Visual walkthrough — Analyze simple AC circuits with reactance

Step 1 — WHAT is a wiggling voltage, in a picture?

Let us earn each symbol on the graph before we ever compute with it.

• $v(t)$ — the height of the curve at time $t$. It is "how hard the source pushes" at this instant.
• $V_m$ — the peak value, the tallest the curve ever reaches (the pale-yellow dashed line). "$m$" for maximum.
• $\omega$ (Greek "omega") — how fast we wiggle, in radians per second. Bigger $\omega$ ⇒ the curve is squeezed left–right ⇒ faster oscillation. It links to everyday frequency by $\omega =2\pi f$ (one full cycle is $2\pi$ radians).
• $\cos$ — the shape that starts at the top ($v=V_m$ when $t=0$) and swings down, an honest smooth back-and-forth.

WHY cosine and not a straight line? Because the whole point of AC is that the push changes direction rhythmically. A capacitor and an inductor only "wake up" when things change — so we must feed them a changing signal, and cosine is the cleanest changing signal there is.

Step 2 — WHAT the inductor law says, drawn as slope

Read the law term by term, on the picture:

• $i$ — the current, the flow of charge through the coil (chalk-blue curve).
• $\frac{di}{dt}$ — the steepness of that current curve: how fast the current is changing right now. On the figure it is the slope of the little tangent line. Steep ⇒ big; flat ⇒ zero.
• $L$ — inductance, a fixed property of the coil (how stubborn it is). Units: henries.
• $v_L$ — the voltage the coil throws back. The law says it is just the slope $\frac{di}{dt}$ scaled by $L$.

WHY a derivative here? Because an inductor doesn't care how much current flows — it cares how fast it changes. That "rate of change" idea is exactly what a derivative measures, so $\frac{di}{dt}$ is the correct tool, not $i$ itself.

Step 3 — WHAT current the inductor lets through (undo the derivative)

We know $v_L=V_m\cos (\omega t)$ (that's our source). We want $i$. The law gives $v$ from $i$; to go the other way we must undo the derivative — that operation is integration.

Term by term:

• $\int \cdots dt$ — "add up the pushes over time." It is the inverse of $\frac{d}{dt}$, exactly the undo we needed.
• $\int \cos (\omega t)dt=\frac{1}{\omega }\sin (\omega t)$ — integrating cosine gives sine, and out pops a factor $\frac{1}{\omega }$. This $\frac{1}{\omega }$ is where the frequency-dependence is born.
• $\frac{1}{L}$ — carried along from dividing the law by $L$.
• Result peak current $I_m=\frac{V_m}{\omega L}$.

WHY does this prove inductors block high frequency? Look at $I_m=\frac{V_m}{\omega L}$: the $\omega$ sits downstairs. Crank $\omega$ up and $I_m$ shrinks — faster wiggling squeezes the current down. On the figure, notice the blue current is a sine: it peaks a quarter-cycle after the cosine voltage. The current lags.

Step 4 — WHAT "reactance" is: Ohm's law in disguise

Compare $I_m=\frac{V_m}{\omega L}$ with the DC statement $I=\frac{V}{R}$. They have the exact same shape. Whatever sits where $R$ used to sit is the AC opposition.

• $X_L$ — inductive reactance, opposition in ohms $\Omega$. It stores energy, never burns it.
• $\omega L$ — grows straight up with frequency (a rising line on the plot).

The capacitor is the mirror image. Its law $i_C=C\frac{dv}{dt}$ needs a derivative (differentiate the voltage), giving $i(t)=C\frac{d}{dt}\left(V_m\cos \omega t\right)=-\omega C{V}_m\sin (\omega t),I_m=\omega C{V}_m$ so $X_C=\frac{V_m}{I_m}=\frac{1}{\omega C}$. Now $\omega$ is downstairs — a falling curve (the "CLiFF"). See both drawn together above: pale-yellow rising $X_L$, pink falling $X_C$. Here the current is $+\sin$-shaped and peaks before the voltage — the current leads. See Capacitors and Inductors for the component physics.

Step 5 — WHY $R$ and $X$ live on perpendicular axes

Here is the deepest picture. A resistor's current is in step with its voltage (peaks together). A reactor's current is a quarter-cycle ($90°$) shifted. Two things separated by a quarter-cycle are, in the rotating-arrow language of Complex Numbers and Phasors, at right angles.

• Horizontal axis (real, chalk-white): everything in phase with the voltage — that's $R$.
• Vertical axis (imaginary, marked with $j$): everything $90°$ out of phase — that's reactance.
• $j$ — the symbol engineers use for $\sqrt{-1}$; geometrically it means "rotate a quarter-turn." Multiply by $j$ ⇒ swing $90°$ counter-clockwise.

So an inductor's impedance points up: $Z_L=j\omega L=j{X}_L$. A capacitor's points down: $Z_C=\frac{1}{j\omega C}=-\frac{j}{\omega C}=-j{X}_C$.

WHY does this matter? Because you cannot add a horizontal arrow to a vertical arrow by simple arithmetic — you must add them as arrows, tip to tail. That is the seed of Pythagoras in the next step.

Step 6 — WHAT the impedance triangle looks like (the payoff)

Put a resistor, an inductor and a capacitor in series (same current through all). Their impedance arrows add tip-to-tail: $R$ points right, $j{X}_L$ points up, $-j{X}_C$ points down. The two vertical arrows partly cancel, leaving a net vertical piece $X_L-X_C$.

Now a right triangle appears with its own name for every side:

• Horizontal leg $R$ — the resistive, energy-burning part.
• Vertical leg $X_L-X_C$ — the net reactive part (up if inductive wins, down if capacitive wins).
• Hypotenuse $|Z|$ — the total opposition, found by Pythagoras because the legs are perpendicular: $|Z|=\sqrt{R^2+(X_L-X_C{)}^2}$
• Angle $\varphi$ between $R$ and the hypotenuse — the phase shift. It is "which angle has this ratio of opposite over adjacent?", i.e. $\varphi ={\tan }^{-1}\frac{X_L-X_C}{R}$

WHY ${\tan }^{-1}$? On this triangle $\tan \varphi =\frac{\text{opposite}}{\text{adjacent}}=\frac{X_L-X_C}{R}$. We know the two legs and want the angle, so we ask the inverse question "what angle gives this tangent?" — that is ${\tan }^{-1}$.

Step 7 — ALL the cases (never leave the reader stranded)

The vertical leg $X_L-X_C$ can be positive, negative, or zero. Each gives a different triangle, and we must draw all three.

The one-picture summary

One flow: a cosine voltage enters; the inductor integrates it (÷$\omega L$), the capacitor differentiates it (×$\omega C$); each ratio $V_m/I_m$ becomes a reactance; because reactances are $90°$ shifted they stand on the vertical axis while $R$ stands on the horizontal; adding the arrows builds the impedance triangle whose hypotenuse is $|Z|$ and whose angle is $\varphi$.

Quick recall

Connections

• Ohm's Law — every reactance was just $V_m/I_m$ in Ohm's-law shape.
• Complex Numbers and Phasors — supplies the "$90°$ = right angle" picture.
• Capacitors, Inductors — the two component laws we integrated / differentiated.
• Resonance in RLC Circuits — the collapsed triangle case.
• RMS and Peak Values — we used peaks $V_m,I_m$; ratios are identical for RMS.
• Power in AC Circuits — the angle $\varphi$ becomes the power factor $\cos \varphi$.