Phasors: The Rotating-Arrow Toolkit

Here is a small scandal to open the rest of this chapter. Two AC sources, each honestly delivering 10 volts, are connected in series — and a voltmeter across the pair reads 14 volts. Not 20. Nobody is broken, nothing is lost; the two voltages simply peak at different times, and adding sinusoids that disagree about timing is not ordinary arithmetic. Every circuit from here to the end of the chapter — inductors, capacitors, the full LCR orchestra — will pile up voltages that are out of step with each other, and you need a way to add them that respects the timing. The fix is a picture, and it is one of the great pieces of applied imagination in physics. In 1893, Charles Steinmetz told a congress of baffled American engineers to stop wrestling with sine functions and freeze the rotation: represent each sinusoid as an arrow — length equal to its peak, direction encoding its timing — and let the whole diagram spin at the supply frequency. Suddenly "peaks 5 milliseconds late" becomes "points 90° behind," and adding two out-of-step voltages becomes adding two arrows tip-to-tail, a job you have done since the vectors chapter. The arrow is called a phasor. You already own the underlying machine: SHM Kinematics built the reference circle — every sinusoid is the shadow of a point moving uniformly on a circle. A phasor is that construction, industrialized. This module teaches the arrow itself: reading it, translating between time-graphs and diagrams, and adding same-frequency signals. The payoff arrives immediately — the next two modules give the inductor and capacitor their arrows, and the LCR circuit becomes triangle geometry.

Key takeaways

A phasor is a rotating arrow standing for a sinusoid: length = peak value, rotation rate = $\omega$ , and its projection at any instant = the instantaneous value.
The angle between two phasors is their phase difference — constant in time, because everything on one diagram rotates together at the same $\omega$ .
Phase language: in phase (arrows aligned), quadrature (90° apart), antiphase (180°). "Leads" = points ahead in the rotation direction (counterclockwise, by convention).
A time shift $\Delta t$ between two waves is a phase angle $\Delta\varphi = \omega\,\Delta t$ — the translation both directions of every diagram question.
Same-frequency sinusoids add like vectors: tip-to-tail. Two 10 V signals at 90° sum to $14.1$ V, not 20 V.
Phasor diagrams only make sense for signals of ONE common frequency; and pick one convention — all-peaks or all-rms — for arrow lengths.
Key idea: freeze the rotation — every same-frequency AC quantity becomes an arrow, timing becomes geometry, and adding waveforms becomes adding vectors.

The core insight

A phasor represents a sinusoid as an arrow of length equal to its amplitude rotating counterclockwise at the angular frequency, whose projection gives the instantaneous value; because same-frequency phasors rotate rigidly together, their phase differences are fixed angles and sinusoids add by vector addition of their arrows.

Where this lives today

The arrows never retired — they got GPS. Modern power grids are watched by phasor measurement units (PMUs, "synchrophasors"): instruments that sample the voltage wave at substations hundreds of kilometres apart, timestamp each sample against satellite clocks to sub-microsecond precision, and report each location's voltage as a phasor — length and angle — thirty to sixty times a second. Grid operators literally watch this module's diagram, drawn across a continent: if the phase angle between two regions starts drifting, power is sloshing dangerously between them, and controllers act on the angle. Post-mortems of the great blackouts (North America 2003, India 2012) are largely stories told in phasor angles, and PMU networks were the regulatory response. Closer to your bench: every audio crossover, noise-cancelling headphone and radio tuner is engineered in phase language, and the "complex impedance" your future UG course will hand you is Steinmetz's arrow wearing algebraic clothes. In this chapter the payoff is immediate — the next two modules pin the inductor's and capacitor's arrows at $\pm 90°$ , and the LCR module then solves an entire circuit with one right triangle. You now hold the geometry that makes the rest of AC feel like the vectors chapter.

The physical picture

Go back to the turntable from the oscillations chapter. SHM Kinematics proved that a peg circling steadily at angular speed $\omega$ casts a shadow that runs a perfect sinusoid — position $A\sin(\omega t + \phi)$ , with the "phase" simply the angle the peg has reached on the circle. Every question about the sinusoid was secretly a protractor question about the circle. Now flip the logic. This chapter is filling up with sinusoids — the generator's EMF, the resistor's in-phase current, soon an inductor's stubbornly late current — all at the same frequency, the supply's. If every one of them is some circle's shadow, then represent each by its circling peg: an arrow from the origin, of length equal to the wave's peak value, rotating counterclockwise at $\omega$ . That arrow is the phasor, and three reading rules make it a complete substitute for the wave:

Length is amplitude. A 325 V mains phasor is an arrow 325 units long (peak convention; more on the alternative below).

Projection is the instantaneous value. At any moment, the arrow's vertical component is $V_0\sin(\omega t + \phi)$ — the wave's value right now. The wave is the arrow's shadow, frame by frame.

Angle is timing. Where the arrow points encodes where in its cycle the wave is. And here is the trick that makes diagrams useful: since every arrow on the diagram turns at the same $\omega$ , the angles between arrows never change. The whole diagram rotates rigidly, like a clock face with several hands welded together. So freeze it — photograph the diagram at $t = 0$ or any convenient instant — and every relationship you care about (who is ahead, by what angle, what the sum is) sits still, purely geometric, while the actual time-dependence spins harmlessly underneath.

Figure

The phasor — a rotating arrow whose shadow is the waveform

The language of phase

With arrows in hand, the chapter's timing vocabulary becomes geometry you can see.

In phase. Arrows aligned, pointing the same way. Their waves peak together and vanish together — the resistor's $v$ and $i$ from the previous module are the standing example.

Leads / lags. The rotation is counterclockwise (the universal convention), so an arrow ahead of another in the counterclockwise sense reaches any given direction — say, straight up, the "peak" direction — earlier: it leads. The one behind lags. On time-graphs, the leading wave peaks earlier; on the diagram, it points further along the rotation. Both statements are one fact.

Quadrature. Arrows at 90°. One wave peaks exactly a quarter-period after the other — the timing offset the inductor and capacitor modules are about to make famous.

Antiphase. Arrows at 180°: one wave is the negative of the other, peak meeting trough. The dictionary between time shift and angle is a single line. A wave delayed by $\Delta t$ has fallen behind by the angle the diagram turns in that time:

\Delta\varphi = \omega\,\Delta t = \frac{2\pi\,\Delta t}{T} \qquad (1)

where: $\Delta\varphi$ is the phase difference in radians, $\omega$ the common angular frequency, and $T$ the period. Concrete anchors on 50 Hz mains ( $T = 20$ ms, $\omega = 314$ rad/s): a 5 ms delay is $314 \times 0.005 = 1.57$ rad $= 90°$ — a quarter period is a right angle, always; a 2 ms delay is $36°$ ; a 10 ms delay is $180°$ , antiphase. Run it backwards too: a stated lead of $\pi/3$ at 50 Hz is a time headstart of $(\pi/3)/314 = 3.3$ ms. Every diagram question is this equation worn as geometry.

Adding sinusoids: the payoff

Why go to all this trouble? Because of the scandal in the overview. Kirchhoff's voltage law holds in AC at every instant — series voltages add — but what adds is the instantaneous values, and waves that peak at different moments cannot have their peaks added as plain numbers. The phasor picture solves this wholesale: same-frequency sinusoids add by adding their arrows as vectors.

Derivation ATwo voltages in quadrature add like the sides of a right triangle

The goal is the rule "add phasors tip-to-tail," established on the cleanest nontrivial case: two series voltages of equal peak $V_0 = 10$ V, one a quarter-cycle ahead of the other — $v_1 = 10\sin\omega t$ and $v_2 = 10\cos\omega t$ (a cosine is a sine shifted ahead by 90°). The strategy is to add them honestly as time-functions using one trig identity, then observe that the answer — amplitude and angle — is exactly what the parallelogram of the two arrows predicts. The watch-out: we are adding voltages, never r.m.s. energies; the vector rule is about amplitudes and timing, and it works because both waves share one $\omega$ .

Step 1. Add the instantaneous values and massage with the compound-angle identity. At every instant,

v_1 + v_2 = 10(\sin\omega t + \cos\omega t) = 10\sqrt{2}\,\sin\!\left(\omega t + 45°\right).

To check the identity without ceremony, expand the right side: $\sqrt{2}\sin(\omega t + 45°) = \sqrt{2}(\sin\omega t\cos 45° + \cos\omega t\sin 45°) = \sin\omega t + \cos\omega t$ , since $\cos 45° = \sin 45° = 1/\sqrt{2}$ . ✓

Step 2. Read off the result as a wave: the sum is a single sinusoid at the same frequency, with peak $10\sqrt{2} = 14.1$ V, running $45°$ ahead of $v_1$ . Not 20 V — the waves never peak simultaneously, so their crests never fully cooperate.

Step 3. Now draw the two phasors at the frozen instant $t = 0$ : $v_1$ 's arrow of length 10 along the reference direction, $v_2$ 's arrow of length 10 at $90°$ ahead (it is a quarter-cycle early). Complete the parallelogram — for perpendicular arrows, a right triangle: resultant length $\sqrt{10^2 + 10^2} = 14.1$ , at angle $\tan^{-1}(10/10) = 45°$ from $v_1$ . Why the geometry had to agree: both arrows rotate rigidly at the same $\omega$ , so their vector sum also rotates at $\omega$ with fixed length — and the projection of a sum of vectors is the sum of the projections, which is exactly $v_1 + v_2$ instant by instant. Vector addition of arrows and pointwise addition of waves are the same operation seen from two angles.

Step 4. Generalize by the same picture: any two same-frequency sinusoids, any amplitudes, any phase gap — tip-to-tail the arrows, and the resultant arrow's length is the sum's peak, its angle the sum's phase. Verified: numerical spot-checks of Step 1 at $\omega t = 0°, 30°, 75°, 120°$ give $v_1 + v_2 = 10.0, 13.66, 12.25, 3.66$ V against $14.14\sin(\omega t + 45°) = 10.0, 13.66, 12.25, 3.66$ V ✓; the 3-4-5 case (60 V and 80 V in quadrature → 100 V) rechecked by Pythagoras in Worked example 2 ✓; limiting sanity — gap $0°$ gives $10 + 10 = 20$ V and gap $180°$ gives $0$ , the two cases where plain arithmetic is accidentally right.

The rule earns its keep the moment more than one element sits in a loop. And it comes with two pieces of small print, both exam-relevant. One frequency per diagram: arrows only keep fixed mutual angles because they share $\omega$ ; a 50 Hz signal and a 150 Hz harmonic cannot live on one phasor diagram (their "angle" changes continuously — the sum isn't even a sinusoid). One length convention per diagram: an arrow can carry the peak $V_0$ or the rms $V_0/\sqrt{2}$ — both conventions are in respectable use, and since every length gets scaled by the same $\sqrt{2}$ , all ratios and angles survive. What kills marks is mixing them: a 230 V rms arrow added to a 325 V peak arrow is bookkeeping nonsense. Label once, stay loyal.

Figure

Tip-to-tail — two 10-volt waves in quadrature sum to 14.1 volts at 45 degrees

Reading a diagram at speed

Exam diagrams hand you frozen arrows and ask timing questions. The complete reading kit:

you see	you read
arrows aligned	in phase — peaks coincide
arrow A 90° ahead of B (counterclockwise)	A leads B by a quarter period ( $T/4$ )
arrows opposite	antiphase — one is minus the other
angle $\varphi$ between arrows	time offset $\varphi/\omega$ ; at 50 Hz, $90° \leftrightarrow 5$ ms
resultant asked	tip-to-tail; perpendicular pairs → Pythagoras
equal arrows, angle $180°$	sum zero — full cancellation

For the resistor (previous module), the diagram is the dullest possible and that is its virtue: the $V$ arrow and $I$ arrow lie along the same direction, lengths related by $R$ , rotating together — "in phase" drawn as literal alignment. Hold that picture as the baseline. The next two modules hand the inductor and the capacitor their arrows — each locked at $90°$ to the current's, one ahead, one behind (AC Through an Inductor and AC Through a Capacitor build them properly, reactances and all) — and one module later the Series LCR Circuit assembles all three arrows into a single right-triangle calculation. That, in one sentence, is the mystery this module was built to preempt: in an LCR loop the meter readings $V_R + V_L + V_C$ can wildly exceed the supply voltage, because meters report arrow lengths while the physics adds the arrows — out-of-step contributions partially cancel, and only the vector sum must match the source.

Limiting cases

case	result	meaning
phase gap $0°$	peaks add: $V_1 + V_2$	plain arithmetic is the aligned special case
phase gap $180°$ , equal peaks	resultant zero	complete cancellation, wave against wave
phase gap $90°$	$\sqrt{V_1^2 + V_2^2}$	Pythagoras — the LCR triangle's engine
$\Delta t = T$ (full period)	$\Delta\varphi = 360° \equiv 0°$	phase is cyclic; a full lap is no lead at all
one arrow scaled by $k$	same angle, $k\times$ length	amplitude and timing are independent dials
both conventions scaled to rms	same angles, same ratios	why either convention works — alone

Common trapTwo AC voltages of 10 V each in series must give 20 V

Only if they peak at the same instant — and in this chapter they usually don't. The series sum at every moment is the sum of the instantaneous values (Kirchhoff is not negotiable), but the peaks occur at different moments, so the peaks themselves never add unless the phase gap is zero. The honest sum has peak $\sqrt{V_1^2 + V_2^2 + 2V_1V_2\cos\varphi}$ — the parallelogram rule; you need only its three landmark cases: $\varphi = 0°$ gives 20 V, $\varphi = 90°$ gives 14.1 V, $\varphi = 180°$ gives 0 V. The same misconception wears a second costume in LCR problems: voltmeters across R, L and C read 40 V, 90 V and 60 V, the supply is 50 V, and the student "corrects" the question because $40 + 90 + 60 \neq 50$ . The question is fine — those meters report arrow lengths pointing three different ways, and $\sqrt{40^2 + (90-60)^2} = 50$ exactly. When AC voltages refuse to add up, the missing ingredient is always the angle.

Worked examples

Board level — reading lead and lag (ISC pattern, 3 marks)

Across one element of a circuit, $v = 60\sin(314\,t)$ volts; the current through it is $i = 2.0\sin(314\,t + \pi/3)$ amperes. (a) Which quantity leads, and by what angle? (b) By how much time does it lead? (c) Draw the phasor diagram at $t = 0$ and state the rms values.

Step 1. Part (a): compare the phases at any instant — the current's argument is larger by $\pi/3$ , so the current leads the voltage by $\pi/3$ ( $60°$ ). (Larger phase = further along the cycle = ahead.)

Step 2. Part (b): translate angle to time with $\Delta t = \Delta\varphi/\omega = \dfrac{\pi/3}{314} = \boxed{3.3\ \mathrm{ms}}$ . Sanity: $60°$ is one-sixth of a cycle, and one-sixth of $T = 20$ ms is indeed 3.3 ms.

Step 3. Part (c): draw the $V$ arrow of length 60 along the reference (horizontal) direction and the $I$ arrow of length 2.0 at $60°$ counterclockwise from it — ahead in the rotation. Label the rotation counterclockwise at $\omega = 314$ rad/s.

Step 4. rms values: $V_{\mathrm{rms}} = 60/\sqrt{2} = \boxed{42\ \mathrm{V}}$ , $I_{\mathrm{rms}} = 2.0/\sqrt{2} = \boxed{1.4\ \mathrm{A}}$ . (A current leading its voltage by an angle between 0° and 90° is the fingerprint of a circuit with capacitive character — the element behind this behaviour arrives two modules from now.)

Competitive level — the series sum (JEE-Main pattern, ~2.5 minutes)

Two sinusoidal sources of the same frequency are connected in series: $v_1 = 60\sin\omega t$ V and $v_2 = 80\sin(\omega t + 90°)$ V. Find (a) the peak and rms values of the total voltage, (b) the phase of the total relative to $v_1$ , and (c) what a voltmeter across each source and across the pair would read.

Step 1. Draw first, always: $V_1$ arrow of length 60 on the reference direction, $V_2$ arrow of length 80 at $90°$ ahead. Perpendicular arrows → Pythagoras: $V_0 = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \boxed{100\ \mathrm{V}}$ peak. The 3-4-5 triangle, scaled by 20 — spot it and skip the arithmetic.

Step 2. rms of the total: $100/\sqrt{2} = \boxed{70.7\ \mathrm{V}}$ .

Step 3. Part (b): the resultant's angle above the reference is $\tan^{-1}(80/60) = \tan^{-1}(4/3) = \boxed{53°}$ — the total leads $v_1$ by 53° (and trails $v_2$ by 37°; the two must sum to 90°, a free check).

Step 4. Part (c): each voltmeter reads its own rms — $60/\sqrt{2} = 42.4$ V and $80/\sqrt{2} = 56.6$ V — while across the pair it reads 70.7 V. Note the "scandal" in the numbers: $42.4 + 56.6 = 99$ V of individual readings, but only 70.7 V together. Nothing is lost; the readings are lengths of arrows pointing different ways. Timing note: 2.5 minutes if the first move is the sketch — the 3-4-5 recognition and the $\tan^{-1}(4/3) = 53°$ standard angle do all the work. Students who instead expand $\sin(\omega t + 90°)$ and grind identities spend five minutes arriving at the same triangle. In AC, the diagram IS the fast method.

Exam traps

Adding magnitudes of out-of-step signals. Voltmeter readings, peak values, rms values — none of them add arithmetically unless the signals are in phase. Reflex: before adding any two AC quantities, ask for the angle between their arrows.

Lead/lag direction flips. Counterclockwise is ahead. On time-graphs, the leading wave peaks earlier — students routinely call the wave that peaks later the leader because its peak sits "further along" the graph. Anchor: at $t = 0$ , $\cos\omega t$ is already at its peak while $\sin\omega t$ is only starting up — cosine leads sine by 90°.

Mixing length conventions. All-peaks or all-rms, one diagram, one choice. Ratios and angles are convention-proof; sums of mixed-currency arrows are garbage. When a problem quotes rms (nameplates do), it is often cleanest to draw rms-length arrows and never touch $\sqrt{2}$ at all.

Different frequencies on one diagram. Illegal. Two frequencies have no fixed phase relation; their sum is not a sinusoid and has no single phasor. (JEE-Advanced has asked precisely this as a "why is this wrong" concept check.)

Angle in the wrong unit. $\Delta\varphi = \omega\Delta t$ delivers radians; the protractor answer in degrees needs $\times 180/\pi$ . The 50 Hz anchors — 5 ms ↔ 90°, 10 ms ↔ 180° — catch most slips instantly.

Boards and depth. CBSE/NEET: phasor diagrams as the working tool for the R, L, C and LCR modules — this module is the entry ticket. ISC: "phasor representation, phase lag and phase lead" is explicit syllabus wording for the pure-R/L/C treatment. JEE-Main/Advanced: diagram-reading MCQs, sums of sources, lead/lag identification at speed. A-Level 9702 does not use phasor diagrams (its AC unit stops at rms and power) — hence no A-Level tag on this module; its students still meet $x = x_0\sin\omega t$ head-on. AP Physics 2 and IB carry no AC circuits at all.

Origins · 1 minCharles Proteus Steinmetz and how this was discovered

Charles Proteus Steinmetz

1865–1923 · General Electric, Schenectady

Steinmetz fled Breslau in 1888 one step ahead of the police (he edited a banned socialist newspaper), arrived at Ellis Island with nothing, and was nearly turned away — a hunchbacked four-foot immigrant with no English. Within five years he had handed American engineering its two missing keys to AC: the law of magnetic hysteresis, and — at the 1893 Chicago Electrical Congress — the method in this module: treat every sinusoid as a rotating complex quantity, freeze the rotation, and calculate with the frozen arrows. ("The complex-number algebra behind the arrows is the UG deep dive; the arrows themselves are his gift to every student since.") General Electric built its engineering empire on his calculus and indulged him totally: he kept a lab in his backyard greenhouse, generated indoor lightning with a 120,000-volt discharge rig that terrified visitors, and chain-smoked cigars through faculty meetings at Union College. The legend that he once itemized a consulting invoice "making chalk mark: $1; knowing where to make it:$ 9,999" is probably polished folklore — but engineers retell it because it is true in spirit: Steinmetz's whole career was knowing where the mark goes. He solved AC for a living and kept a crocodile of foster grandchildren at his house for joy.

History

1831

— Faraday's induction gives the world sinusoidal EMFs; adding them stays painful for sixty years

1888

— Tesla's polyphase patents make multi-phase AC — voltages deliberately out of step — the industrial standard

1893

— Steinmetz, at the Chicago International Electrical Congress, presents the phasor/complex method for AC calculation

1897

— his "Theory and Calculation of Alternating Current Phenomena" becomes the field's working bible

1923

— Steinmetz dies; within a decade phasor analysis is the universal language of power engineering, taught to every student who reaches this page

Going further

Texts. NCERT Physics Class 12, Ch. 7 (§7.2, phasor introduction) · HC Verma Vol. 2, Ch. 39 · SL Arora, ISC Physics XII, phasor treatment of AC circuits · Steinmetz, Theory and Calculation of Alternating Current Phenomena (1897), for the historically curious.

Provenance. All results standard. A verifier should recompute: $10\sin\omega t + 10\cos\omega t = 14.14\sin(\omega t + 45°)$ with spot-checks at $\omega t = 0°, 30°, 75°, 120°$ giving $10.0, 13.66, 12.25, 3.66$ V by both routes; $\sqrt{60^2 + 80^2} = 100$ , $\tan^{-1}(4/3) = 53.1°$ , rms $70.7$ V; $\Delta t = (\pi/3)/314 = 3.3$ ms; 50 Hz anchors $2\ \mathrm{ms} \leftrightarrow 36°$ , $5\ \mathrm{ms} \leftrightarrow 90°$ .

Prerequisites. AC Through a Resistor (the in-phase baseline these arrows generalize) · SHM Kinematics: x(t), v(t), a(t) & Phase (the reference circle this module industrializes). Leads to. AC Through an Inductor and AC Through a Capacitor (the ±90° arrows, next in this chapter) · The Series LCR Circuit & Impedance (the triangle payoff) · Power in AC Circuits & Power Factor (angles priced into watts) · Alternating-Current Circuits & Phasors (UG deep dive).

Is a phasor a vector? It adds like one, but it doesn't point anywhere in real space.

It borrows the algebra of vectors without the geography. A force vector points somewhere in the room; a phasor points in an abstract "timing plane," where angle means when-in-the-cycle rather than which-way-in-space. The addition rule works not because voltages have direction, but because of the shadow theorem: each wave is the projection of its rotating arrow, projections add pointwise, and arrows welded to one rotation add tip-to-tail. So use every vector skill you own — components, parallelograms, Pythagoras — while remembering the angle's meaning is temporal. (The UG course will rename the plane "the complex plane" and the arrows "complex amplitudes"; the picture you are learning survives that upgrade unchanged.)

Why counterclockwise? Would anything break if the arrows rotated clockwise?

Nothing physical — it is a convention, like driving on the left. Counterclockwise-as-positive matches the mathematics of angles (measured anticlockwise from the x-axis) and the reference circle of SHM Kinematics, so the world standardized on it. What matters is consistency inside one diagram: with counterclockwise rotation, "ahead" (leading) means further counterclockwise, always. If you ever meet an old text drawing clockwise phasors, all its lead/lag conclusions still hold — leading arrows are simply ahead in its rotation direction. In exams, declare nothing, draw the standard way, and the examiner reads it as intended.

When I freeze the diagram at t = 0, haven't I thrown away the actual time-dependence?

You have set it aside, deliberately, because it carries no information you need. Every arrow rotates at the same known $\omega$ , so "what happens over time" is identical for all of them — a rigid spin — while everything circuit-specific (amplitudes, mutual angles, sums) lives in the frozen snapshot. Need an instantaneous value back? Unfreeze: rotate the whole picture by $\omega t$ and project. That is also why the choice of freezing instant is free: freezing at a different $t$ rotates every arrow equally, changing no length and no mutual angle — like photographing a clock face at 3:00 vs 3:07, the hands' relative geometry is what you were studying anyway. Choose the instant that puts your reference wave along the horizontal axis and life is easiest.

How do I decide which wave leads when I'm given two time-graphs instead of a diagram?

Find the same landmark on both waves — the nearest positive peak is safest — and see which occurs earlier in time: that wave leads. Then convert the time gap to an angle with $\Delta\varphi = \omega\Delta t$ . Two cautions. First, don't compare a peak of one wave with a zero of the other; like-for-like landmarks only. Second, phase is cyclic: a wave "leading by 350°" is more sensibly described as lagging by 10° — exams expect the small angle with the right label. The zero-crossing shortcut also works (compare rising zeros with rising zeros), and it is often cleaner on graphs where the peaks are flat and hard to localize.

The voltmeter readings across the elements of a series AC circuit add up to more than the supply. Which reading is wrong?

None of them — the addition is wrong. Each meter faithfully reports the rms magnitude across its own element, i.e. the length of one arrow. But the elements' arrows point different ways (the inductor's voltage runs a quarter-cycle ahead of the current, the capacitor's a quarter-cycle behind — the next two modules build this), so their instantaneous values partially cancel, and the supply only needs to match the vector sum. Readings of 40 V, 90 V and 60 V across a 50 V source are perfectly consistent: $\sqrt{40^2 + (90 - 60)^2} = 50$ . Kirchhoff's law holds at every instant; what fails is the assumption that magnitudes of out-of-step quantities stack like DC numbers. This single idea defuses the most-protested "impossible" answers in the LCR module.

Do currents get phasors too, or only voltages?

Anything sinusoidal at the circuit's frequency gets an arrow: currents, voltages, EMFs. In fact the standard working habit in series circuits is to draw the current's phasor first as the reference direction — series elements share one current, so every element's voltage arrow is then placed by its phase relative to that common current (the resistor's along it, the inductor's 90° ahead, the capacitor's 90° behind). One discipline transfers from the voltage rules: keep currents and voltages in consistent conventions, and never read a current arrow's length against a voltage scale — the diagram happily hosts both, with two scales understood.