac{1}{100\times 100\times {10}^{-6}}=100 \mathrm{\Omega }$

$Z=\sqrt{{\left(10-100\right)}^2+R^2}=\sqrt{{90}^2+{120}^2}$$=30\times 5=150 \mathrm{\Omega }$

$i_{\mathrm{peak} }=\frac{V\mathrm{peak}}{z}=\frac{30}{150}=\frac{1}{5} \mathrm{amp}=0.2 \mathrm{amp}$

$\&$ For resonant frequency

$\Rightarrow \omega L=\frac{1}{\omega C}\Rightarrow {\omega }^2=\frac{1}{LC}\Rightarrow \omega =\frac{1}{\sqrt{LC}}$

$\& f=\frac{1}{2\pi \sqrt{LC}}\Rightarrow \frac{1}{2\pi \sqrt{100\times {10}^{-3}\times 100\times {10}^{-6}}}$

$=\frac{100\sqrt{10}}{2\pi }=\frac{100\pi }{2\pi }=50 \mathrm{Hz}$

as $\sqrt{10}\approx \pi$