Phasor diagrams

An AC circuit's voltages and currents are all oscillating at the same frequency, yet a resistor's drop, an inductor's drop and a capacitor's drop never peak at the same instant. Trying to track that timing with sine-wave algebra is painful; a phasor diagram sidesteps it by freezing every sinusoid into a single rotating arrow, drawn with the correct length and the correct angle relative to a common reference. Once the arrows are down on paper, adding voltages becomes ordinary vector addition — a right triangle for R-L or R-C, a triangle with a partial cancellation for R-L-C — and the answer drops out with Pythagoras and an inverse tangent instead of calculus. This article builds that picture from the ground up: how a phasor represents a sinusoid, why current is the natural reference in a series circuit, how the $90^\circ$ leads and lags of L and C combine with R, and how to read supply voltage, phase angle and inductive-or-capacitive behaviour straight off the diagram.

The core insight

A phasor diagram freezes every AC voltage and current in a circuit into a rotating arrow of the right length and angle, so their algebraic sum (Pythagoras plus an inverse tangent, using current as the reference in a series circuit) gives the supply voltage, the phase angle, and whether the circuit behaves as net inductive or net capacitive.

The core idea

A phasor is a straight arrow of fixed length, spinning counter-clockwise at the angular frequency $\omega$ of the AC supply, whose length equals the magnitude (normally RMS) of a sinusoidal voltage or current and whose instantaneous angle equals that sinusoid's phase. Because every phasor in a circuit spins at the same $\omega$ , the whole bundle of arrows rotates together like a rigid body — so only the angles between the phasors matter, and the diagram can be drawn as one frozen snapshot rather than an animation.

Which quantity is the reference matters. In a series circuit the current $I$ is common to every element, so $I$ is drawn along the horizontal axis at $0^\circ$ and every voltage drop is measured relative to it. (In a parallel circuit the roles swap: voltage is common, so voltage becomes the horizontal reference instead — a distinction worth remembering even though this article stays with series circuits.)

The three building blocks follow directly from each element's own current-voltage relationship. For a pure resistor, $V_R$ is in phase with $I$ — both arrows point the same way. For a pure inductor, $V_L$ leads $I$ by $90^\circ$ — drawn rotated a quarter-turn counter-clockwise ahead of $I$ . For a pure capacitor, $V_C$ lags $I$ by $90^\circ$ — equivalently, $I$ leads $V_C$ , so $V_C$ is drawn a quarter-turn clockwise behind $I$ .

Combining them in a series R-L-C circuit stacks $V_R$ along $I$ , $V_L$ ahead, and $V_C$ behind — so $V_L$ and $V_C$ sit exactly $180^\circ$ apart and partially cancel, leaving one net reactive phasor $V_L - V_C$ . The supply voltage is the vector sum of the resistive arm and this net reactive arm, a right triangle whose hypotenuse is $V$ and whose base angle is the phase angle $\theta$ between $V$ and $I$ :

V = \sqrt{V_R^2 + (V_L - V_C)^2}, \qquad \theta = \tan^{-1}\!\left(\frac{V_L - V_C}{V_R}\right)

where: $V_R = IR$ , $V_L = IX_L$ and $V_C = IX_C$ are the RMS voltage drops across the resistor, inductor and capacitor, $I$ is the common series current, and $\theta$ is the angle by which $V$ leads $I$ . A positive $V_L - V_C$ (net inductive) gives current lagging voltage; a negative $V_L - V_C$ (net capacitive) gives current leading voltage. Setting $V_C = 0$ collapses this to the pure R-L result $V = \sqrt{V_R^2+V_L^2}$ ; setting $V_L = 0$ gives the pure R-C result $V = \sqrt{V_R^2+V_C^2}$ .

Figure

A phasor diagram with current along the horizontal axis, the resistive voltage drop in phase with current, the inductive drop rotated ninety degrees ahead, the capacitive drop rotated ninety degrees behind, and the resultant supply voltage as the diagonal sum with the phase angle marked between it and the current

Derivation ASupply voltage and phase angle of a series R-L-C circuit

The goal of this derivation is to obtain the two headline results of phasor-diagram analysis — the magnitude and phase angle of the supply voltage of a series R-L-C circuit — starting from nothing more than the three element voltage phasors and elementary right-triangle trigonometry, so that the diagram and the algebra agree exactly.

Step 1. Fix the reference and place the three element phasors. Since $R$ , $L$ and $C$ are in series, the current $I$ is common to all three, so take $I = I\angle0^\circ$ as the horizontal reference phasor. By each element's own phase relationship, $V_R = IR$ is drawn at $0^\circ$ (in phase with $I$ ), $V_L = IX_L$ is drawn at $+90^\circ$ (leading $I$ ), and $V_C = IX_C$ is drawn at $-90^\circ$ (lagging $I$ ).

Step 2. Combine the two reactive phasors first, since they lie on the same line. $V_L$ points along $+j$ and $V_C$ points along $-j$ , so they are exactly $180^\circ$ apart and add algebraically along that single perpendicular axis: the net reactive phasor is $V_L - V_C$ (inductive if positive, capacitive if negative), pointing along $+j$ or $-j$ according to its sign.

Step 3. Add the resistive phasor to the net reactive phasor. $V_R$ lies along the horizontal ( $0^\circ$ ) axis and the net reactive phasor $(V_L-V_C)$ lies along the vertical ( $90^\circ$ ) axis, so the two are perpendicular. The resultant supply voltage $V$ is the diagonal of the rectangle these two arms complete — equivalently, the hypotenuse of a right triangle with legs $V_R$ and $(V_L-V_C)$ .

Step 4. Apply Pythagoras and basic trigonometry to that right triangle. The hypotenuse gives the magnitude, $V = \sqrt{V_R^2+(V_L-V_C)^2}$ , and the angle between the hypotenuse and the horizontal leg $V_R$ (which is also the angle between $V$ and $I$ , since $I$ is horizontal) is $\theta = \tan^{-1}\!\left(\dfrac{V_L-V_C}{V_R}\right)$ . This is exactly the phase angle by which $V$ leads $I$ : positive when $V_L>V_C$ (net inductive, current lags voltage) and negative when $V_C>V_L$ (net capacitive, current leads voltage).

Worked example

Problem. A series R-L circuit carries a current of $5\ \text{A}$ . The resistor drop is $V_R = 40\ \text{V}$ and the inductor drop is $V_L = 30\ \text{V}$ . Find the supply voltage and phase angle from the phasor diagram.

Step 1. Set the reference. Take current $I = 5\angle0^\circ\ \text{A}$ as the horizontal reference phasor, since $R$ and $L$ are in series and share the same current.

Step 2. Place the element voltages. $V_R = 40\angle0^\circ\ \text{V}$ lies in phase with $I$ ; $V_L = 30\angle90^\circ\ \text{V}$ lies perpendicular, leading $I$ by $90^\circ$ . There is no capacitor here, so this is the $V_C=0$ special case of the general R-L-C formula.

Step 3. Apply the derivation's result with $V_C=0$ : $V = \sqrt{V_R^2+V_L^2} = \sqrt{40^2+30^2} = \sqrt{1600+900} = \sqrt{2500} = 50\ \text{V}$ .

Step 4. Find the phase angle: $\theta = \tan^{-1}(30/40) = \tan^{-1}(0.75) = 36.87^\circ$ . The supply voltage is $V = 50\angle36.87^\circ\ \text{V}$ , leading the current by $36.87^\circ$ — the current lags the voltage, the expected behaviour for an inductive circuit.

Competitive stretch

Extend the same circuit to a full R-L-C case by adding a capacitor drop $V_C = 20\ \text{V}$ and a second resistor arm so that $V_R = 20\ \text{V}$ , $V_L = 50\ \text{V}$ , current still $I = 2\ \text{A}$ . The net reactive drop is $V_L - V_C = 50 - 20 = 30\ \text{V}$ (net inductive, since $V_L>V_C$ ), so $V = \sqrt{V_R^2+(V_L-V_C)^2} = \sqrt{20^2+30^2} = \sqrt{400+900} = \sqrt{1300} = 36.06\ \text{V}$ , and $\theta = \tan^{-1}(30/20) = \tan^{-1}(1.5) = 56.3^\circ$ . Since $V_L>V_C$ , the diagram's net reactive arrow points in the $+j$ (inductive) direction, confirming the circuit is net inductive — matching the earlier impedance-triangle result for the same R, L, C values.

Common traps

Choosing voltage instead of current as the reference phasor for a series circuit — current is common to all series elements and must be the horizontal reference; voltage is the natural reference only for a parallel circuit.
Adding $V_R$ , $V_L$ , $V_C$ by ordinary scalar (arithmetic) addition instead of as perpendicular or opposed phasors — only the Pythagorean vector sum gives the correct supply voltage.
Drawing $V_L$ and $V_C$ in the same direction instead of $180^\circ$ apart — this single slip is the most common phasor-diagram mistake in R-L-C circuits, and it silently turns a subtraction into an addition.
Mixing up which way lead and lag point: remember "ELI the ICE man" — in an inductor (L), EMF (E) leads current (I); in a capacitor (C), current (I) leads EMF (E).

Ask the tutor: if you find it hard to see which way $V_L$ and $V_C$ should point relative to the current, ask the AI tutor to sketch a few phasor diagrams step by step for different R-L, R-C and R-L-C combinations.

Why is current chosen as the reference phasor for series circuits but voltage for parallel circuits?

The reference phasor should always be the quantity that is physically shared by every element being combined, since that keeps the diagram consistent and the angles meaningful. In a series circuit, the same current flows through every element one after another, so current is common and becomes the $0^\circ$ reference against which each element's voltage drop is measured. In a parallel circuit the roles reverse: every branch shares the same voltage across it, while the branch currents differ, so voltage becomes the natural $0^\circ$ reference and the individual branch currents are drawn relative to it.

Why do V_L and V_C end up exactly 180 degrees apart instead of at some other angle?

Each element's phase relationship to the current is fixed by its own physics: an inductor's voltage always leads the current by exactly $90^\circ$ , and a capacitor's voltage always lags the current by exactly $90^\circ$ , regardless of the actual values of $L$ , $C$ or frequency. Leading by $90^\circ$ and lagging by $90^\circ$ from the same reference direction places the two phasors on exactly opposite sides of that reference line, which is a $180^\circ$ separation. That is why, no matter what R-L-C values you are given, $V_L$ and $V_C$ always partially cancel rather than add.

Does the phasor diagram give a different answer from the impedance-triangle method taught alongside it?

No — they must always agree, because they describe the same physical circuit from two equivalent viewpoints. The impedance triangle uses $R$ , $X_L-X_C$ and $Z$ ; the phasor diagram uses $V_R$ , $V_L-V_C$ and $V$ . Since $V_R=IR$ , $V_L=IX_L$ and $V_C=IX_C$ all share the same common factor $I$ , dividing every side of the voltage triangle by $I$ reproduces the impedance triangle exactly, angle for angle. If your two methods ever disagree on $\theta$ , the arithmetic — not the physics — has gone wrong somewhere.

Phasor diagrams

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads