Phasor representation of AC signals

Try to add two alternating voltages by hand — $v_1 = 30\sin\omega t$ and $v_2 = 40\sin(\omega t + 90°)$ — using pure trigonometry, and you are stuck expanding $\sin(A+B)$ identities before you have even started the real physics. Every AC circuit in this chapter forces the same headache: a resistor's voltage is in step with the current, an inductor's voltage runs a quarter-cycle ahead, a capacitor's voltage runs a quarter-cycle behind, and the source voltage is whatever combination of all three actually reaches the wall socket. Adding sinusoids algebraically to find that combination is technically possible but painfully slow, and it is exactly the kind of calculus-adjacent bookkeeping the CBSE syllabus deliberately avoids for Class 12. The escape hatch is geometric. Freeze each sinusoidal quantity as an arrow — a phasor — of length equal to its peak value, pointing in a direction that encodes its phase. Because every phasor in a single-frequency AC circuit spins at the same angular speed $\omega$ , the angles between the arrows never change, so one static diagram, drawn at any convenient instant, captures the entire phase relationship for all time. Adding alternating voltages then becomes adding vectors — a ruler-and-protractor problem, or at worst a Pythagoras-and-tangent problem, rather than a trigonometric identity hunt. This article builds that geometric toolkit from scratch: what a phasor actually is, how its rotation reproduces $v = V_0\sin\omega t$ , how to draw phasor diagrams for R, L, and C elements, and how to add phasors correctly — including the special case that gives the resistor-inductor impedance triangle its "3-4-5" flavour. Every later article on reactance, impedance, and resonance in the LCR circuit leans on exactly this machinery, so getting comfortable with rotating arrows now pays off through the rest of the chapter.

Key takeaways

A phasor is an arrow of fixed length, rotating counter-clockwise at angular speed $\omega$ ; the instantaneous AC value is its projection onto a reference axis.
For $v = V_0\sin\omega t$ , the phasor has length $V_0$ and starts along the reference axis at $t=0$ ; at time $t$ it has swept through angle $\omega t$ .
A phasor diagram is one frozen snapshot: because all phasors in the circuit share the same $\omega$ , their relative angles (phase differences) stay fixed forever, so the snapshot is valid at any instant.
Phasors add like vectors, by the parallelogram/triangle law: $V_{res}=\sqrt{V_1^2+V_2^2+2V_1V_2\cos\phi}$ , where $\phi$ is the angle between them.
When two phasors are at $90°$ (the resistor-inductor or resistor-capacitor case), the addition collapses to plain Pythagoras, $V_{res}=\sqrt{V_1^2+V_2^2}$ — the geometric origin of the LCR impedance triangle.
CBSE prescribes the LCR series circuit to be solved only by this phasor method — the differential-equation route is explicitly out of syllabus for Class 12.
Key idea: a phasor is a rotating arrow whose length is the peak value and whose angle is the phase, so that adding AC voltages or currents of the same frequency reduces to ordinary vector addition of a frozen diagram rather than trigonometric algebra.

The core insight

Represent each sinusoidal AC quantity by a rotating arrow of length equal to its peak value and angle equal to its phase, so that combining AC voltages or currents becomes vector addition on a single frozen diagram instead of trigonometric calculation.

Where it fits in the syllabus

Phasors are the load-bearing tool of the Alternating Current unit. CBSE's 2025-26 prescription is explicit that the series LCR circuit must be treated by the phasor (graphical) method only — the formal differential-equation solution used in older or engineering-level treatments is out of scope. That single restriction is why this topic exists as a standalone article: before a student can meet reactance, impedance, or resonance, they need to know what a phasor is, how its projection reproduces a sine wave, and how to add two or more of them geometrically. Every worked LCR problem in this chapter — finding impedance, finding the phase angle between current and voltage, finding resonance — is, underneath, a phasor-diagram exercise, so this article is best read as the geometric prerequisite that the resistor, inductor, capacitor, and LCR-circuit articles all cite rather than re-derive.

The physical picture

Picture an arrow of length $V_0$ pinned at the origin, spinning counter-clockwise at a constant rate $\omega$ (the angular frequency of the AC source, in rad/s). At time $t=0$ the arrow lies along a chosen reference direction; after time $t$ it has swept through angle $\omega t$ . The projection of the arrow onto the reference axis at any instant is defined to be the instantaneous value of the physical quantity — voltage or current — that the arrow represents. This is exactly why the phasor reproduces the sinusoid without any calculus.

v(t) = V_0\sin(\omega t)

where: $v(t)$ is the instantaneous voltage (V) at time $t$ , $V_0$ is the peak (maximum) voltage (V), and $\omega$ is the angular frequency (rad/s), related to the ordinary frequency by $\omega = 2\pi f$ . Read physically: as the arrow of length $V_0$ rotates, its vertical projection traces out exactly the sine curve you would plot on an oscilloscope — the phasor is a geometric machine for generating the graph, not an abstract trick layered on top of it. If a second alternating quantity has a phase offset, its phasor is simply drawn at a fixed angle away from the first one:

i(t) = I_0\sin(\omega t - \phi)

where: $I_0$ is the peak current (A) and $\phi$ is the phase angle (rad or degrees) by which the current lags the reference voltage phasor (a negative sign inside the bracket means "lags"; a positive sign would mean "leads"). Because both the voltage phasor and the current phasor rotate at the identical $\omega$ , the angle $\phi$ between them is frozen for all time — the two arrows spin together like the hands of a clock locked at a fixed angular separation. This is the entire reason a single static phasor diagram is enough to describe a circuit that is, in reality, changing every instant: what changes is the common angle $\omega t$ that both arrows share, while the relative angle $\phi$ between them stays put.

Figure

A single arrow of length V naught rotating counter clockwise about the origin, with its vertical projection at angle omega t equal to V naught sine omega t, plotted alongside the resulting sine wave graph so the rotation and the curve are seen to match point for point

Two arrows drawn on the same diagram, at a fixed angle $\phi$ apart, add exactly like displacement vectors. This is the machinery behind every impedance calculation in the chapter:

V_{res} = \sqrt{V_1^2 + V_2^2 + 2V_1V_2\cos\phi}

where: $V_1$ and $V_2$ are the peak magnitudes of the two phasors being combined and $\phi$ is the angle between them on the diagram. Physically, this is just the law of cosines applied to the triangle formed by the two arrows and their resultant — the same formula used to add any two vectors of known magnitude and angle in mechanics, now applied to voltage or current amplitudes instead of forces or velocities.

Derivation AWhy a rotating arrow's projection is a sine wave

Goal: starting only from the definition of a phasor as an arrow of fixed length $V_0$ rotating at constant angular speed $\omega$ about the origin, show that its projection onto a fixed reference axis is precisely the alternating voltage $v(t) = V_0\sin(\omega t)$ that an AC source actually produces, so that the geometric picture of "spinning arrows" can be trusted as an exact stand-in for the trigonometric functions used to describe AC quantities, and not merely a loose analogy or mnemonic.

Step 1. Place the tail of the phasor at the origin of an $x$ – $y$ plane, with the reference axis chosen along the vertical ( $y$ ) direction — the axis along which the physical quantity is read off. At time $t=0$ , by convention, let the arrow coincide with the horizontal (reference-perpendicular) direction, so its instantaneous projection on the vertical axis is zero, matching the fact that a sine wave starts at zero.

Step 2. As time advances, the arrow rotates counter-clockwise at the constant rate $\omega$ (rad/s), so at time $t$ it has turned through the angle $\theta = \omega t$ from its starting position. The tip of the arrow is now at coordinates $(V_0\cos\omega t,\ V_0\sin\omega t)$ , using ordinary trigonometry for a point on a circle of radius $V_0$ at angle $\omega t$ .

Step 3. The projection onto the chosen reference (vertical) axis is, by definition of projection, the $y$ -coordinate of the tip: $v(t) = V_0\sin(\omega t)$ . This is identical, term for term, to the standard expression for an alternating voltage of peak value $V_0$ and angular frequency $\omega$ , so the geometric construction and the algebraic sinusoid are the same object viewed two ways.

Step 4. Because this equivalence holds at every instant $t$ (the argument used no special value of $t$ ), it holds for the entire waveform, not just a snapshot — the rotating arrow does not merely resemble a sine wave at one moment, it generates the whole curve as it turns through a full $360°$ (one period $T = 2\pi/\omega$ ).

Step 5. The same logic extends immediately to a second quantity with a phase offset $\phi$ : if a phasor starts rotation already displaced by angle $\phi$ from the reference arrow (rather than starting from the reference direction itself), its projection is $V_0\sin(\omega t + \phi)$ , since the total angle swept at time $t$ is now $\omega t + \phi$ instead of just $\omega t$ . This justifies drawing "leading" and "ahead" phasors at a fixed positive angular offset and "lagging" phasors at a fixed negative offset, exactly as used throughout the resistor, inductor, and capacitor AC articles.

Worked examples

Board level - resistor and inductor voltage phasors

A resistor and an inductor are connected in series to an AC source. At a given instant, the voltage phasor across the resistor has magnitude $V_R = 30\ \text{V}$ along the reference direction (phase $0°$ ), and the voltage phasor across the inductor has magnitude $V_L = 40\ \text{V}$ , leading $V_R$ by $90°$ (inductor voltage phasors are always drawn $90°$ ahead of the current, and hence $90°$ ahead of the in-phase resistor voltage). Find the magnitude and phase, relative to $V_R$ , of the resultant (supply) voltage phasor.

Step 1. Since $V_L$ is at $90°$ to $V_R$ , set up $V_R$ along the x-axis and $V_L$ along the y-axis of the phasor diagram, so the resultant is the hypotenuse of a right triangle with legs $30\ \text{V}$ and $40\ \text{V}$ .

Step 2. Apply Pythagoras (the $\phi = 90°$ special case of the general phasor-addition formula):

V_{res} = \sqrt{V_R^2 + V_L^2} = \sqrt{30^2+40^2} = \sqrt{900+1600} = \sqrt{2500} = 50\ \text{V}

Step 3. Find the phase angle of the resultant relative to $V_R$ using the tangent of the triangle:

\tan\phi = \frac{V_L}{V_R} = \frac{40}{30} = 1.333 \implies \phi = \tan^{-1}(1.333) \approx 53.1°

The supply voltage phasor has magnitude 50 V and leads the resistor-voltage (and hence the current) phasor by about 53.1 degrees. This is the standard "3-4-5" impedance-triangle calculation that recurs throughout LCR problems, because $30$ , $40$ , $50$ is a Pythagorean triple that CBSE examiners use often precisely because the arithmetic comes out clean.

Competitive level - two equal phasors at an oblique angle

Two AC voltages of equal peak value $V_0 = 20\ \text{V}$ and the same frequency are combined, with a phase difference of $60°$ between their phasors. Find the peak value of the resultant voltage using the general phasor (vector) addition rule, and compare it with the naive scalar sum.

Step 1. Apply the general phasor-addition formula with $V_1 = V_2 = 20\ \text{V}$ and $\phi = 60°$ :

V_{res} = \sqrt{V_1^2+V_2^2+2V_1V_2\cos\phi}

Step 2. Substitute the values, using $\cos60° = 0.5$ :

V_{res} = \sqrt{20^2+20^2+2(20)(20)\cos60°} = \sqrt{400+400+800(0.5)} = \sqrt{400+400+400} = \sqrt{1200}

Step 3. Evaluate the square root:

V_{res} = \sqrt{1200} \approx 34.6\ \text{V}

The resultant peak voltage is approximately 34.6 V — noticeably less than the naive scalar sum $20+20=40\ \text{V}$ , because the two phasors are not aligned and their $60°$ separation causes partial cancellation, exactly as with any two non-parallel vectors added by the parallelogram rule. Timing note: in a competitive-exam setting, recognise the $\phi=60°$ case quickly: since $\cos60°=1/2$ is one of the standard trigonometric values examiners favour (along with $0°$ , $30°$ , $45°$ , $90°$ ), this calculation should take under a minute — do not waste time re-deriving the law of cosines from scratch each time; keep $\cos0°=1,\ \cos30°=\frac{\sqrt3}{2},\ \cos45°=\frac{1}{\sqrt2},\ \cos60°=\frac12,\ \cos90°=0$ memorised so the substitution step is instant.

Common traps

Students lose marks on phasor questions in a few very predictable ways. The most common is treating phasor addition as ordinary scalar addition ( $V_1+V_2$ ) whenever the phasors are not perfectly in phase — this always overestimates the resultant unless $\phi=0°$ , and Worked Example 2 above shows the gap can be substantial ( $40\ \text{V}$ scalar sum versus the correct $34.6\ \text{V}$ ). A second frequent slip is confusing the phasor's length with the RMS value rather than the peak value: either convention (peak-length phasors or RMS-length phasors) can be used consistently throughout a diagram, but mixing the two within the same calculation silently introduces a factor of $\sqrt2$ error. A third trap is forgetting that the phasor diagram is a frozen snapshot: it correctly captures the phase and amplitude relationships between quantities, but the actual instantaneous voltage or current values are still changing continuously as all the phasors rotate together at $\omega$ — a phasor diagram is not a picture of the circuit "at rest." Finally, students often get the direction of the phase shift backwards: a current that lags a voltage must be drawn rotated clockwise (behind) the voltage phasor on the standard counter-clockwise-rotation convention, not counter-clockwise (ahead) of it — this reversal silently flips inductive circuits into capacitive-looking diagrams and vice versa.

Common trapA phasor diagram is not a picture of the circuit at one frozen moment in the everyday sense

It is tempting to read a phasor diagram as "the circuit at $t=0$ , and nothing changes after that." In reality, every phasor on the diagram is continuously rotating at the same angular speed $\omega$ ; what the diagram captures and preserves is only the relative angle between phasors (the phase difference), not their absolute direction at any one instant. The instantaneous voltage and current values are constantly sweeping through their full sinusoidal range — it is only the geometry of how the arrows sit relative to each other that stays fixed. Treating the diagram as a static, unchanging state of the circuit — rather than as a snapshot of relative phase — is the single biggest misconception to unlearn before moving on to reactance and impedance.

You should be able to…

Explain what a phasor is and how its counter-clockwise rotation and projection onto a reference axis reproduce a sinusoidal AC quantity, $v(t)=V_0\sin\omega t$ .
Justify, using Step-by-step reasoning, why the projection of a rotating arrow is exactly a sine function and not merely similar to one.
Draw a simple phasor diagram showing the correct relative phase of voltage and current for a resistor, inductor, or capacitor, including the correct clockwise/counter-clockwise convention for lag and lead.
Add two or more phasors using the vector (parallelogram or triangle) rule, including the simplified right-angle special case that reduces to Pythagoras.
Use $\tan\phi = (\text{opposite component})/(\text{adjacent component})$ to find the phase angle of a resultant phasor relative to a reference phasor.
Connect the phasor picture to why LCR-circuit impedance is found by vector (not arithmetic) addition of resistance and reactances — the geometric idea this whole chapter is built on.

Ask the tutor: if the "rotating arrow becomes a sine wave" idea still feels abstract, ask the AI tutor to walk through the rotation-and-projection picture step by step for a few specific values of $\omega t$ , such as $0°$ , $90°$ , and $180°$ .

Why do we even need phasors if we already have the equation v equal to V naught sin omega t?

The equation is perfectly correct on its own, but combining two or more such equations algebraically — say, adding the resistor and inductor voltages in a series circuit — requires expanding trigonometric identities like $\sin(A+B)$ , which gets messy fast and is exactly the calculus-adjacent bookkeeping CBSE keeps out of Class 12. Phasors let you do the same combination geometrically, with a ruler-and-protractor style vector addition, which is quicker, less error-prone, and matches how the LCR circuit is meant to be solved on this syllabus.

Does the phasor actually rotate in real life, or is it just a mathematical trick?

It is a mathematical construction, not a physical arrow spinning somewhere inside the wire. Nothing in the real circuit is literally rotating. The rotation is a bookkeeping device: choosing an arrow that spins at the signal's angular frequency $\omega$ is simply the geometric way of encoding "this quantity oscillates sinusoidally with this particular amplitude and phase." The payoff is that the geometry reproduces every property of the real oscillating quantity exactly, which is why it is trusted as more than just an analogy.

Should phasor length represent the peak value or the RMS value?

Either choice works, as long as you are consistent throughout a single diagram and a single calculation. Most textbook diagrams, including the ones in this chapter, use the peak value ( $V_0$ or $I_0$ ) as the phasor length, since peak values appear directly in $v=V_0\sin\omega t$ . If a question already gives you RMS values, you may draw RMS-length phasors instead — the angles and the shape of the diagram are unaffected either way — but never mix peak-length and RMS-length phasors on the same diagram, since that silently introduces a factor of $\sqrt2$ error into any addition you do.

How is the phasor method different from just drawing a graph of voltage against time?

A graph of $v$ against $t$ shows you the full time history of one signal, which is useful, but comparing the phase and amplitude of two or three overlapping sinusoids on the same time-axis graph quickly turns into a visual tangle. A phasor diagram throws away the explicit time axis and instead freezes the relative geometry — lengths for amplitude, angles for phase — into one clean static picture, which is exactly what you need to add or compare multiple AC quantities without reading off intersecting curves.