physics-class-12-unit-07-chapter-05-lcr-circuit-graphical-solution-alternating-currents-l-5-10-eek-fqsfm4

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> AC Circuit With Capacitive Load </primary>
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- A.C circuit with capacitive load.

- Purely Resistive load: Current is in phase with voltage

- Inductive load (L): Current lags behind voltage

- Capacitive load (C) : Current leads the voltage

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<footer style = "font-size: 18px"> $\rightarrow$ LCR Circuit Graphical Solution Alternating Currents $\rightarrow$ <span style = "color:blue"> AC Circuit With Capacitive Load </span> $\rightarrow$ LC Circuit $\rightarrow$ LC Circuit </footer>

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> LC Circuit </primary>
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- E - voltage

- L - inductor

- I - current

- c - capacitive

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<footer style = "font-size: 18px">LCR Circuit Graphical Solution Alternating Currents $\rightarrow$ AC Circuit With Capacitive Load $\rightarrow$ <span style = "color:blue"> LC Circuit </span> $\rightarrow$ LC Circuit $\rightarrow$ Phase </footer>

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> LC Circuit </primary>
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- The Ratio of Voltage Maximum to Current Maximum.

- Resistive circuit $I_M = \frac {V_m}{R}$

- Inductive load : Inductive Resistance

- $\Chi_L = \omega L$

- $I_m = \frac {V_m}{\Chi_L} = \frac {V_m}{\omega L}$

- Capacitive Load : $\Chi_C = \frac {1}{\omega C}$

- $I_m = \frac{V_m}{\Chi_C} = V_m . \omega C$

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<footer style = "font-size: 18px">AC Circuit With Capacitive Load $\rightarrow$ LC Circuit $\rightarrow$ <span style = "color:blue"> LC Circuit </span> $\rightarrow$ Phase $\rightarrow$ Impedance </footer>

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> Impedance </primary>
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- $V(t)  =V_ m e^{i \omega t} =V_m[\cos \omega t + i \sin \omega t] $

- I(t)  =$\operatorname{I_m} e^{i(\omega t+\Phi)}$

- Complex Impedance: $Z=\frac{V(t)}{I(t)}=\frac{V_m}{I_m} \cdot e^{-\iota \varphi}$

- For resistive circuit: $\varphi=0, \quad Z=R$

- For inductive circuit: $\varphi=-\pi / 2$, $\quad Z=\omega L e^{i  \pi / 2}=i \omega L$

- For a capacitive: $Z=\frac{1}{\omega c} e^{-i \pi / 2 } = (-i) \frac{1}{\omega c}=\frac{1}{i \omega c}$.

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<footer style = "font-size: 18px">LC Circuit $\rightarrow$ Phase $\rightarrow$ <span style = "color:blue"> Impedance </span> $\rightarrow$ Impedance $\rightarrow$ Power </footer>

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> Impedance </primary>
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- $Z  = R+i\left[\omega_L-\frac{1}{\omega C}\right] $

- $\quad =R+i\left[\Chi _ L - \Chi_ C\right]$

- |Z|  =$\sqrt{R^2+\left(\Chi_ L - \Chi_C\right)^2}$

- $ \tan \varphi=\frac{\Chi_ C - \Chi_L}{R}$

- Impedance triangle: $|\Chi_ C - \Chi_L|$

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<footer style = "font-size: 18px">Phase $\rightarrow$ Impedance $\rightarrow$ <span style = "color:blue"> Impedance </span> $\rightarrow$ Power $\rightarrow$ LCR Circuit (Series) </footer>

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> LCR Circuit (Series) </primary>
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- $ V(t)-V_R - V_L - V_c = 0 $

- $ V(t) = I R + L \frac{d I}{d t}+\frac{Q}{C}$

- Current must be the same through the three elements.

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

- Voltage is in series

- $V = V_m \sin{\omega t}$

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<footer style = "font-size: 18px">Impedance $\rightarrow$ Power $\rightarrow$ <span style = "color:blue"> LCR Circuit (Series) </span> $\rightarrow$ LCR Circuit (Series) $\rightarrow$ Impedance </footer>

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> LCR Circuit (Series) </primary>
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- $I= I_m \sin (\omega t+\phi)$.

- $ |\vec{OA}|= (\Chi_C-\Chi_L) I_{m }$

- $ V_m^2 = (V_m^m)^2+(V_C^m-V_L^m)^2 $

- $ =(I_m R)^2+I_m^2(\Chi_C-\Chi_L)^2 $

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

- $ Z = \sqrt{R^2+ (\Chi_C-\Chi_L)^2} $
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<footer style = "font-size: 18px">Power $\rightarrow$ LCR Circuit (Series) $\rightarrow$ <span style = "color:blue"> LCR Circuit (Series) </span> $\rightarrow$ Impedance $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> Impedance </primary>
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- $ Z \longrightarrow R$

- $ Z \longrightarrow \Chi_C , \Chi_L$

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

- Example:  r.m.s current of 2A is passing through the circuit.

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<div style='float: right; width:50%; current of 2A is passing'>

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<footer style = "font-size: 18px">LCR Circuit (Series) $\rightarrow$ LCR Circuit (Series) $\rightarrow$ <span style = "color:blue"> Impedance </span> $\rightarrow$ Problem $\rightarrow$ Impedance </footer>

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
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- Z  =$\sqrt{R^2+\left(\Chi_C-\Chi_L\right)^2} $

- =$\sqrt{80^2+60^2} = 100\Omega $   
 
- $V^{r m s}  =I^{r m s} \times Z =200 V-r m s $

- $V_R = I R = 2 \times 80 = 160 V(r m s)$

- $V_C  =2 \Chi_C=200 V(r m s)$

- $V_L  =2 \Chi_L=80 V(r m s)$

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<footer style = "font-size: 18px">LCR Circuit (Series) $\rightarrow$ Impedance $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Impedance $\rightarrow$ Inductive circuit </footer>

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> Impedance </primary>
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- $ V_R$ = 160 V(rms)

- $V_L$ = 80 V

- $V_C$ = 200 V

- Resulting voltage lags by $\varphi$

- $tan \varphi =  \frac {120}{160} = \frac{3}{4}$

- $\varphi = 37^0$ (0.64 rods)

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

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> Inductive circuit </primary>
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- Time Lag Between Current Max and Voltage Max

- $\frac {\varphi}{\omega} = \frac{0.64}{400}$ = 1.6 ms.

- what happens when $\omega$ increases.

- $\varphi \rightarrow 0$ for capacitive circuit.

- $tan \varphi = \frac {\Chi_C - \Chi_L}{R}$ capacitive behaves lika a conductor.

- For inductive circuit

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

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> Average Power Delivered </primary>
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- $\langle P\rangle  =I_{r m s}^2 \cdot R $

- $ = 4 \times 80 $

- $ =320 $ watts
 
- **Example:-**  (RC Circuit)

- $R=3 \Omega, C=2.5 \times 10^{-4} \mathrm{F}$

- $\omega=1000 \mathrm{rad} / \mathrm{s}$

- $v_{max }= 5 \mathrm{~V}$

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### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> Average Power Delivered </primary>
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- $ \Chi_C = \frac{1}{\omega C} = \frac{1}{1000 \times 2.5 \times 10^{-4}} $

- $ V=5 \sin \omega t = 4\Omega \$
 
- $ I=I_m \sin (\omega t+\varphi) $
 
- $ Z=\sqrt{R^2 + \Chi_C^2} = 5 \Omega, \quad I_m = 1 A$

- $ v_n^{\max }=3 \mathrm{V}, \mathrm{v}_C^{\max }=\frac{I}{\omega C}=4 \mathrm{V} $

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

### <Success style="font-size:25px"> Lcr Circuit Graphical Solution Alternating Currents L-5 </Success>
### <primary style="font-size:24px"> Example </primary>
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- $\Chi_L = \omega L = 2000\Omega$

- $\Chi_C = \frac {1}{\omega C} = 5000\Omega$

- $ Z = \sqrt {R^2 + (\Chi_C -\Chi_L)^2} = 1800\Omega$

- $ I_m = \frac {140}{1800} = 0.078 A , I_{rms} = 55 MA $

- $ V_{R}^{max} = 78 V $

- $ V_{C}^{max} = I^{max} . \Chi_C = 39 V $

- $V_{V}^{max} = 156 V$

- $tan \varphi = \frac {\Chi_C -\Chi_L}{R} = \varphi = - 56^0$

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