$L \frac{d I}{d t}+I R+\frac{Q}{C}=V_m \sin \omega t$

$I= \frac{dQ}{d t}$

$L \frac{d^2 I}{d t^2} + R \frac{d I}{dt}+\frac{I}{C}=V_m \omega cos \omega t$

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

$\varphi$ Amount of phase by which current leads 1hz voltage.

$I_m = \frac{V_m}{Z}: Z = \sqrt {R^2 + (X_c - X_L)}$

$tan \varphi = \frac {X_c - X_L}{R}$

Maximum current $I_m$ has a frequency dependence.

$I_m$ has a maximum value as a function of $\omega$

$I_m$ has a maximum as a function of $\omega$ when $X_c - X_L$ $\rightarrow$ $\omega L = \frac {1}{\omega C}$

$\omega = \omega_0 = \frac{1}{\sqrt{\omega c}}$ (Resonance)

$I_m^{max} = \frac {V_m}{R}$

Inductive (L$\omega$)

Resistive R

Capacitive ($\frac {1}{\omega c}$)

At $\omega = \omega_0$, the circuit absorbs maximum power.

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

$I-m^{2} \propto \frac {1}{Z}$ maximum power

$\rightarrow$ Z is minimum

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

$I_m =\frac {I_m^{max}}{\sqrt 2}$

1hz power absorbed is half the possible maximum $I_{m}^{max}$

Full width at half max $2\Delta \omega$

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

Band width : BW = $2\Delta \omega = \frac{R}{L}$

Quality factor:

Q = $\frac {\omega_0}{2\Delta \omega} = \frac {\omega_0 L}{R}$

Typical circuit application Q $\sim$ 10 to 100

LCR circuit R = 5 $\Omega$, C= 20 $\mu$ F, L = 200 mH

(i) Resonant frequency

$\omega_0 = \frac{1}{\sqrt LC} = \frac {1}{\sqrt{ 2 \times 10^{-1} \times 2 \times 10^{-5}}}$

= 500 rad/s (f $\approx$ 80 Hz)

(ii) Value of $\omega$ for which half power maximum occurs.

$\frac {I_m^{max}}{\sqrt 2} \Rightarrow (\omega L - \frac{1}{\omega c})^2 = R^2$

$\omega L-\frac {1}{\omega C} = \pm R$

$\omega^2 L C \mp R \omega C-1=0$

$\omega= \frac{ \pm R C+\sqrt{R^2 C^2+4 L C}}{2 L C}$

$\omega = 512.5 \mathrm{rad} / \mathrm{s} \text { or } 487.5 \mathrm{rad} / \mathrm{s}$

$\text { spread of } \pm 12.5 \mathrm{rad} / \mathrm{s} \text {. }$

$\text { BW }=2 \Delta W=25 \mathrm{rad} / \mathrm{s}$

$B W=2 \Delta W=\frac{R}{L}=\frac{5}{2 \times 10^{-1}}=25 \mathrm{rad} / \mathrm{s}$

High pass filter (RL)

For D.C supply $V_{out}$ = 0

$\omega$ increases

$X_L = \omega_L$ will increase.

What in $\omega$ such that

$\frac{V_{\text {out }}}{V^{\text {in }}}=\frac{1}{2} \quad \Rightarrow$ $V_out = 5 \sin \omega t$.

$Z_{i n}=\sqrt{50^{\circ}+.04 \omega^2}$

$I_m=\frac{10}{Z_{i n}}=\frac{10}{\sqrt{50^2+0.04 \omega^2}}$

${ Z_{out} }=\sqrt{10^2+0.04 \omega^2}$

$V_{out }^{m}={I_m} \times Z_{out}$

=$\frac{10}{\sqrt{50^2+0.04 \omega^2}} \times \sqrt{10^8+0.04 \omega^2}$

$X_{L}^{2} = 700 = \omega^2 \times 0.04$

$\omega$ = 132 rad/s

$f = 21 \mathrm{Hz}$

$I_m = \frac {V_in^{m}}{\sqrt {R^2 + L^2 \omega^2}}$

$V_{out} = \frac {V_in^{m}}{\sqrt{R^2 + L^2 \omega^2}}$ $\times L \omega$

=$V_{in}^{m}$ .$\frac {L \omega} {\omega L (1 + \frac {R^2}{L^2 \omega^2})^{1/2}}$

$\approx V_{in}^{m}[1+ \frac{1}{2} . \frac {R^2}{L^2 \omega^2}]$

As $\omega$ increases. $V_{out}$ $\rightarrow$ $V_{in}$

$\omega \rightarrow$ 0 $V_{out}$ = 0

$V_{out} = \frac {V_{in} R}{\sqrt{R^2 + L^2 \omega^2}}$

= $\frac {V_{in} R} {L \omega (1 + \frac {R^2}{L^2 \omega^2})^{1/2}}$

For large $\omega$ $V_{out}$ decreases.

For $\omega \rightarrow 0$, $V_{out} \rightarrow V_{in}$

$I_m = \frac {V_{in}^{m}}{Z} = \frac {10} {\sqrt{1o^4 + 0.04\omega^2}}$

$V_{m}^{out} = \frac {10 \times 10^3} {\sqrt{1o^4 + 0.04\omega^2}}$

$\omega$ = 500 rad/s : $V_{out}^m = 9.95 V$

$\omega$ = 5000 rad/s : $V_{out}^m = 7.07 V$

$\omega$ = 50000 rad/s : $V_{out}^m = 0.995 V$

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

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

$I_m = \frac{V_m}{Z}$ : $\varphi = tan^{-1} \frac{X_c - X_L} {R}$

P(t) = I(t) V(t)

= $V_m I_m sin \omega t . sin (\omega t + \varphi)$

=$V_m I_m [sin^2 \omega t cos \varphi + sin \omega t + cos \omega t sin \varphi)]$

$<P(t)> = \frac {V_m I_m}{2} . cos \varphi = \frac {I_m^2}{2} . cos \varphi = \frac {V_m^2}{2Z} cos \varphi.$

$< P> = \frac {I_m^2 Z}{2} . Cos \phi$

$tan \phi = \frac {X_c - X_L}{R}$

$cos \phi = {R} {\sqrt {R^2 + (X_c - X_L)^2}} = \frac {R}{Z}$

$< P > = \frac {I_m^2 Z}{2} .\frac {R}{Z} = \frac {I_m^2 .R}{2} = I_{rms}^2. R$

= $\frac {V_{rms}^2}{R^2 + (X_c - X_L)^2} . R$

$< P>$ is maximum at $X_L = X_c$

$< P> = \frac {V_{rms}^2}{R}$

For a purely resistive circuit.

$cos \phi = \frac {R}{Z} \Rightarrow 1 : \phi = 0$

$\phi = \pm \frac {\pi}{2} : cos \phi = 0$

watt-loss circuits.

L-C-R circuit

$tan \phi = \frac {\Chi_C - \Chi_L}{R} \rightarrow \phi \neq 0$

Power dissiparon through resistances only

Circuit at resonance $\Chi_L = \Chi_C : \phi = 0$

Maximum power dissipation (through resistance only)

$<P> = \frac {V_{rms}^2}{R^2 + (X_L - X_c)^2}$

$<P_{max}> = \frac {V_{rms}^2}{R}$ ,

When $X_L = X_c \rightarrow \omega = \omega_0$

$<P> = \frac {V_{rms}^2}{2R} \rightarrow (X_L - X_c)^2 = R^2$

$\omega= \pm \frac{R}{2 L} \pm \frac{1}{2} \sqrt{\frac{R^2}{L^2}+4 \omega_0^2}$

$\omega_1=\frac{R}{2 L}+\frac{1}{2} \sqrt{\frac{R^2}{L^2}+4 \omega_0^2}$

$\omega_2=-\frac{R}{2 L}+\frac{1}{2}\sqrt{ \frac {R^2}{L^2}+4 \omega_0^2}$

$2 \Delta \omega=\omega_1-\omega_2=\frac{R}{L}$

$Q=\frac{\omega_0}{2 \Delta \omega}=\frac{\omega_0 L}{R}$