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

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

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

$\phi = \tan^{-1} \frac{\Chi_C - X _L}{R}$

$\omega = \omega _0 = \frac{1}{\sqrt{LC}} \Rightarrow I_m (\omega)$ reaches a maximum

$\frac{1}{\sqrt{LC}}$ = Resonant Frequency

Sharpness of resonance Bandwidth = $2 \Delta \omega = \frac{R}{L}$

Quality Factor Q = $\frac{\omega _0 L}{R}$

Average Power absorbed by circuit

$< P(t)>=\frac{I_m^2 Z}{2} \cos \phi=\frac{V _m^2}{2 Z} \cos \phi$

$\cos \phi = \frac{R}{Z}$ is "Power Factor" of circuit

For a purely resistive circuit Z = R $\cos {\phi = 1} \Rightarrow {\phi = 0}$ : Circuit absorbed maximum power

Maximum Power is also absorbed at resonance $(\Chi_L = \Chi_C)$.

For pure capacitive or inductive circuit $\phi = \frac{\pi}{2} : \cos \phi = 0$ : watt-less circuits

For LCR circuit, $\phi \neq 0$.

$< P > = \frac{V_m^2}{2Z} \cos \phi$

$= \frac{V_m^2}{2Z} \frac{R}{Z}=\frac{V_{rms}^2.R}{Z^2}$

If purely resistive Load, voltage & current in phase $V I> 0. \Rightarrow$ Power flows to Load.

If not in Phase $V I$ may become $< 0$ for some part of the cycle

0 to $\frac{T}{4}$ absorbs

$\frac{T}{4}$ to $\frac{T}{2}$ returned

$\frac{T}{2}$ to $\frac{3T}{4}$ absorbs

$\frac{3T}{4}$ to T returns

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

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

Current Leads (...)

Voltage by cp = $\pi/4$

(time Lead by T/8)

From $\quad t=0$ to $\frac{3 T}{8} \rightarrow V>0 ; I>0$ absorbs power

$t=\frac{3 T}{8}$ to $\frac{T}{2} \quad V>0; I<0 \quad$ Power is returned

$t=\frac{T}{2}$ to $\frac{7 T}{8} \quad V<0 ; I<0$ Power is absorbed

$t=\frac{3 T} {8}$ $to$ $\quad V<0 ; I>0$ Power is returned

$\Chi_L$ (Inductive Reactance)

$\Chi_L ^2 + R^2 = Z^2$

$\cos \phi = \frac{R}{Z}$

$\sin \phi = \frac{\Chi_L}{Z}$

$I^2R$ : Power Consumed by resistive Load (Watts : True Power)

= $I. V. \frac{R}{Z} = I ~V \cos \phi.$ (True) Active Power

$I^2 \Chi_L = I.\frac{V}{Z}. \Chi_L = I ~V \sin \phi$ (Watt-less or Reactive Power) V Amp

$I ~V$ = Apparent Power (V A)

$V I \cos \phi$ (Watts)

$V I \sin \phi$ (V Amp)

Power Factory = $\frac{\text{True Power}}{\text{Apparent Power}}$

$= \cos \phi$

For a given value of Power current drawn

$I = \frac{P}{V \cos \phi}$

If $\cos \phi$ is small $\Rightarrow$ I drawn is Large Power Loss $I^2 R_c \propto \frac{1}{\cos ^2 \phi}$

Compensation for reducing apparent Power $\rarr$ Neutralize the Wattless Component

$I_p =I \cos \phi$ (True)

$I_q = I \sin \phi$ (Watt-less Power)

$\theta < \phi$

$\cos \theta > \cos \phi$.

$250 V - 60Hz$ source supplies 1.5 kw of Power.

r.m.s. current drawn by Load is 10 A

(a) Power Factor.

True Power = 1.5KW.

Apponent Power $= 250\times10$

$= 2.5 KVA$

Reactive Power = $\sqrt{(2.5)^2 - (1.5)^2}$

$\text{ = 2 KVA}$

Power Factor $\cos \phi = \frac{1.5}{2.5} = 0.6$

(b) For full compensation

$2\times10^3 = \frac{V^2}{\Chi_C} = \frac{(250)^2}{\Chi_C}$

$\Chi_C = 31.25 \Omega$

$C = \frac{1}{\Chi_C \omega} = \frac{1}{31.25\times(60\times 2\pi)}$

$= 84 ~\mu F$

(c) $\Chi_C = \frac{1}{\omega C}$

$33.15 ~\Omega$

$I = \frac{250}{33.15} = 7.54 ~A$

$I ~V = 250 \times 7.54 = 1.885 ~\text{K V Amp}$

Reactive Power is reduced by 1.885 K V Amp is it becomes

$2 - 1.885 = 0.115$ K V Amp

$\text{Apparent Power} = \sqrt{(.115)^2 + (1.5)^2}$

$=1.5044 ~{kVA} \approx 1.5 ~{kVA}$

230 V, 50 Hz supply given to a Load results in 280 K V Amp. Power factor 0.86, find the value of C to compensate Phase lag.

$\cos \phi \approx \frac{\sqrt{3}}{2}$

apparent power

$V_R = \frac{V_q}{\sin \phi} = \frac{280}{1/2} = 560$ K V A

Active Power

$V_P = V_R .\cos \phi = 484.4 ~\text{K W}$

We require

$V_R^\prime = 484.4 ~\text{K V A}$

For full Compensation:

$V_{q^\prime} = 280 \times 10^3$

$= \omega c ~V_{rms}^2$

= $314.16 c \times (230)^2$

= $1.66 \times{10^7} .c$

$\omega$ for $f$ = 50 Hz

$= 314.16$

$C = \frac{280\times 10^3}{1.66\times 10^7}$

$= 16.86 ~\text{m F}$