In North America the electric voltage of wall outlets looks like this: \[ v(t) = 150\cos\left( 120\pi t \right), \] where $\omega=2\pi f$ is the angular frequency.

Some smart people invented phasor notation because they found it annoying to always write down the cos part each time they had to solve calculate something in an AC circuit. In DC circuits, you could just write $150V$ next to a voltage source, but in AC circuits you need to write the whole function

In an AC circuit all quantities (voltages and currents) are oscillating at a frequency of 60Hz so we might as well skip the cos part! Instead of writing the whole expression for $v(t)$ we will write just: \[ \vec{V} = \vec{150}, \] where any quantity with an arrow is called a phasor. The oscillating part (cos term, possibly with some phase) is implicit in the phasor notation.

First, let's introduce the main players. The objects of study.

• $i(t)$: Current the circuit as a function of time. Measured in Amperes $[A]$.
• $v(t)$: Voltage as a function of time. Measured in Volts [V].
• $v_{\mathcal{E}}(t)$: Voltage provided by some AC voltage source like the wall power outlet.

In North America and Japan $v_{\mathcal{E}}$ is 110 [V]rms.

  The symbol $\mathcal{E}$ comes from the name //Electromotive force//. 
  This name has since decreased in usage. Do you see yourself asking your friend "Hey Jack, please plug in my laptop power supply into the electromotive force source"?
* $v_{R}(t)$: The voltage as a function of time across some //resistive load// $R$, like an electric heater. 
  Measured in Ohms [$\Omega$]s. The //current-voltage relationship// for a resistive element is 
  $v_{R}(t)=Ri(t)$, where $i(t)$ is the current flowing through the resistor.
* $P(t)=v(t)i(t)$: power consumed/produced by some component. Measured in Watts [W].

Every current and voltage in an AC circuit is oscillating. Current flows in one direction, then it stops, then it flows in the other direction, stops and turns around again, and does this 60 times per second. This is why we call it alternating.

• $f$: Frequency of the AC circuit. Measured in $[Hz]=[1/s]$.
• $\omega=2\pi f$: Angular frequency = the coefficient in front ot $t$ inside $\cos$.

We will use the phasor notation in order to deal with cos terms and sin terms in all our equations. The name phasor comes from the fact that this notation allows us to represent not only the magnitude but also the phase of the sinusoid. The examples below are shown correspond to a frequency of $f=60$ [Hz], which gives us an angular frequency of $\omega \equiv 2\pi f = 120 \pi$.

• $150 \equiv 150\angle 0$: The phasor notation for $150\cos\left( 120\pi t \right)$
• $150\angle 90$: The phasor notation for $150\sin\left( 120\pi t \right)=150\cos\left( 120\pi t -90 \right)$.
• $A\angle 0$: The phasor notation for $A\cos\left( 120\pi t - 0 \right)$
• $A\angle -90$: Is the negative sin function $-A\sin\left( 120\pi t \right)=A\cos\left( 120\pi t +90 \right)$

The following shorthands are very useful:

• $j \equiv 1\angle{90}$: The shorthand notation for $1\sin\left( \omega t \right)\equiv\cos\left( \omega t - 90 \right)$
• $-j\equiv 1\angle-90=-1\angle 90$: The shorthand notation for $-1\sin\left( \omega t \right)\equiv\cos\left( \omega t + 90 \right)\equiv-\cos\left( \omega t - 90 \right)$
• $150j\equiv 150\angle 90$: Equivalent to $150\sin\left( 120\pi t \right)=150\cos\left( 120\pi t -90 \right)$.

You can think of phasors as vector-like, and like vectors they have the component notation and the magnitude-direction notation.

• $\vec{A}$: Some phasor. Could be a voltage $\vec{V}$[V], a current$\vec{I}$[A], or the impedance $\vec{Z}$[$\Omega$] of an inductor, a capacitor or a resistor.
• $\vec{A}=(A_r,A_j)$: The phasor which corresponds to $A(t) = A_r+A_jj = A_r\cos(\omega t)+A_j\sin(\omega t)$.
• $A_r$: The resistive part of $\vec{A}$.
• $A_j$: The rective part of $\vec{A}$.
• $\vec{A}= A\angle \phi = |\vec{A}| \angle \phi_{A}$: The phasor $\vec{A}$ expressed in magnitude-direction notation.
• $|\vec{A}|=A$: The magnitude of the phasor $\vec{A}$. $|\vec{A}| = \sqrt{ A_r^2 + A_j^2}$. Sometimes we will simply refer to it as $A$, so that writing $\vec{A}=A \angle \phi_A$.
• $\phi_{A}=\tan^{-1}\left(A_j/A_r\right)$: The phase of $\vec{A}$

The circuit components which are commonly used in circuits are the following.

• $R$: Resistor. Measured in Ohms [$\Omega$]. We can think of resistance as a physical property of some device or as an operational interpretation: the voltage to current ratio of some device. Indeed, we have $R=V/I$.
• $\vec{Z}$: Impedance is a generalization of the concept of resistance that allows us to treat resistors, capacitors and inductors on the same footing.

The impedance of some device is defined as its voltage to current ratio

  \[
    \vec{Z}\equiv \frac{ \vec{V} }{ \vec{I} }.
  \]
* $\vec{Z}=(Z_r,Z_j)=R+Xj$: Every impedance has a //resistive// part $R$ and a //reactive// part $X$. You can think of these as the two components of the vector $\vec{Z}$: $Z_r$ is the first component, $Z_j$ is the second component.
* $\vec{Z}=|\vec{Z}|\angle \theta=Z\angle \theta$: An can also impedance has a //resistive// part $R$ and a //reactive// part $X$. You can think of these as the two components of the vector $\vec{Z}$: $Z_r$ is the first component, $Z_j$ is the second component.
* $\vec{Z}_C=\frac{-j}{\omega C}$: The impedance of a capacito.  $\vec{Z}_C=(0,X_C)=0-\frac{1}{\omega C}j$
* $\vec{Z}_{eq}$: The equivalent impedance of several circuit elements taken together. Similar to $R_{eq}$, but a little more complicated since we now have resistive and reactive parts. Adding p like vectors.   more general n  how difficult it is to get current through this device. Recall the  The definition of impedance  the type of resistance that a load  circuit element to refits 

impedance

For most purposes in AC circuits we will not talk about the amplitude of the voltage source but its power-average.

• $\vec{V}=150$: Amplitude phasor which corresponds to $v(t)=150\cos\left( 120\pi t \right)$
• $\vec{V}_{rms}=110$: Root-mean squared phasor which corresponds the voltage $v(t)=150\cos\left( 120\pi t \right)$

\[ V_{rms} \equiv \frac{V}{\sqrt{2}}. \]

They are brothers in arms. Every time sin is in the house, cos is there too.

I want to also give you some intuition about the units we normally see for circuit quantities.

Suppose the voltage source was actually source: Provides you with $v(t) = A\sin(2\pi f t)$ [V]

check for yourself if you don't trust me.

The derivative of $\sin(\omega t)$ is $\cos(\omega t)$ \[ \frac{d}{dt}\sin(\omega t) = \omega \cos(\omega t) \]

\[ \frac{d}{dt} \vec{j} = \omega 1 \]

adding

\[ (a + bj) + (c+dj) = (a+c) + (b+d)j \]

subtract

multiply \[ (aj)(bj) = ab(j^2) = -ab \]

divide \[ \frac{1}{j} = \frac{j}{j^2} = \frac{j}{-1} = -j \]

Resistor: Can carry any current $I$, and has a voltage accross its terminals of $V=RI$ where $R$ is the resistance measured in Ohmns [$\Omega$]. The energy of the electrons (the voltage) right before netering the resistor is $IR$ [V] higher than when they leave the resistor. It is important to label the positive and negative terminals of the resistor. The positive temrinal is where the current enters, the negative where the current leaves.

Then there are the energy storing ones:

Two plates \[ \vec{Z}_C = 0 + X_Cj = \frac{-j}{\omega C} \]

\[ q(t) = \int i(t) dt = Cv(t), \qquad \Leftrightarrow \qquad \frac{dq(t)}{dt} = i(t) = C \frac{dv(t)}{dt} \]

The impedance of an inductor is \[ \vec{Z}_L = 0 + X_Lj = j \omega L \] An inductor is given by the \[ \int v(t) dt = Li(t), \qquad \Leftrightarrow \qquad \frac{d\Phi(t)}{dt} = v(t) = L \frac{di(t)}{dt} \]

There is only one really

\[ \vec{V} = \vec{Z} \vec{I}, \] where $Z$ is called the impedance.

It used to be $V=RI$, and actually it still is for resistors, but now we have also capacitors and inductors to deal with.

The way we defined $Z_C$ and $Z_L$ above, however, we can deal with them just as if they were resistors.

\[ Z_{eq} = Z_1 + Z_2 \]

So for example if you have a resistor of 200[$\Omega$] in series with a capacitor of 60[$\mu F$] in an AC circuit at $f=60$[Hz], then the equivalent impedance will be:

\[ \vec{Z}_{eq} = Z_R + Z_C = Z_R + X_Cj = R + \frac{-1}{\omega C}j = 200 - 44.2j. \]

The magnitude of the impedance is: \[ Z = |\vec{Z}_{eq}| = \sqrt{200^2 + 44.2^2} = 204.83. \]

Thus, if I told you that the amplitude of the current is 1[A], and I asked you to find the maximum value of the voltage, then you could use:

\[ V_{max} = Z I_{max} \] and tell me that the amplitude of the voltage is 204.83[V].

\[ Z_{eq} = \frac{Z_1Z_2}{Z_1 + Z_2} \]

alskj \

The resulting phasor $\vec{Z}_{eq}$

Power is voltage times current \[ P=VI. \] For a resistive element, we can also write $P=VI=RI^2=V^2/R$.

In DC we were just muliplying numbers, but in AC circuits the current and the veoltage are oscillating funcitons, so the power will also be oscillating \[ p(t) = v(t)i(t) = V_{mac}\cos(\omega t) I_{max}\cos(\omega t) = V_{mac}I_{max}\cos^2(\omega t). \]

Two observations:

1. The power is oscillating—humming. There is no clean continuous flow of electric energy, but a shaky 120Hz bumpy transfer. This shakyness in the power delivery would be a serious problem for devices that consume a lot of power. This is why Nicola Tesla had to follow up his invetion of alternating currents on two wires into an upgraded version: alternating currents on three wires or three-phase AC.
2. The maximum power is $P_{max} = V_{mac}I_{max}$, but this is not what Hydro will charge you for. Hydro calculates the average power consmption over some time period. At $t=0,t=\pi,t=2\pi,\ldots$ the power consumption is maximum $P_{max}$, but at times $t=\frac{\pi}{2},t=\frac{3\pi}{2}$ the power consumption is zero. To calculate the average power consumption we need to integrate $\cos^2(x)$ over some representative interval, say $[0,2\pi]$:

\[ \frac{1}{2\pi}\int_0^{2\pi} |\cos(t)|^2 dt = \frac{1}{2}, \]

This means that \[ P_{avg} = \frac{P_{max}}{2 } = \frac{V_{max} I_{max}}{2} =\frac{V_{max}}{\sqrt{2}}\frac{I_{max}}{\sqrt{2} } \]

\[ V_{rms} \equiv \sqrt{ \mathop{avg}_t \{ |v(t)|^2 \} } \]

This is what Hydro charges you for.

The voltage amplitude of the wall outlets in North America is 150V, but the effective voltage as far as power is concerened is $\frac{1}{\sqrt{2}}$ of the maximum amplitude of the sine wave, which gives $V_{rms}=110 \approx \frac{150}{\sqrt{2}}$.

Note that it is not high voltage that kills, it is high current.

E before I in L=inductor
I before E in C=capacitor

[ Alternating currents module in the US Navy Training Series ]
http://jacquesricher.com/NEETS/14174.pdf