Every circuit so far has been DC: constant sources, constant currents, algebra with real numbers. AC changes the source, not the toolkit you'd expect to need — $v(t) = V_m\sin(\omega t+\theta)$ replaces a battery, and suddenly every voltage and current in the circuit is a function of time, and KVL/KCL applied directly turn into differential equations (a resistor's $v=iR$ is still instant, but an inductor's $v=L\,di/dt$ and a capacitor's $i=C\,dv/dt$ are not). Solving a multi-loop circuit that way, by hand, every single time, would be miserable. The phasor is the trick that makes it unnecessary: it repackages a sinusoid's magnitude and phase into one static complex number, and — because differentiation of a complex exponential is just multiplication by $j\omega$ — every derivative in the circuit's equations turns into ordinary multiplication. The differential equations disappear, and you're back to solving simultaneous algebraic equations with node-voltage or mesh-current analysis, exactly like DC, just with complex numbers standing in for real ones. This chapter builds that transform from scratch: what a sinusoid's parameters mean, why the transform is legal, and how to move fluently between the time-domain picture and the phasor picture. Chapters 9–10 then spend that machinery on impedance and full AC circuit analysis — this chapter is the toll you pay once so the rest of AC is DC-shaped.

8.1  Anatomy of a sinusoid

Every AC voltage or current in this course is written in the general form

with three parameters that fully describe it:

• $V_m$ — the peak amplitude (volts or amps), the maximum value the signal reaches.
• $\omega$ — the angular frequency (rad/s), related to the ordinary frequency $f$ (Hz) and period $T$ (s) by $\omega = 2\pi f = 2\pi/T$.
• $\theta$ — the phase angle (degrees, occasionally radians), a horizontal shift: it tells you how far the waveform is shifted earlier in time relative to a plain $\sin(\omega t)$.

8.2  RMS value

Multimeters and power calculations don't use the peak $V_m$ — they use the root-mean-square (RMS) value, the DC-equivalent value that would deliver the same average power to a resistor. Squaring makes the signal positive everywhere, so a plain time-average of $v(t)$ (which is zero for any pure sinusoid) won't do; you need the average of $v(t)^2$, then undo the square with a root:

For a pure sinusoid this integral collapses to a clean constant. Using $\sin^2 x = \tfrac{1}{2}(1-\cos 2x)$:

The $\cos(2\omega t + 2\theta)$ term oscillates twice per period and integrates to exactly zero over one full period $T$, leaving only the constant $\tfrac12$ term:

8.3  The phasor transform

Here is the actual trick. Euler's formula says $e^{j\phi} = \cos\phi + j\sin\phi$ for any real $\phi$ — a single complex exponential secretly contains both a cosine and a sine. This course's phasors use the cosine as the reference (matching every worked solution you'll see in the sources), so first put any signal into cosine form, then read Euler's formula backwards: the real part of a rotating complex exponential is exactly a cosine wave.

Take $v(t) = V_m\cos(\omega t+\theta)$. Then:

Split the exponent into two factors: $e^{j\theta}$ (fixed — depends only on the signal's own phase) and $e^{j\omega t}$ (spinning — depends only on time, and is the same for every voltage and current in a fixed-frequency circuit). Since every quantity you'll ever compare shares that same spinning factor, it carries no information once you're comparing things to each other — so it's dropped, and what's left, $\mathbf{V} = V_m e^{j\theta} = V_m\angle\theta$, is the phasor: a single static complex number that fully encodes the signal's magnitude and phase, with time removed entirely.

8.4  Why $d/dt \leftrightarrow j\omega$

This is the entire payoff, and it follows in two lines. Differentiate $v(t)$ using the rotating-exponential form directly — $\omega$ and $\theta$ are constants, so the derivative only ever touches the $t$ in the exponent:

Read the right-hand side the same way you read the original definition: it's a sinusoid at the same frequency $\omega$, whose phasor is $j\omega\mathbf{V}$. So differentiating a sinusoid in the time domain is the same as multiplying its phasor by $j\omega$ — a real, second-order, messy operation (differential equations) has become a single complex multiplication (algebra). Integration runs the same argument backwards, dividing instead of multiplying:

This is exactly what makes an inductor's $v_L = L\,di/dt$ and a capacitor's $i_C = C\,dv/dt$ collapse into plain Ohm's-law-shaped equations in the phasor domain — the derivative just becomes a multiplying factor of $j\omega L$ or $j\omega C$. That's Chapter 9's impedance; this chapter only needs you to trust the mechanism.

8.5  Sine ↔ cosine, and rectangular ↔ polar

Two mechanical conversions come up constantly.

Sine to cosine. The phasor transform above assumed a cosine reference, but sources — and this course's own slides — are often written with $\sin$. Convert first, using $\sin x = \cos(x-90°)$:

Rectangular ↔ polar. A phasor $\mathbf{V}=V_r+jV_i$ (rectangular) and $\mathbf{V}=|V|\angle\theta$ (polar) describe the same number two ways — pick whichever form the next step needs (rectangular for addition/subtraction, polar for multiplication/division):

8.6  Lead and lag from a phasor diagram

Once two signals are phasors at the same $\omega$ (comparison is meaningless otherwise — the shared $e^{j\omega t}$ factor is what made phasors comparable in the first place), their relative timing reads straight off the angle between the two vectors on a phasor diagram: whichever phasor sits at the larger angle reaches its peak earlier in time, i.e. it leads.

Formally: for $\mathbf{V}=|V|\angle\alpha$ and $\mathbf{I}=|I|\angle\beta$, the phase difference is $\beta-\alpha$. If it's positive, $\mathbf{I}$ leads $\mathbf{V}$; if negative, $\mathbf{I}$ lags. This is exactly the comparison you'll make constantly in Chapter 9 to read off whether a load is inductive (current lags) or capacitive (current leads).

8.7  Worked examples

8.8  Draggable phasor widget

The static phasor and the moving sinusoid are the same object viewed two ways — the derivation in §8.3 said so, but it's worth watching happen. Drag the tip of the vector (or use the sliders) to set $V_m$ and $\theta$, then press Animate to spin the phasor at rate $\omega$ and watch its real-axis projection trace out $v(t)$ live on the right.

8.9  Self-check

8.10  Practice problems

Each solution below is cross-solved two independent ways — the sine-to-cosine identity, and direct coefficient matching against the $A\cos\omega t - B\sin\omega t$ expansion of $\text{Re}[\mathbf{V}e^{j\omega t}]$ — and shipped only because both agree.