Physics 2 Lecture 19

Exponential Analysis of AC circuits

Root mean square

• Square instantaneous current/voltage
• Take the average and square root Given by:

$$i(t)-I\cos (\omega t)$$ $$⟹I_{rms}=\sqrt{(i^2{)}_{avg}}=\sqrt{\frac{1}{T}{\int }_0^T{i}^2(t)dt}$$ $$\dots$$ $$I_{rms}=\frac{I}{\sqrt{2}};V_{rms}=\frac{V}{\sqrt{2}}$$

Power in AC circuit

$$P(t)=i(t)v(t)$$

Resistor

$$P_r(t)=i_R{v}_R=\frac{V_r^2}{R}\sin (\omega t)$$ $$P_{Ravg}=\frac{1}{2}\frac{V_R^2}{R}=\frac{V_{rms}^2}{R}$$

capacitors

$$P_c(t)=i_c{v}_c=V_c^2\omega t\sin (\omega t)\cos (\omega t)$$ $$P_{cavg}=0$$

• As the charge oscillates, energy flows into and then out of a capacitor.
• Ideal capacitors and inductors don’t dissipate any energy.

Inductors

$$p_L(t)=\frac{V_L^2}{X_L}=\sin (\omega t)\cos (\omega t)$$ $$P_{Lavg}=0$$

• As current oscillates, energy flows into and out of an inductor.

Complex Exponential analysis

Resistor

$$v_{in}(t)=V_0\cos (\omega t)=\mathcal{R}e(V_0{e}^{j\omega t})$$

Where $\mathcal{R}e$ is the real part. The relation between the voltage across a resistor and the current across is given by Ohm’s Law.

$$v_R(t)-i(t)R$$ $$i(t)-\frac{v_0(t)}{R}=\frac{V_0{e}^{j\omega t}}{R}=I_0{e}^{j\omega t}$$ $$Z_R=\frac{v_r}{i_r}=R$$

Impedance across a resistor is given by the resistance (what is impedance) In general, the impedance is given by $Z=R+jX$ The resistance is in phase and the reactance is out of phase part of current/voltage

$$v(t)=Zj(t)$$

$R \to$ "in phase" part of $\frac{v}{i}$
$X\to$ "out of phase" part of $\frac{v}{i}$

Capacitors

$$v_{in}(t)=v_0\cos (\omega t)=V_0{e}^{j\omega t}$$ $$v_c(t)=\frac{q_c(t)}{C}$$ $$⟹q_c(t)=Q_0{e}^{j\omega t}$$ $$i(t)-\frac{dq}{dt}=j\omega V_{0}C{e}^{j\omega t}$$ $$Z_c=\frac{v(t)}{i(t)}=\frac{V_0{e}^{j\omega t}}{j\omega V_{0}C{e}^{j\omega t}}=\frac{1}{j\omega c}$$ $$X_c=\frac{1}{\omega C}$$

Phase change is given by $\frac{1}{j\omega c}$

Inductors

$$v_{in}(t)=V_0\cos (\omega t)=V_0{e}^{j\omega t}$$ $$v_{in}+v_2=0⟸v_{in}=-v_2=L\frac{di}{dt}$$ $$i(t)=\frac{1}{L}\int V_L(t)dt=\frac{V_0}{L}\int e^{j\omega t}dt=\frac{V_0}{j\omega L}{e}^{j\omega t}$$ $$Z_L=\frac{v}{i}=\frac{V_0{e}^{j\omega t}}{\frac{V_0}{j\omega L}{e}^{j\omega t}}=j\omega L$$

Reactance across an inductor is given by $X_L=\omega L$ Multiply by $j$ = advancing the phase. Multiplying by $-j=\frac{1}{j}$ - delays the phase by $\frac{\pi }{2}$.

RC Circuit

$$v_{in}=V_0{e}^{j\omega t}$$

Loop rule:

$$v_{in}-v_R-v_c=v_{in}-i{Z}_R-i{Z}_C=0$$ v_{in} = i\left( \frac{1}{j\omega c} + R \right) $$The [[Impedance]] in the circuit is given by the content in the parentheses above

i(t) = \frac{v_{in}}{R_{t} + \frac{1}{j\omega t}}

$$Takingthisequationandmultiplyingthetopandbottombythecomplexconjugateyields:$$

\frac{v_{in}\left( R-\frac{1}{j\omega c} \right)}{R^{2}+\left( \frac{1}{\omega} \right)^{2}} \implies v_{in} \frac{R+\frac{j}{\omega c}}{R^{2} + \left( \frac{1}{\omega c} \right)^2}

$$andforcurrent:$$

I = (i \cdot i^*)^{1/2} = v_{in} \frac{1}{\left[ R^{2} + \frac{1}{\omega c} ^{2}\right]^{1/2}}

V_{c} gain = g = | \frac{V_{c}}{V_{in}}| = | \frac{IX_{c}}{V_{in}}|

## RL Circuit ![[Pasted image 20231102110156.png]]

V_{in} = V_{0}e^{j\omega t}

$$LoopRule:$$

v_{in}=[R+j\omega L]

i(t) = \frac{v_{in}}{R+j\omega L} = \frac{v_{in}}{R} \frac{1}{1 + j \frac{\omega L}{R}}

fill this in later... ## RLC Circuit ![[Pasted image 20231102110638.png]]

v_{in} = V_{0}e^{j\omega t}

v_{in} = i[Z_{i} + Z_{L} + R] = i\left( \frac{1}{j\omega c} +j\omega L + R \right)

i(t) = \frac{v_{in}}{R+j\left( \omega L-\frac{1}{\omega c} \right)}

v_{out} = Ri = v_{in} \frac{R}{R+j\left( \omega L - \frac{1}{\omega c} \right)}

g_{R} = | \frac{V_{out}}{V_{in}}| = | \frac{1-j\left( \omega \frac{L}{R} - \frac{1}{\omega RC} \right)}{}

$$finishabovelater$$