2 - doing algebra with rotating arrows#1 - your introduction to ac circuit analysis

introduction

now that the idea of phasors is on the table, this post has been all about manipulating them. in other words: phasor algebra - or, as I like to think of it, doing math with rotating arrows.

phasors are, at their core, complex numbers. and that opens the door to a neat little toolbox of operations that make ac analysis surprisingly straightforward.

phasor notation in practice

in general, we can write a phasor either in polar form:

$\vec{V} = A \angle \theta$

or in rectangular form:

$\vec{V} = a + jb$

most of the time, i’ll be using the polar form since we as engineers usually have our calculators in hand and its easier to see polar form. but it’s good to know that both forms are equivalent.

visualizing what’s happening

the first thing to note is that phasors are just a way of representing sinusoids. they’re not some magical new thing. they’re just a different way of looking at the same data. the important thing is that you have to pick $\sin$ or $\cos$ as your base function. since $\sin$ is $\cos$ shifted by $90^{\circ}$, you have to consider this phase shift when you’re converting between the two. i usually use $\cos$ as my base function, but it doesn’t really matter which one you pick. just be consistent.

sinusoids to phasors - polar form

polar form is one of the way of representing the sinusoids. so, if you have a sinusoid like this:

$v(t) = 5 \cos(100t + 30^\circ)$

you can simply ignore the angular frequency and just put its amplitude and phase into polar form:

$\vec{V} = 5 \angle 30^\circ$

this is a phasor representation of the sinusoid. it’s like a snapshot of the sinusoid at a specific moment in time, but it also captures the amplitude and phase shift. to visualize this, you can think of the phasor as a rotating arrow in the complex plane. the length of the arrow represents the amplitude, and the angle represents the phase shift.

sinusoids to phasors - rectangular form

the rectangular form is a bit different. it’s like taking the phasor and breaking it down into its real and imaginary components. so, for the same sinusoid:

$v(t) = 5 \cos(100t + 30^\circ)$

you can convert it to rectangular form like this:

$\vec{V} = 5 \cos(30^\circ) + j5 \sin(30^\circ)$

$\vec{V} \approx 4.33 + j2.5$

and it is easier to visualize this in the complex plane. the real part is along the x-axis, and the imaginary part is along the y-axis. so, if you were to plot this phasor in rectangular form, it would look like this:

common pitfalls

there are a few moments where people trip:

• forgetting angle units. degrees vs radians. always worth double-checking before plugging anything into a calculator. i usually convert everything to degrees, but radians are also fine. just be consistent.
• misusing different forms. trying to add rectangular form directly to a polar form doesn’t work. you’ve got to convert to polar first and vice versa.
• missing the physical meaning. some get so deep into the math that they forget what the phasor actually represents - a sinusoid with a certain amplitude and phase. if the phase is off, the wave’s peak shifts in time. simple as that.

i try to emphasize that the math is only half the picture - the other half is intuition. what does it mean when two phasors add up to a smaller result? It means they’re out of phase - partially cancelling each other.

the math

polar form is great for multiplication and division, but rectangular form is better for addition and subtraction. so, we’ll be switching between the two forms as needed. but since we as engineers usually have our calculators in hand, we can do most of the math in just one form - i’m still going to show you both.

addition and subtraction

the most straightforward operations are addition and subtraction. you can use either form, but it’s usually easier to do it in rectangular form.

• rectangular form: add/subtract the real and imaginary parts separately. so, for example:

$(3 + j4) + (1 + j2) = (4 + j6)$

• polar form: convert to rectangular form, add/subtract, and convert back. so, for example:

$5 \angle 50^\circ + 2 \angle 60^\circ \approx 6.97 \angle 52.87^\circ$

$5 \angle 50^\circ \approx 3.21 + j3.83$

$2 \angle 60^\circ = 1 + j\sqrt{3}$

$(3.21 + j3.83) + (1 + j\sqrt{3}) \approx (4.21 + j5.56)$

$(4.21 + j5.56) \approx 6.97 \angle 52.87^\circ$

it is indeed easier to do addition/subtraction in rectangular form, but if you have a calculator that can handle complex numbers, you can do it all in polar form - i always prefer to stay in one form to be consistent.

multiplication

multiplication is a bit different. you can do them in either form, but it’s usually easier to do it in polar form.

• rectangular form: follow the distributive property, don’t forget that $j^2 = -1$. so, for example:

$(3 + j4) \cdot (1 + j2) = (3 \cdot 1 - 4 \cdot 2) + j(3 \cdot 2 + 4 \cdot 1) = -5 + j10$

• polar form: multiply the magnitudes and add the angles. so, for example:

  $5 \angle 50^\circ \cdot 2 \angle 60^\circ = (5\cdot 2) \angle (50^\circ + 60^\circ) = 10 \angle 110^\circ$

division

division is similar to multiplication. you can do them in either form, but it’s usually easier to do it in polar form.

• rectangular form: Multiply the numerator and denominator by the complex conjugate of the denominator to eliminate the imaginary part in the denominator. so, for example:

$\frac{3 + j4}{1 + j2} = \frac{(3 + j4)(1 - j2)}{(1 + j2)(1 - j2)} = \frac{11 - j2}{5} = 2.2 - j0.4$

• polar form: divide the magnitudes and subtract the angles. so, for example:

$\frac{5 \angle 50^\circ}{2 \angle 60^\circ} = \frac{5}{2} \angle (50^\circ - 60^\circ) = 2.5 \angle -10^\circ$

and that’s it! you can do all the math you need with phasors using these simple rules. just remember to keep track of your units and be consistent with your forms.

where we’re headed

in the next post we move into the physical side of this: how phasors interact with r, l, and c components, and how impedance plays into the picture.

but right now, we’re building fluency - learning to treat phasors like second nature. because once you’re comfortable with the algebra, the rest of ac circuit analysis becomes a matter of pattern recognition.

it’s kind of like learning to read music. at first, every note takes effort. then suddenly, one day, your brain stops “decoding” and just sees the melody.

that’s the goal here.