Physics Alternating Current

R = $2\Omega$

L = 2 mH

E = 9V

$i=\frac{\epsilon }{2R}=\frac{9v}{4\Omega }=2.25A$

Just after the switch 'S' is closed, the inductor acts as open circuit.

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

  $=\left(l_L+l_C\right)$  

Therefore current through R circuit at resonance will be zero

 

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = $\frac{1}{2\pi \sqrt[]{LC}}=\frac{1}{2\pi \sqrt[]{\left(0.5*10^{-3}\right)*\left(200*10^{-6}\right)}}$

$\Rightarrow f=\frac{10^4}{2\pi \sqrt[]{10}}\approx 5*10^{2}Hz$ $\left[Taking\pi =\sqrt[]{10}\right]$

$x_R=120\Omega$

$x_L=\omega L=100*100*10^{-3}10\Omega$

$x_C=\frac{1}{\omega c}=\frac{1}{100*100*10^{-6}}=100\Omega$

$z=\sqrt[]{120^2+{\left(100-10\right)}^2}=\sqrt[]{120^2+90^2}$

= 150  $\Omega$

$\Rightarrow 32=\frac{160}{75}*t\Rightarrow t=\frac{32*75}{160}=15$

X? = X? ⇒ ωL = R ⇒ 2 * 3.14 * 50 * L = 20 ⇒ L = 63.7mH

Purely inductive circuit

              $\theta =\frac{\pi }{2}$  

              $cos\frac{\pi }{2}=0$  

Average power = 0

$\omega =2\pi \mathrm{f}\Rightarrow \omega =100\pi$

$\mathrm{Z}=\sqrt[]{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2}$

$=\sqrt[]{{10}^2+{\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)}^2}$

$=\sqrt[]{100+{\left(100\pi *\frac{50}{\pi }*{10}^{-3}-\frac{1}{100\pi *\frac{{10}^3}{\pi }*{10}^{-6}}\right)}^2}$

$=\sqrt[]{100+(5-10{)}^2}$

$=\sqrt[]{100+25}$

$\mathrm{Z}=5\sqrt[]{5}\mathrm{\Omega }$

Capacitive reactance $=\frac{1}{\omega \mathrm{C}}={\mathrm{X}}_{\mathrm{c}}$ (say) on decreasing the operating frequency $\omega$  reduces

As ${\mathrm{X}}_{\mathrm{c}}$ is inversely proportional to $\omega$  the value of ${\mathrm{X}}_{\mathrm{c}}$  increase

$\therefore {\mathrm{I}}_{\mathrm{C}}={\mathrm{I}}_{\mathrm{D}}$

$=\frac{{\mathrm{V}}_{\mathrm{o}}}{{\mathrm{X}}_{\mathrm{c}}}$

As $X_c$  increases, therefore displacement current Id decreases.

For transformer $\left(\frac{i_p}{i_s}\right)=\left(\frac{V_s}{V_P}\right)=\left(\frac{N_s}{N_P}\right)$

For bulb, $\mathrm{V}\mathrm{s}=12\mathrm{V}$

$\begin{array}{rr}& {\mathrm{i}}_{\mathrm{s}}=\left(\frac{60}{12}\right)=5\mathrm{A}\\ & \Rightarrow \frac{{\mathrm{i}}_{\mathrm{p}}}{5}=\left(\frac{12}{220}\right)\\ & \mathrm{}{\mathrm{i}}_{\mathrm{P}}=\frac{60}{220}=\left(\frac{3}{11}\right)=0.27\mathrm{A}\end{array}$

For resonance frequency

$\begin{array}{rr}& \omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}\\ & \Rightarrow \omega =\frac{1}{\sqrt[]{\mathrm{L}\mathrm{C}}}=\frac{1}{\sqrt[]{10*{10}^{-3}*1*{10}^{-6}}}\\ & =\frac{1}{\sqrt[]{{10}^{-8}}}={10}^4\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\mathrm{e}\mathrm{c}\\ & \mathrm{f}=\left(\frac{\omega }{2\pi }\right)=\frac{{10}^4}{2\pi }=1.59\mathrm{k}\mathrm{H}\mathrm{z}\end{array}$