$Z=\sqrt{R^2+(\Chi_C-\Chi_L)^2}$

$\Chi_C=\frac{1}{\omega C};\Chi_L=\omega L$

$\omega C \rightarrow$ capacitive Reactance

$\Chi_L \rightarrow$ Inductive Reactance

$I(t)=I_m \sin (\omega t + \phi)$

$\phi \rightarrow$ phase by which current leads the voltage

$I_m = \frac{V_m}{2}$

$\phi$ Angle by which supply voltage lags the current

$\therefore$ circuit is capacitive in nature

$100 \mu F$ capacitor is series with a $400\Omega$ resistance, connected to a $110V (rms)/60Hz$ supply.

What is the time lag between current maximum and voltage maximum?

$60Hz \rightarrow \omega=2 \pi \times 60 \approx 377 rad/s$

Capacitive reactance

$\frac{1}{\omega C}=\frac{1}{377 \times 10^{-4}}=26.5 \Omega$

$Z=\sqrt{40^2 (26.5)^2} \approx 48 \Omega$

$I_{vms} = \frac{110}{48}=2.29A \rightarrow I_m = 2.29 \sqrt{2} =3.24 A$

Angle by which Current Leads voltage

$\frac{\phi}{\omega}=1.55 ms$

$f=1KHZ \rightarrow \omega = 6263 rad/s$

$\omega C=6263 \times 10^{-4} \approx 0.63 \Omega^{-1}$

$\Chi_C = \frac{1}{\omega C}=1.59 \Omega$

$Z=\sqrt{40^2+(1.59)^2}=40.03 \Omega$

$\frac{\Chi_C}{R}=\frac{1.53}{40}=0.039 \rightarrow \phi = \tan^{-1} \frac{\Chi_C}{R}\approx 0.039 rad$

$\frac{dI}{dl}+IR+\frac{\phi}{C}=V_m \sin \omega t$

$I=\frac{d \phi}{dt}$

$\frac{d^2I}{dt^2}+R\frac{dI}{dt}+\frac{1}{C}.I=V_m \omega \cos (\omega t)$

$I=I_m \sin (\omega t +\phi)$

$\frac{dI}{dt}=I_m \omega \cos (\omega t + \phi)$

$\frac{d^2 I}{dt^2}=-I_m \omega^2 \sin (\omega t + \phi)$

$I_m \omega [(-L \omega + \frac{1}{\omega C}) \sin (\omega t + \phi)+R \cos (\omega t + \phi )]$

$=V_m \omega \cos \omega t$

$I_m \omega [(-L \omega + \frac{1}{\omega C})\sin (\omega t + \phi)+R \cos (\omega t + \phi)]$

$=V_m \omega \cos \omega t$

$R=A \cos \theta$

$\Chi_C - \Chi_L = A \sin \theta$

$A^2 = R^2 + (\Chi_C - \Chi_L)^2 = Z^2 \rightarrow A = z$

$tan \theta = \frac{\Chi_C - \Chi_L}{R}$

$I_m Z \cos (\omega t + \phi - \theta)=V_m \cos (\omega t)$

$I_m z = V_m \rightarrow I_m =\frac{V_m}{z}$

$I_m =\frac{V_m}{\sqrt{R^2 +(\Chi_C - \Chi_L)^2}}$

$\Chi_C = \frac{1}{\omega C}; \Chi_L= \omega L$

Z is minimum when $\frac{1}{\omega_C}= \Delta \omega L$

$\rightarrow \omega = \omega_0 = \sqrt{\frac{1}{LC}}$

$\omega_0$ = Resonant frequency

At resonance $I_m=\frac{V_m}{R};$ phase

Both L and C must be Present

If $\omega^2 > \omega_0^2$

$\omega^2 > \frac{1}{LC}$

$I \propto \frac{1}{Z}$

$I^2 Z$ is maximum at $\omega = \omega_0$

Increase or decrease frequency (standing walk $\omega_0$) till power absorbed is half the maximum(at $\omega_0)$

$20 \omega =(\omega_1 - \omega_2)$

Bandwitdh (BW)

$P=0.5 I^3 R = (\frac{I}{\sqrt{2})^2}.R \rightarrow$ At half power point

Let, $\omega_1 = \omega_0 + \Delta \omega$

$\omega_2 = \omega_0 - \Delta \omega$

$2 \Delta \omega =$ Band width

Sharper resonance $\rightarrow$ Lower bandwidth

At, $\omega = \omega_1 \rightarrow I_m$

$= \frac {V_ m}{\sqrt{R^2 +(\omega_ 1 L - \frac{1}{\omega_ 1 c^2})}} \equiv \frac{I_ m^{max}}{\sqrt{2}}=\frac{V_m}{R.\sqrt{2}}$

$R^2 (\omega_1 L - \frac{1}{\omega_1 C})^2 = 2R^2$

$\omega_1 L - \frac{1}{\omega_1 C}=R$

$(\omega_0 + \Delta \omega)L-\frac{1}{\omega_0 + \Delta \omega)C}=R$

$\omega_0 L[1+\frac{\Delta \omega}{\omega_0}]-\frac{1}{\omega_0 C}[1-\frac{\Delta \omega}{\omega_0}]=R$

$L. \Delta \omega +\frac{1}{\omega_0 C}.\frac{\Delta \omega}{\omega_0}=R$

$(L).2 \Delta \omega = R$

$\Delta \omega =\frac{R}{2 L}$

In a series LCR circuit resonance freq. $f_0=1KHZ$; Q=100

If R, L, C are doubled. What would Q be?

$\omega_0=\frac{1}{\sqrt{LC}}$

$\omega_0$ reduces by a tactor of 2 If L and C are doubled

$Q=\frac{\omega_0 L}{R}$

$\therefore Q=50$

$V(t)=(240 V) \sin \omega t$

$L=10mH$

$C=1 \mu F$

$R=40 \Omega$

(a) Resonance Frequency

$\omega_0 = \frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10^{-3} \times 10^{-6}}}$

$=10^{-4} rad/s$

(b) What is the amplitude of current at resonance>

$I_m^{max} =\frac{V_m}{R} = \frac{240}{R}=6A.,R=40 \Omega$

(c) $Q=\frac{\omega_0 L}{R}=\frac{10^{4} \times 10^{-2}}{40}=2.5$

(d) Voltage across inductor at resonance

$V_L^{max} = I_m^{max}.\omega_0L$

$=6 \times [10^4 \times 10^{-2}]$

$L = 3H$

$C = 27 \mu F$

$R=7.4 \Omega$

What should we do to reduce FWHM by factor of 2?

$\omega_0 = \frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{8 \times 27 \times 10^{-6}}}=111 rad/s$

$Q=\frac{\omega_0L}{R}=\frac{111 \times 3}{7.4} \approx 45$