AC Circuits

Harmonic Signals

Harmonic EMF occurs when a coil is spun in a magnetic field.
Sinusoidal voltages can be transformed with inductors
Any periodic signal can be represented with a Fourier Series as a series of sines and cosines
Any pulse can be represented as the integral of sine and cosine signals Fourier transform

Given Harmonic quantity:

v = V_{0} cos (ω t + ϕ)

it can be represented as a rotating vector known as a Phasor. Phasors follow the following rules:

They rotate in the CCW direction with a given angular speed $ω$ .
The length of each phasor is proportional to the AC amperage.
The projection of the phasor on the x-axis yields an instantaneous value for the phasor. When we want to add instantaneous voltages around a loop, we need to know the phase.

v_{1} (t) = V_{1} cos (ω t)

v_{2} (t) = V_{2} cos (ω t + ϕ)

Phasors can be added just like vectors, adding two AC voltages that are 90 degrees yields the following:

v_{1} (t) = V_{1} cos (ω t)

v_{2} (t) = V_{2} cos (ω t + \frac{π}{2})

v_{1} (t) + v_{2} (t) = V_{T} cos (ω t + φ)

Where $φ$ represents the phase angle

V_{T} = V_{1}^{2} + V_{1}^{2}

and

φ = tan^{- 1} (\frac{V _{1}}{V _{2}})

- The [[Phasor]]s for current and voltage are in phase with one another. - Functionally very similar to a DC circuit. # Capacitors in an AC circuit ![[Attachments/Pasted image 20231030162019.png]]

v_{c}(t) = V_{c}\sin(\omega t)

\implies q_{c} = Cv_{c}= CV_{c}\sin(\omega t)

i_{c} = \frac{dq_{c}}{dt} = cv_{c}\cos(\omega t) =\frac{V_{c}}{I/C\omega }\cos (\omega t)

\implies i_{c}= I_{c}\sin(\omega t+\phi)

C a p a c i t or C o nn ec t e d t o an A C so u rces v o lt a g ec u rre n t am pl i t u d esre l a t e d b y

V_{c} = I_{c}X_{c}

X_{c}=\frac{1}{\omega c}

Where $X_{c}$ is known as the *reactance* - The voltage curve "lags" behind the current curve by a quarter cycle, so $\phi=-\frac{\pi}{2}$. # Inductors in AC circuit ![[Attachments/Pasted image 20231030162411.png]]

v_{L}(t)= V_{c}\sin(\omega t) = +L \frac{di}{dt}

\implies i(t) = -\frac{V_{L}}{\omega L} \cos(\omega t)

V o lt a g e inin d u c t ors i s 90 d e g rees ah e a d o f c u rre n t .! [[A tt a c hm e n t s / P a s t e d ima g e 20231030162520. p n g]] Wh e nanin d u c t or i sco nn ec t e d t o an A C so u rce, t h e v o lt a g e an d c u rre n t a rere l a t e d b y :

V=IX_{L}

X_{L} \equiv \omega L

**eLi** the **iCe** man Where in an inductor (L), voltage leads current and vice versa for capacitors. # [[Complex number]] Review Complex numbers are important because they can represent [[Phasor]]s, which have both a real and imaginary component. The complex number $j = \sqrt{ -1 }$. [[Complex number]]s have a real and imaginary part, and can be represented graphically.

\tilde{z} = x + jy

a lt er na t e f or m o f :

\tilde{z} = \mathrm{re}^{i\theta}

r=\sqrt{ \vec{r} \cdot \vec{r} } = \sqrt{ x^{2}+y^{2} }

\theta = \tan ^{-1}\left( \frac{y}{x} \right)

A n o t h er a lt er na t e f or m u s in g [[E u l e r^{'} s f or m u l a]]

\tilde{z}=r(\cos \theta_+j\sin \theta)

Multiplying a harmonic function by $j$ increases the phase by $\frac{\pi}{2}$. Increasing the phasor by 90* ccw.

\tilde{A}(t) = jae^{j\omega t} = ae^{j(\omega t + \pi/2)}

\sin(\omega t) = \cos\left( \omega t - \frac{\pi}{2} \right)

\tilde{z}{i} + \tilde{z}{l} = e^{j_{4}}[(A_{1}+A_{2}e^{j
})]

## Basic complex arithmetic If $\tilde{z}_{1} =A_{1}e^{j}$ and $\tilde{z}_{2}=A_{2}e^{j(\varphi+\delta)}$ then

\tilde{z}{1} + \tilde{z}{2} = e^{j\varphi} (A_{1}+A_{2}e^{j\delta})

\dots

=e^{j\varphi}Me^{j\beta} = Me^{j(\beta+\varphi)}

w h ere

M = \sqrt{ (A_{1}+A_{2}\cos \delta)^{2} + (A_{2}\sin\delta)^{2} }

If $\tilde{z}_{2} = r_{1}e^{j\theta_{1}}$ and $\tilde{z}_{2} = r_{2}e^{j\theta_{2}}$ then

\tilde{z}{1}\tilde{z}{2} = r_{1}e^{j\theta_{1}}r_{2}e^{j\theta_{2}} = r_{1}r_{2}e^{j(\theta_{2}+\theta_{1})}

an d

\frac{\tilde{z}{1}}{\tilde{z}{2}} = r_{1}e^{j(\theta_{1}-\theta_{2})}

Complex Conjugate: invert complex components (put negative sign in front of j) Complex ratios: $$\frac{a+jb}{c+jd} = \frac{(a+jb)(c-jd)}{(c+jd)(c-jd)}$$

=\left( \frac{ac+bd}{c^{2}+d^{2}} \right) + j\left( \frac{-ad+bc}{c^{2}+d^{2}} \right)