Phasor diagram for LCR circuit

This is the geometric heart of the Alternating Current chapter's series-LCR-circuit topic. CBSE's 2025-26 syllabus explicitly restricts the LCR circuit to the phasor (graphical) method — the older differential-equation solution is out of scope — so the ability to construct this diagram from first principles, not merely quote the final impedance formula, is itself an examinable skill. Board papers regularly phrase this as "using the phasor diagram, obtain an expression for..." rather than asking for the formula outright, which means a student who has only memorised $Z=\sqrt{R^2+(X_L-X_C)^2}$ without knowing where the square root comes from will lose the derivation marks even while getting the numerical answer right.

Picture the series LCR circuit driven by an AC source $V=V_0\sin\omega t$. Because the three elements are in series, exactly the same instantaneous current $I=I_0\sin\omega t$ flows through all of them — there is one, and only one, current phasor in the whole diagram, and everything else is measured relative to it. Draw this current phasor $I_0$ along a convenient reference direction, taken here as horizontal (angle $0°$), and let it rotate counter-clockwise at angular frequency $\omega$; the instantaneous current at any moment is the vertical projection of this arrow. Each element's voltage phasor is now added using its own known phase relationship with the current — a relationship you should already recall from the previous topic on AC applied to each element separately:

where: $R$ is the resistance in ohms, $X_L=\omega L$ is the inductive reactance, $X_C=1/(\omega C)$ is the capacitive reactance, both also in ohms, and the signed phase angles are measured relative to the current phasor, positive meaning "rotated counter-clockwise, i.e. leading." Physically, $V_R$ is in step with the current because a resistor has no memory — the voltage across it depends only on the instantaneous current. $V_L$ leads because an inductor opposes any change in current, so its voltage peaks a quarter-cycle before the current that produces it reaches its own peak. $V_C$ lags because a capacitor's voltage depends on the accumulated charge, which takes a quarter-cycle to catch up with the current that is charging it. Because $V_L$ and $V_C$ sit on the very same vertical line (one straight up, one straight down from the tail of the current phasor), they are not independent directions to be combined by Pythagoras — they are simply subtracted, giving one net reactive phasor of magnitude $(V_L-V_C)$ pointing along whichever of the two is larger. This is the single most important simplification in the whole construction: instead of adding three phasors at various angles, the diagram reduces to just two mutually perpendicular phasors, $V_R$ (horizontal) and $(V_L-V_C)$ (vertical).

The resultant of these two perpendicular phasors — found the way you would find the resultant of any two perpendicular vectors — is the applied source voltage phasor $V_0$:

where: the square root is the Pythagorean-theorem hypotenuse of a right triangle whose legs are $V_R$ and $(V_L-V_C)$. Physically, this single equation is the whole reason phasor diagrams are worth learning — it replaces "add up three sinusoids that peak at different times," which needs calculus or careful trigonometric identities, with "find the hypotenuse of a right triangle," which needs nothing beyond Class 10 geometry. The angle $\phi$ between the resultant $V_0$ and the current phasor $I_0$ is read straight off the same triangle and is exactly the phase by which the source voltage leads (or, if $(V_L-V_C)$ is negative, lags) the circuit current.

Reading the sign of $(X_L-X_C)$ off the diagram immediately tells you the circuit's character without any further calculation: if $X_L>X_C$, the net reactive phasor points along $V_L$'s direction, $\phi$ is positive, and the applied voltage leads the current — the circuit behaves as inductive. If $X_C>X_L$, the net phasor points along $V_C$'s direction, $\phi$ is negative, and voltage lags current — the circuit is capacitive. If $X_L=X_C$ exactly, the reactive phasor vanishes, $Z=R$, $\phi=0°$, and the circuit is purely resistive — this special condition is resonance, occurring at $\omega_0=1/\sqrt{LC}$, and is the natural next application of the very same diagram.

Board level - reading V0, Z, and phi off the phasor diagram

A series LCR circuit carries a peak current $I_0=2\ \text{A}$ through $R=25\ \Omega$, $X_L=60\ \Omega$, and $X_C=20\ \Omega$. Using the phasor diagram method, find $V_R$, the net reactive voltage $(V_L-V_C)$, the resultant applied voltage $V_0$, the impedance $Z$, and the phase angle $\phi$, and state whether the circuit is inductive or capacitive.

Step 1. Find each individual voltage magnitude from $V=I_0\times(\text{resistance or reactance})$: $V_R=I_0R=2\times25=50\ \text{V}$; $V_L=I_0X_L=2\times60=120\ \text{V}$; $V_C=I_0X_C=2\times20=40\ \text{V}$.

Step 2. Combine the antiparallel reactive phasors first, as the diagram requires: $V_L-V_C=120-40=80\ \text{V}$, directed along $V_L$'s side since it is the larger of the two.

Step 3. Apply Pythagoras to the two remaining perpendicular phasors, $V_R$ and $(V_L-V_C)$, to get the resultant applied voltage: $V_0=\sqrt{V_R^2+(V_L-V_C)^2}=\sqrt{50^2+80^2}=\sqrt{2500+6400}=\sqrt{8900}\approx94.3\ \text{V}$.

Step 4. Get the impedance either directly from $Z=V_0/I_0=94.3/2\approx47.2\ \Omega$, or, matching the derivation above, from $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{25^2+40^2}=\sqrt{625+1600}=\sqrt{2225}\approx47.2\ \Omega$ — both routes agree, which is exactly the point of the impedance triangle being the same picture rescaled.

Step 5. Get the phase angle from the triangle: $\tan\phi=(V_L-V_C)/V_R=80/50=1.6\implies\phi\approx58.0°$. Since the net reactive phasor lies along $V_L$'s direction, the applied voltage leads the current, so the circuit is inductive. Answer: $V_R=50\ \text{V}$, $(V_L-V_C)=80\ \text{V}$, $V_0\approx94.3\ \text{V}$, $Z\approx47.2\ \Omega$, $\phi\approx58.0°$ (inductive).

Competitive level - working backward from a given resultant to verify the diagram and locate resonance

A series LCR circuit's phasor diagram shows the resultant voltage $V_0=130\ \text{V}$ at $\phi=22.6°$ to the current, with $V_R=120\ \text{V}$. (a) Verify this is geometrically consistent and find $(V_L-V_C)$. (b) If instead $R=25\ \Omega$ stays fixed but the frequency is tuned until $X_L=X_C$, what do $V_0$, $Z$, and $\phi$ become for the same peak current $I_0=2\ \text{A}$?

Step 1. Check the given data against $\cos\phi=V_R/V_0$: $\cos(22.6°)\approx0.923$, while $V_R/V_0=120/130\approx0.923$. The two match, confirming the numbers describe a valid right-triangle phasor diagram.

Step 2. Find the net reactive voltage from $\sin\phi=(V_L-V_C)/V_0$: $\sin(22.6°)\approx0.384$, so $V_L-V_C=V_0\sin\phi=130\times0.384\approx50\ \text{V}$. A Pythagoras check confirms it: $\sqrt{120^2+50^2}=\sqrt{14400+2500}=\sqrt{16900}=130\ \text{V}=V_0$, as required.

Step 3. Now consider part (b) — tuning to $X_L=X_C$. On the phasor diagram, the two reactive phasors $V_L$ and $V_C$ become equal in magnitude and, being antiparallel, cancel completely: the net reactive phasor shrinks to zero length, leaving only $V_R$ standing alone on the diagram.

Step 4. With the reactive phasor gone, the resultant collapses onto $V_R$ itself: $V_0=V_R=I_0R=2\times25=50\ \text{V}$, so $Z=R=25\ \Omega$ and $\phi=0°$. This is exactly the resonance condition read straight off the geometry — no separate formula needed, only the observation that a zero-length phasor cannot contribute to the resultant. Answer: (a) consistent, $(V_L-V_C)\approx50\ \text{V}$; (b) at $X_L=X_C$: $V_0=50\ \text{V}$, $Z=25\ \Omega$, $\phi=0°$ (purely resistive). Timing note: about three minutes for both parts. Part (a) is a pure geometry check — trust $\sin^2\phi+\cos^2\phi=1$ rather than recomputing every value from scratch. Part (b) rewards recognising resonance as a picture (the reactive phasor vanishing) rather than reaching for the resonance formula $\omega_0=1/\sqrt{LC}$, which needs $L$ and $C$ values that were never given here — the phasor argument sidesteps that entirely.

Board answers most often lose marks on this topic for reasons that have nothing to do with arithmetic — they are errors in the diagram itself, which then propagate into an otherwise correctly-computed wrong answer. Drawing $V_L$ and $V_C$ as two separate arrows and then attempting to add all three phasors ($V_R$, $V_L$, $V_C$) simultaneously by resolving components is far more error-prone than the intended two-step method: combine the antiparallel pair first into $(V_L-V_C)$, then apply Pythagoras to just two perpendicular phasors. Skipping the combination step is the single most common source of wrong numerical answers even when every individual voltage was computed correctly. Rotating $V_L$ and $V_C$ the wrong way is the second common slip: $V_L$ must lead the current (rotated $90°$ counter-clockwise from it), and $V_C$ must lag the current (rotated $90°$ clockwise). Swapping these two flips the sign of $(X_L-X_C)$ and hence the sign of $\phi$, silently turning an inductive circuit's answer into a capacitive one, or vice versa. Applying the Pythagorean theorem to the wrong pair of quantities — for instance, using $I_0$ together with a voltage, instead of two voltages (or, equivalently, two impedances) — is a dimensional-consistency trap; every leg of a given triangle in this construction must have the same units, either all voltages or, after dividing by $I_0$, all resistances/reactances. Finally, some students think the phasor diagram must be "redrawn" at every instant because AC quantities keep changing. In fact the diagram's shape — the relative lengths and angles of $V_R$, $V_L$, $V_C$, and their resultant — is fixed for a given $R$, $L$, $C$, and $\omega$; only the whole rigid picture rotates together at angular speed $\omega$ as time passes, so the relative geometry you compute from once is valid throughout.

• Construct the series-LCR phasor diagram from scratch, starting from the common current phasor and correctly placing $V_R$, $V_L$, and $V_C$ at their prescribed phase angles.
• Explain why $V_L$ and $V_C$ must be combined into a single net reactive phasor $(V_L-V_C)$ before applying Pythagoras with $V_R$.
• Derive $Z=\sqrt{R^2+(X_L-X_C)^2}$ and the three equivalent forms of $\phi$ directly from the geometry of the diagram, on demand, without reciting them from memory alone.
• Read off, from the sign of $(X_L-X_C)$ alone, whether a given LCR circuit is inductive, capacitive, or (at resonance) purely resistive.
• Use the same diagram to locate the resonance condition $X_L=X_C$ as the special case where the reactive phasor vanishes and $Z$ reaches its minimum value $R$.
• Distinguish between the diagram's fixed shape (set by $R$, $L$, $C$, $\omega$) and its rotation with time (the whole rigid picture turning at angular speed $\omega$).

Ask the tutor: if you would like the AI tutor to sketch this phasor diagram step by step for your own values of $R$, $X_L$, and $X_C$, or to help you keep straight which way $V_L$ and $V_C$ rotate, ask it to walk through the construction with you.