Physics Alternating Current

$\omega =2\pi f=2\pi *50=100\pi$

$X_L=wL=100\pi *100*10^{-3}$

$X_C=\frac{1}{wC}=\frac{1}{100\pi }*100*10^{-6}=\frac{100}{\pi }$

$z=10.0086\Omega$

$l=\frac{v}{z}=\frac{220}{10.008}=22A$

V = 210 sin 3000 t

$\omega$ = 3000

$X_L=\omega L=30$                

$X_C=\frac{1}{\omega c}=\frac{40}{3}$                

$tan?=\frac{V_L-V_C}{V_R}=\frac{X_L-X_C}{R}$                

$\begin{array}{l}tan?=0.17\\ ?=ta{n}^{-1}\left(0.17\right)\end{array}$                

The current lags the voltage by an angle of 90 Degree (? /2 radians). This is due to the inductor opposing the change in current.

The inductor generates induced electromotive force (EMF) and opposes the changes in the current flow. Due to this the current lag the voltage by 90 degrees.

for lamp,   $P=\frac{v^2}{R}$

$\Rightarrow R=\frac{25*25}{5}=125\Omega$              

$l_{max}=\frac{v}{R}=\frac{25}{125}=\frac{1}{5}A$    

$\Rightarrow I_{max}=\frac{220}{125+R}=\frac{1}{5}$

$\Rightarrow 1100=125+R$        

R = 975  $\Omega$

Given, I = 5 sin (120πt), $\omega$ $$ = 120

$T=\frac{2\pi }{\omega }$

$T=\frac{2\pi }{120\pi }=\frac{1}{60}sec,\frac{T}{4}=\frac{1}{60*4}=\frac{1}{240}sec$

Current in capacitor l = $\frac{V}{X_C}$ $\frac{}{{}_{}}$

I = v * $\left(\omega c\right)$ $$

$C=\frac{1}{v\omega }=\frac{6.9*10^{-6}}{230*600}=50pF$

$f=\frac{1}{2\pi \sqrt[]{LC}}$

$f\alpha \frac{1}{\sqrt[]{C}}$

Another capacitor should be added in series with the first capacitor.

$\begin{array}{rr}\mathrm{i}& =\frac{\mathrm{V}}{\mathrm{r}}\left(1-{\mathrm{e}}^{-\mathrm{R}\mathrm{t}/\mathrm{L}}\right)\\ & =4\left(1-{\mathrm{e}}^{-500\mathrm{t}}\right)\end{array}$

At  $\mathrm{t}=\mathrm{\infty }$

${\mathrm{i}}_1=4\mathrm{A}$

at  $\mathrm{t}=40\mathrm{s}$

${\mathrm{i}}_2=4\left(1-{\mathrm{e}}^{-20000}\right)$

$=4\left[1-\frac{1}{{\left({\mathrm{e}}^2\right)}^{10000}}\right]$

$=4\left[1-\frac{1}{(7.389{)}^{10000}}\right]$

$\frac{{\mathrm{i}}_1}{{\mathrm{i}}_2}$  is slightly greater than 1 .