physics-class-12-unit-07-chapter-08-lcr-circuit-power-factor-l-8-10-epsspkwqis8

### <Success style="font-size:25px"> Lcr Circuit Power Factor L-8 </Success>
### <primary style="font-size:24px"> LCR Circuit </primary>
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- $V(t) = V_m \sin \omega t$    
   
- $ 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 = $\frac{\omega _0 L}{R}$

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<footer style = "font-size: 18px"> $\rightarrow$ LCR Circuit Power Factor $\rightarrow$ <span style = "color:blue"> LCR Circuit </span> $\rightarrow$ Average Power $\rightarrow$ Average Power </footer>

### <Success style="font-size:25px"> Lcr Circuit Power Factor L-8 </Success>
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- 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

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<footer style = "font-size: 18px">LCR Circuit Power Factor $\rightarrow$ LCR Circuit $\rightarrow$ <span style = "color:blue"> Average Power </span> $\rightarrow$ Average Power $\rightarrow$ Capacitive circuit </footer>

### <Success style="font-size:25px"> Lcr Circuit Power Factor L-8 </Success>
### <primary style="font-size:24px"> Average Power </primary>
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- 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
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<footer style = "font-size: 18px">LCR Circuit $\rightarrow$ Average Power $\rightarrow$ <span style = "color:blue"> Average Power </span> $\rightarrow$ Capacitive circuit $\rightarrow$ Capacitive circuit </footer>

### <Success style="font-size:25px"> Lcr Circuit Power Factor L-8 </Success>
### <primary style="font-size:24px"> Capacitive circuit </primary>
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- 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

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<footer style = "font-size: 18px">Average Power $\rightarrow$ Average Power $\rightarrow$ <span style = "color:blue"> Capacitive circuit </span> $\rightarrow$ Capacitive circuit $\rightarrow$ Inductive Reactance </footer>

### <Success style="font-size:25px"> Lcr Circuit Power Factor L-8 </Success>
### <primary style="font-size:24px"> Capacitive circuit </primary>
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- $ 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

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<footer style = "font-size: 18px">Average Power $\rightarrow$ Capacitive circuit $\rightarrow$ <span style = "color:blue"> Capacitive circuit </span> $\rightarrow$ Inductive Reactance $\rightarrow$ Power Factory </footer>

### <Success style="font-size:25px"> Lcr Circuit Power Factor L-8 </Success>
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- $\Chi_L$ (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)

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<footer style = "font-size: 18px">Capacitive circuit $\rightarrow$ Capacitive circuit $\rightarrow$ <span style = "color:blue"> Inductive Reactance </span> $\rightarrow$ Power Factory $\rightarrow$ Wattless Component </footer>

### <Success style="font-size:25px"> Lcr Circuit Power Factor L-8 </Success>
### <primary style="font-size:24px"> Power Factory </primary>
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- $ V I \cos \phi$ (Watts)

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

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

- $ = \cos \phi $

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<footer style = "font-size: 18px">Capacitive circuit $\rightarrow$ Inductive Reactance $\rightarrow$ <span style = "color:blue"> Power Factory </span> $\rightarrow$ Wattless Component $\rightarrow$ Wattless Component </footer>

### <Success style="font-size:25px"> Lcr Circuit Power Factor L-8 </Success>
### <primary style="font-size:24px"> Example </primary>
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- $250 V - 60Hz$ 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 $\cos \phi = \frac{1.5}{2.5} = 0.6$

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<footer style = "font-size: 18px">Wattless Component $\rightarrow$ Wattless Component $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Lcr Circuit Power Factor L-8 </Success>
### <primary style="font-size:24px"> Example </primary>
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- **(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} $

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<footer style = "font-size: 18px">Wattless Component $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Lcr Circuit Power Factor L-8 </Success>
### <primary style="font-size:24px"> Example </primary>
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- **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} $

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Thank You </footer>