Physics - EMI/AC Question with Solution | TestHub

Given : $\text{R=40Ω , L= 20 mH , C=50 μF}$

Comparing the given equation for voltage with general equation $\mathrm{V}={\mathrm{V}}_0 \sin \mathrm{\omega t}$

$\mathrm{\omega }=340 \mathrm{rad} {\mathrm{s}}^{-1}$        and                           peak voltage   ${\mathrm{V}}_0=10 \mathrm{V}$

Inductive reactance ${\mathrm{X}}_{\mathrm{L}}$$\mathrm{\omega }\mathrm{L}=340\times 20\times {10}^{-3}=68\times {10}^{-1}=6.8 \mathrm{\Omega }$
Capacitive reactance ${\mathrm{X}}_{\mathrm{C}}$
$\frac{1}{\mathrm{\omega }\mathrm{C}}=\frac{1}{340\times 50\times {10}^{-6}}=\frac{{10}^4}{34\times 5}=\frac{2}{34}\times {10}^3$

$=0.0588\times {10}^3=58.82 \mathrm{\Omega }$

$\text{ Z=}\sqrt{{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2{\text{+R}}^2}\text{⇒Z}=\sqrt{{\left(\mathrm{\omega }\mathrm{L}-\frac{1}{\mathrm{\omega C}}\right)}^2+{\mathrm{R}}^2}$
$\text{ Z=}\sqrt{{\left(58.8-6.8\right)}^2\text{+}{\left(40\right)}^2}\text{ ⇒ Z}=\sqrt{2704+1600}\approx 65.6 \mathrm{\Omega }$

${\mathrm{I}}_{\mathrm{rms}}=\frac{{\mathrm{V}}_{\mathrm{rms}}}{\text{Z}}=\frac{10}{65.6\times \sqrt{2}}({\mathrm{V}}_{\mathrm{rms}}=\frac{{\mathrm{V}}_0}{\sqrt{2}})$

Power $={\mathrm{E}}_{\mathrm{rms}}{\mathrm{I}}_{\mathrm{rms}} \cos \mathrm{\varphi }={\mathrm{I}}_{\mathrm{rms}}^2\mathrm{R}$ 

$\left(\because \cos \mathrm{\varphi }=\frac{\mathrm{R}}{\mathrm{Z}}, {\mathrm{V}}_{\mathrm{rms}}={\mathrm{I}}_{\mathrm{rms}}\mathrm{Z}\right)$

$=\frac{100\times 40}{{\left(65.6\right)}^2\times 2}=\frac{2000}{{\left(65.6\right)}^2}$

$=0.51\mathrm{ }\mathrm{W}$