physics-class-12-unit-07-chapter-03-circuits-with-resistance-and-inductance-l-3-10-tevjdg2zwv8

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Circuits with Resistance and Inductance </primary>
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<footer style = "font-size: 18px"> $\rightarrow$  $\rightarrow$ <span style = "color:blue"> Circuits with Resistance and Inductance </span> $\rightarrow$ Faraday's Law of Electromagnetic Induction $\rightarrow$ Lenz's Law </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Faraday's Law of Electromagnetic Induction </primary>
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- **Faraday's Law**

- $\varepsilon=-N\frac{dQ}{dt}$

- $ N = \text{Number of turns} $

- $\Phi = \text{Flux through each turn}$

- $\varepsilon = \int\vec{B}.\vec{ds}$, **Lenz's Law**.

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<footer style = "font-size: 18px"> $\rightarrow$ Circuits with Resistance and Inductance $\rightarrow$ <span style = "color:blue"> Faraday's Law of Electromagnetic Induction </span> $\rightarrow$ Lenz's Law $\rightarrow$ Alternating Voltage </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Lenz's Law </primary>
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- The direction of current is always such that the magnetic field produced by it, opposes the change that produced it.

- $ \Phi_B= \text{B A}\cos\theta (t)$

- = $\text{B A} \cos (\omega t)$

- $\varepsilon=-N \frac{d \phi_B}{d t} =$N B A $~\omega \sin \omega t$

- $=\epsilon_0 \sin \omega t .$

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<footer style = "font-size: 18px">Circuits with Resistance and Inductance $\rightarrow$ Faraday's Law of Electromagnetic Induction $\rightarrow$ <span style = "color:blue"> Lenz's Law </span> $\rightarrow$ Alternating Voltage $\rightarrow$ Alternating Voltage </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Alternating Voltage </primary>
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- Potential $ V = V_m \sin {\omega t} $

- T : Period

- $ \text{V(t + T )} = V (t)$

- Linear Frequency, $f = \frac{1}{T} $

- $\omega = 2\pi f = \frac{2\pi}{T}.$

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<footer style = "font-size: 18px">Faraday's Law of Electromagnetic Induction $\rightarrow$ Lenz's Law $\rightarrow$ <span style = "color:blue"> Alternating Voltage </span> $\rightarrow$ Alternating Voltage $\rightarrow$ Alternating Voltage </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Alternating Voltage </primary>
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- $ V = V_m \sin\omega t$

- $I=\left(\frac{V_m}{R}\right) \sin \omega t$

- $ = I_m \sin \omega t.$

- ($I_m =$ Maximum Value of Current)

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<footer style = "font-size: 18px">Lenz's Law $\rightarrow$ Alternating Voltage $\rightarrow$ <span style = "color:blue"> Alternating Voltage </span> $\rightarrow$ Alternating Voltage $\rightarrow$ Phasor Diagram </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Phasor Diagram </primary>
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- t = 0 (Reference Line)

- $|\overrightarrow{O A}|=V_m$

- $|\overrightarrow{O C}|=I_m$

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

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

- **For a purely resistive circuit, the current is in Phase with the voltage.**

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<footer style = "font-size: 18px">Alternating Voltage $\rightarrow$ Alternating Voltage $\rightarrow$ <span style = "color:blue"> Phasor Diagram </span> $\rightarrow$ Average Force $\rightarrow$ Average Force </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Average Force </primary>
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- The average of F (t) over a cycle,

- $\langle F(t) \rangle = \frac{1}{T} \int_{0}^{T} F(t) dt$
 
- $I (t) = {I_m}. \sin (\omega t)$

- $\langle I(t) \rangle = {I_m}.\frac{1}{T} \int_{0}^{T} \sin \omega t dt$

- = ${I_m} \frac{1}{\omega T} [1 - \cos \omega T]$

- $\omega~T = 2\pi = 0 $

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<footer style = "font-size: 18px">Alternating Voltage $\rightarrow$ Phasor Diagram $\rightarrow$ <span style = "color:blue"> Average Force </span> $\rightarrow$ Average Force $\rightarrow$ Root Mean Square Current </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Root Mean Square Current </primary>
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- **Average Current is Zero**

- $P_{av} = \langle I^2 R \rangle $

- = $ I_ m^2 R \langle \sin^2 \omega t \rangle $ = $\frac{I_m^2 R}{2}$

- Root mean square current

- $I_ {rms}=\sqrt{\langle I^2(t) \rangle} = \frac{I_m}{\sqrt{2}}.$

- $ {V_ {rms}} = \frac{V_m}{\sqrt{2}}$

- ${P _ {av}} = I^2_ {rms}~R$

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<footer style = "font-size: 18px">Average Force $\rightarrow$ Average Force $\rightarrow$ <span style = "color:blue"> Root Mean Square Current </span> $\rightarrow$ House Hold AC $\rightarrow$ Current Amplitude </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Current Amplitude </primary>
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- Alternating Source applied to a purely inductive load No Resistance.

- Back emf = $-L\frac{dI}{dt}$

- V(t) $-L\frac{dI}{dt}=0$

- $\frac{dI}{dt}=\frac{V(t)}{L}=(\frac{V_m}{L})\sin \omega t$

- I (t)= $\int\frac{V_m}{L} \sin \omega t = -\frac{V_m}{L\omega} \cos \omega t$

- =$\frac{V_m}{L\omega} \sin(\omega t-\frac{\pi}{2})$

- **$\frac{V_m}{L\omega}$= Current Amplitude**

- $(\omega t-\frac{\pi}{2})$ **= Phase lag of $\pi/2$ with respect to voltage.**

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<footer style = "font-size: 18px">Root Mean Square Current $\rightarrow$ House Hold AC $\rightarrow$ <span style = "color:blue"> Current Amplitude </span> $\rightarrow$ Inductive Reactance $\rightarrow$ Inductive Circuit </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Inductive Reactance </primary>
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- I(t)=$\frac{V_m}{(\omega L)} \sin (\omega t-\frac{\pi}{2})$

- $(\omega L)$: Inductive Reactance

- $X_L = \omega L$

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<footer style = "font-size: 18px">House Hold AC $\rightarrow$ Current Amplitude $\rightarrow$ <span style = "color:blue"> Inductive Reactance </span> $\rightarrow$ Inductive Circuit $\rightarrow$ Phasor Diagram (Inductive  Circuit) </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Inductive Circuit </primary>
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- For inductive circuit, current Lags the voltage (in phase by $\frac{\pi}{2})$ by T/4

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<footer style = "font-size: 18px">Current Amplitude $\rightarrow$ Inductive Reactance $\rightarrow$ <span style = "color:blue"> Inductive Circuit </span> $\rightarrow$ Phasor Diagram (Inductive  Circuit) $\rightarrow$ Instantaneous Power </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Phasor Diagram (Inductive  Circuit) </primary>
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- t = 0 (Reference)

- $|\overrightarrow {OB}| = V_m$

- $|\overrightarrow {OC}| = I_m.$

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<footer style = "font-size: 18px">Inductive Reactance $\rightarrow$ Inductive Circuit $\rightarrow$ <span style = "color:blue"> Phasor Diagram (Inductive  Circuit) </span> $\rightarrow$ Instantaneous Power $\rightarrow$ Inductive Circuit </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Instantaneous Power </primary>
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- $ P = IV$

- = $ I_m \sin (\omega t - \frac{\pi}{2}) V_m \sin (\omega t) $

- = $-I_m ~ V_m. \cos \omega t \sin \omega t$

- = $ - \frac{I_m V_m}{2} \sin 2 \omega t$

- $ \langle P \rangle$ = 0 over a cycle

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<footer style = "font-size: 18px">Inductive Circuit $\rightarrow$ Phasor Diagram (Inductive  Circuit) $\rightarrow$ <span style = "color:blue"> Instantaneous Power </span> $\rightarrow$ Inductive Circuit $\rightarrow$ Phasor Diagram </footer>

### <Success style="font-size:25px"> Circuits With Resistance And Inductance L-3 </Success>
### <primary style="font-size:24px"> Phasor Diagram </primary>
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- **V (t) in Volts**

- **I (t) in ampere**

- **From $T / 4$ to $T / 2 \quad I>0, \frac{d I}{d t}>0 \quad V(t)>0$**

- $P = I V > 0$ : Energy is absorbed from the source.

- **From $T / 2\text{ to}\quad 3 T / 4 \quad I>0$, $\frac{d I}{d t}<0 ; V(t)<0$**

- $P = I V < 0$ : Energy is returned back.

- **From $\frac{3 T}{4}$ to $T \quad I<0, \frac{d I}{d t}<0: V(t)<0$**
- $P > 0$ : Energy is absorbed

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<footer style = "font-size: 18px">Instantaneous Power $\rightarrow$ Inductive Circuit $\rightarrow$ <span style = "color:blue"> Phasor Diagram </span> $\rightarrow$ Example $\rightarrow$ Thank You </footer>