Alternating Current Overview

At t = 0, no current flows through inductor.

$i=\frac{E}{\frac{6*9}{6+9}}=\frac{5E}{18}$

At t = $\infty$ $$ , current flows through circuit as if inductor is shorted.

$i=\frac{E}{\frac{5}{2}+\frac{4}{5}}=\frac{10E}{33}$

Key is open.

$i_{rms}=\frac{15}{60}=\frac{1}{4}A$

$20=\frac{1}{4}*100*L\Rightarrow L=0.8H$

$10=\frac{1}{4}.\frac{1}{100C}\Rightarrow C=2.5*10^{-4}F=250\mu F$

For capacitor : current leads emf

by 90°.

Power dissipated in cycle = P = $\frac{V_0{l}_0}{2}cos?=\frac{V_0^2}{2z}*\frac{R}{z}=\frac{V_0^{2}R}{2z}$  

For resonance, z = R, so

$\Rightarrow P=\frac{V_0^2}{2R}=\frac{250\sqrt[]{2}*250\sqrt[]{2}}{2*5}=12500Watt=125*10^{2}Watt$        

  $\omega =100\pi ,R=100\Omega ,X_L=L\omega =0.5*10^{-3}*100\pi =5\pi *10^{-2}\Omega ,$ and

  $X_C=\frac{1}{C\omega }=\frac{1}{0.1*10^{-12}*100\pi }=\frac{10^{11}}{\pi }\Omega$            

Since X_C is approximately infinite, so the phase angle between current and supplied voltage and the nature of the circuit is $\approx 90°,$ predominantly capacitive circuit.

$X_c=\frac{1}{\omega c},X_L=\omega L$ so at very high frequencies capacitor behaves as conduct and inductor behaves as open circuit. The effective impedance will be $2\Omega .$

$f_{rms}^2={\left(\frac{\sqrt[]{42}}{\sqrt[]{2}}\right)}^2+10^2=121$

$\Rightarrow f_{rms}=11A$

$\Delta V_{D1}=\Delta V_{D2}=l*R=0$

Since resistance is zero in forward bias.

=> V₀ = 5V

Let, current through inductor,

$i_L=l_{L}sin\left(\omega t+{?}_1\right)$                

$l_L=\frac{200\sqrt[]{2}}{\sqrt[]{50^2+{\left(100*0.50\right)}^2}}=4A$              

${?}_1=ta{n}^{-1}\left(\frac{-100*0.50}{50}\right)=-\frac{\pi }{4}$             

Let, current through capacitor,

As  ${?}_2-{?}_1=\frac{\pi }{2}$  

$l=\sqrt[]{{\left(I_L\right)}^2+{\left(I_C\right)}^2}$

$=\sqrt[]{4^2+2^2}=4.47A$