AC applied to a series LCR circuit

Plug a resistor, an inductor, and a capacitor into the same AC socket, one after another in series, and something interesting happens that none of the three elements shows on its own. A pure resistor lets current and voltage rise and fall in perfect step. A pure inductor makes the current lag the voltage by a quarter cycle; a pure capacitor makes it lead by a quarter cycle. Wire all three together in series and the circuit has to referee a three-way disagreement about timing — and the outcome depends on the tug-of-war between the inductor's and the capacitor's opposite phase habits. The referee's verdict is a single number called impedance, $Z$ , which plays the role resistance played for a plain resistor but now depends on the AC frequency itself. Get the frequency right and the inductor's and capacitor's effects can cancel almost completely, letting a startlingly large current flow — the resonance phenomenon that makes an LCR circuit the working heart of every radio tuner. Get it wrong and the impedance is large, the current small, and the voltage and current fall wildly out of step. This article builds that single formula for $Z$ using nothing more than the phasor (rotating-arrow) method already used for L and C separately — CBSE deliberately keeps the algebra of differential equations out of the picture. By the end you should be able to look at any $R$ , $L$ , $C$ , and frequency and predict, with a short calculation, exactly how much current flows and by how many degrees it lags or leads the source voltage.

Key takeaways

In a series LCR circuit the same current flows through all three elements; their voltage phasors combine as vectors, not as plain numbers, because they point in different directions in time.
$V_R$ is in phase with the current; $V_L$ leads the current by $90°$ ; $V_C$ lags the current by $90°$ — so $V_L$ and $V_C$ point in exactly opposite directions and partially cancel.
The impedance is $Z=\sqrt{R^2+(X_L-X_C)^2}$ , with $X_L=\omega L$ and $X_C=1/(\omega C)$ , and $V_{rms}=I_{rms}Z$ .
The phase angle between voltage and current obeys $\tan\phi=(X_L-X_C)/R$ : the circuit is inductive if $X_L>X_C$ , capacitive if $X_C>X_L$ , and purely resistive ( $Z=R$ , $\phi=0$ ) when $X_L=X_C$ — the resonance condition.
Impedance is frequency-dependent, unlike ordinary resistance — the same LCR circuit behaves completely differently at $50\ \text{Hz}$ than at $5000\ \text{Hz}$ .
CBSE restricts this topic strictly to the phasor/graphical method; no differential-equation solution is required or expected.
Key idea: in a series LCR circuit the phasor sum of $V_R$ , $V_L$ , and $V_C$ (with $V_L$ and $V_C$ opposing each other) gives a frequency-dependent impedance $Z=\sqrt{R^2+(X_L-X_C)^2}$ that fixes both the current magnitude and its phase lag or lead relative to the source.

The core insight

Because $V_L$ and $V_C$ point opposite to each other on the phasor diagram while $V_R$ points along the current, their vector sum gives $Z=\sqrt{R^2+(X_L-X_C)^2}$ , a single frequency-dependent number that fixes both how much current flows and by how many degrees it lags or leads the voltage.

Where it fits in the syllabus

This is the synthesis topic of the CBSE Class 12 Alternating Current chapter: it combines the separate resistor, inductor, and capacitor behaviours from earlier sections into one working circuit, using strictly the phasor method — the board explicitly keeps the formal differential-equation solution out of scope, so nothing here goes beyond vector addition of rotating arrows. It sits between the individual-element articles that come before it and the phasor-diagram and resonance articles that follow, both of which zoom in on details of the same picture built here.

The physical picture

In a series circuit the same instantaneous current, $i=I_0\sin\omega t$ , flows through $R$ , $L$ , and $C$ at every moment, because they share a single loop. What differs is how each element's voltage lines up with that shared current in time. A resistor's voltage rises and falls exactly with the current. An inductor fights any change in current, so its voltage phasor is thrown $90°$ ahead of the current. A capacitor's voltage depends on the charge built up so far, so its voltage phasor falls $90°$ behind the current. Representing each voltage as an arrow (phasor) rotating counter-clockwise at angular frequency $\omega$ , with length equal to its peak value, gives three arrows anchored to the same current arrow but pointing in three different directions.

V_R=I_0R,\qquad V_L=I_0X_L,\qquad V_C=I_0X_C

where: $V_R$ , $V_L$ , $V_C$ are the peak voltages across the resistor, inductor, and capacitor; $I_0$ is the shared peak current; $X_L=\omega L$ is inductive reactance and $X_C=1/(\omega C)$ is capacitive reactance, both in ohms. Physically, $X_L$ grows with frequency (a coil resists rapid current changes more at high frequency) while $X_C$ shrinks with frequency (a capacitor charges and discharges more easily when the current reverses quickly). Because $V_L$ (at $+90°$ ) and $V_C$ (at $-90°$ ) point in exactly opposite directions, they do not add — they subtract, leaving a net reactive voltage of $(V_L-V_C)$ standing at right angles to $V_R$ . Adding this net reactive arm to $V_R$ by the Pythagorean rule (since they are perpendicular) gives the peak applied voltage:

V_0=\sqrt{V_R^2+(V_L-V_C)^2}=I_0\sqrt{R^2+(X_L-X_C)^2}

where: the quantity multiplying $I_0$ is the circuit's total opposition to current, called impedance:

Z=\sqrt{R^2+(X_L-X_C)^2}\ \ (\text{ohm}),\qquad V_0=I_0Z,\qquad V_{rms}=I_{rms}Z

The phase angle $\phi$ by which the applied voltage leads the current is read off the same right triangle of arrows (the impedance triangle, with $R$ along the base and $(X_L-X_C)$ up the side):

\tan\phi=\frac{X_L-X_C}{R}

where: if $X_L>X_C$ , the circuit behaves inductively and current lags voltage ( $\phi>0$ ); if $X_C>X_L$ , it behaves capacitively and current leads voltage ( $\phi<0$ ); if $X_L=X_C$ , the reactances cancel exactly, $Z=R$ is at its smallest possible value, and current and voltage are back in phase — this special frequency is resonance, developed fully in its own article.

Figure

A phasor diagram showing the current arrow along the horizontal axis, the resistor voltage arrow parallel to it, the inductor voltage arrow pointing straight up, the capacitor voltage arrow pointing straight down, and the resultant applied voltage arrow tilted above the current by the phase angle phi

Derivation

Derivation AThe impedance of a series LCR circuit from its phasor diagram

The goal is to combine the three separate voltage-current relationships already known for a resistor, an inductor, and a capacitor into one formula for the total impedance $Z$ of a series LCR circuit, using nothing beyond the fact that phasors add like vectors and that perpendicular vectors combine by the Pythagorean rule — exactly the method CBSE prescribes in place of solving a differential equation.

Step 1. Fix the common current. In a series circuit all three elements carry the same instantaneous current $i=I_0\sin\omega t$ , since there is only one loop and one path for charge to flow. Every voltage phasor below is therefore drawn relative to this single current phasor, taken as the reference direction (along the horizontal axis).

Step 2. Place each element's voltage phasor using its known phase behaviour. The resistor's voltage, $V_R=I_0R$ , is in phase with the current, so its phasor lies along the same horizontal axis. The inductor's voltage, $V_L=I_0X_L$ , leads the current by $90°$ , so its phasor points straight up. The capacitor's voltage, $V_C=I_0X_C$ , lags the current by $90°$ , so its phasor points straight down — directly opposite the inductor's.

Step 3. Combine the two vertical phasors first, since they lie on the same line but point opposite ways. Being collinear and opposed, $V_L$ and $V_C$ combine by plain subtraction, not the Pythagorean rule, giving a single net vertical phasor of magnitude $(V_L-V_C)$ — pointing up if $X_L>X_C$ , down if $X_C>X_L$ .

Step 4. Add the net vertical phasor to the horizontal $V_R$ phasor. These two are now perpendicular to each other (one horizontal, one vertical), so their resultant — the applied voltage $V_0$ — is found by the Pythagorean rule: $V_0=\sqrt{V_R^2+(V_L-V_C)^2}$ . Substituting $V_R=I_0R$ , $V_L=I_0X_L$ , $V_C=I_0X_C$ and factoring out $I_0$ gives $V_0=I_0\sqrt{R^2+(X_L-X_C)^2}$ .

Step 5. Name the bracketed quantity impedance. Since $V_0=I_0Z$ defines $Z$ by direct analogy with $V_0=I_0R$ for a pure resistor, comparison with Step 4 gives $Z=\sqrt{R^2+(X_L-X_C)^2}$ , in ohms. The same ratio holds for RMS values, $Z=V_{rms}/I_{rms}$ , because peak and RMS values differ from each other by the same factor of $\sqrt2$ throughout and that factor cancels in the ratio.

Step 6. Read off the phase angle from the same right triangle used in Step 4. The angle $\phi$ that the resultant $V_0$ makes with the reference current phasor satisfies $\tan\phi=\dfrac{\text{opposite side}}{\text{adjacent side}}=\dfrac{X_L-X_C}{R}$ , which is the standard impedance-triangle formula used to decide whether the circuit is inductive, capacitive, or (at resonance) purely resistive.

Worked examples

Board level - impedance, current, and phase angle for a given LCR circuit

A series LCR circuit has $R=30\ \Omega$ , $L=0.5\ \text{H}$ , and $C=40\ \mu\text{F}$ , connected to a $200\ \text{V}$ (RMS), $50\ \text{Hz}$ AC source. Find (a) $X_L$ , (b) $X_C$ , (c) the impedance $Z$ , (d) the RMS current, and (e) the phase angle $\phi$ .

Step 1. Angular frequency: $\omega=2\pi f=2\pi(50)=314\ \text{rad/s}$ .

Step 2. Reactances: $X_L=\omega L=314\times0.5=157\ \Omega$ , and $X_C=\dfrac{1}{\omega C}=\dfrac{1}{314\times40\times10^{-6}}=\dfrac{1}{0.01256}\approx79.6\ \Omega$ .

Step 3. Impedance: $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{30^2+(157-79.6)^2}=\sqrt{900+5991}=\sqrt{6891}\approx83.0\ \Omega$ .

Step 4. RMS current: $I_{rms}=\dfrac{V_{rms}}{Z}=\dfrac{200}{83.0}\approx2.41\ \text{A}$ .

Step 5. Phase angle: $\tan\phi=\dfrac{X_L-X_C}{R}=\dfrac{77.4}{30}=2.58\implies\phi\approx68.8°$ . Since $X_L>X_C$ , the circuit is inductive and the current lags the applied voltage. Answer: $X_L\approx157\ \Omega$ , $X_C\approx79.6\ \Omega$ , $Z\approx83.0\ \Omega$ , $I_{rms}\approx2.41\ \text{A}$ , $\phi\approx68.8°$ (current lagging).

Competitive level - working backwards from current to an unknown reactance

A series LCR circuit with $R=40\ \Omega$ carries an RMS current of $2\ \text{A}$ when connected to a $220\ \text{V}$ (RMS) AC source. At the operating frequency, $X_C=120\ \Omega$ and $X_C>X_L$ . Find the impedance $Z$ and the value of $X_L$ , and state whether the circuit is inductive or capacitive.

Step 1. Find $Z$ directly from $V_{rms}=I_{rms}Z$ : $Z=\dfrac{V_{rms}}{I_{rms}}=\dfrac{220}{2}=110\ \Omega$ .

Step 2. Use $Z^2=R^2+(X_L-X_C)^2$ to isolate the reactive part: $(X_L-X_C)^2=Z^2-R^2=110^2-40^2=12100-1600=10500$ , so $|X_L-X_C|=\sqrt{10500}\approx102.5\ \Omega$ .

Step 3. Since the problem states $X_C>X_L$ , the circuit is capacitive, so $X_C-X_L=102.5\ \Omega$ , giving $X_L=X_C-102.5=120-102.5=17.5\ \Omega$ . Answer: $Z=110\ \Omega$ , $X_L\approx17.5\ \Omega$ ; the circuit is capacitive, so the current leads the applied voltage. Timing note: about ninety seconds. The shortcut is to get $Z$ straight from $V_{rms}/I_{rms}$ without touching $R$ , $L$ , or $C$ individually first, then use $Z^2-R^2$ to isolate $(X_L-X_C)^2$ — squaring first and taking the square root only at the very end avoids sign confusion. Always re-check which reactance is larger from the problem statement before assigning the final sign to $X_L$ or $X_C$ ; the magnitude $|X_L-X_C|$ alone cannot tell you that.

Common traps

Adding $R$ , $X_L$ , $X_C$ arithmetically to get impedance — this is never correct; they must be combined by the impedance-triangle (Pythagorean) formula, because $V_R$ is out of phase with $V_L$ and $V_C$ .
Forgetting the subtraction $(X_L-X_C)$ inside the square root and instead adding $X_L+X_C$ — since the inductive and capacitive voltage phasors point in opposite directions, they partially cancel, not add.
Assuming the phase angle always makes current lag voltage — true only when $X_L>X_C$ ; when $X_C>X_L$ , current actually leads voltage, and $\phi$ should be reported as negative or explicitly stated as capacitive.
Mixing peak and RMS values partway through a calculation — if $V_{rms}$ is given, use $I_{rms}=V_{rms}/Z$ , not $I_0=V_{rms}/Z$ , and never divide a peak voltage by an RMS current.
Trying to derive $Z$ using calculus or a differential equation for exam credit — CBSE grades this topic strictly by the phasor method; a correctly drawn phasor triangle earns full marks, calculus does not add anything and wastes time.

Common trapThe impedance formula is not simple addition of resistance and reactance

The single most common mark-losing error on this topic is writing $Z=R+X_L+X_C$ or $Z=R+(X_L-X_C)$ as if impedance behaved like resistors in series. It does not, because $R$ , $X_L$ , and $X_C$ are not measured along the same direction in time — $V_R$ is in phase with the current while $V_L$ and $V_C$ are $90°$ out of phase with it in opposite senses. Only vector (phasor) addition, which reduces to the Pythagorean rule for the perpendicular components $R$ and $(X_L-X_C)$ , gives the correct impedance $Z=\sqrt{R^2+(X_L-X_C)^2}$ . Skipping the square root and the square, or adding the three quantities directly, is an easy way to lose full marks on an otherwise correct worked example.

You should be able to…

Draw the phasor diagram for a series LCR circuit, correctly placing $V_R$ , $V_L$ , and $V_C$ relative to the common current phasor.
Derive the impedance formula $Z=\sqrt{R^2+(X_L-X_C)^2}$ from the vector sum of the three voltage phasors, without appealing to calculus.
Calculate $X_L$ , $X_C$ , $Z$ , RMS current, and the phase angle $\phi$ for a given series LCR circuit at a specified frequency.
Decide whether a series LCR circuit is inductive, capacitive, or purely resistive from the relative sizes of $X_L$ and $X_C$ , and state whether current leads or lags voltage accordingly.
Explain, in words, why impedance depends on frequency even though resistance $R$ does not.
Work backwards from measured $V_{rms}$ , $I_{rms}$ , and one known reactance to find the other unknown quantity in the circuit.

Ask the tutor: if the phasor-triangle geometry or the sign of $\phi$ feels shaky, ask the AI tutor to redraw the diagram for your own $R$ , $L$ , $C$ , and $f$ values step by step.

Why do V_L and V_C subtract instead of add if both are reactances?

Reactance measures opposition to current, but the direction in time matters just as much as the size. An inductor's voltage phasor points ninety degrees ahead of the current, while a capacitor's points ninety degrees behind — that is, on opposite sides of the current arrow. When you add two arrows that point in exactly opposite directions, the result is their difference, not their sum. Physically this makes sense too: at the same instant, the inductor is trying to increase the voltage in one sense while the capacitor is trying to increase it in the opposite sense, so they work against each other rather than reinforcing each other. Only if one of them were absent would the full reactive voltage show up.

Is impedance the same thing as resistance, just for AC?

Not quite — impedance includes resistance but adds the frequency-dependent reactive effects of L and C on top of it. Resistance $R$ is fixed regardless of frequency and dissipates energy as heat. Reactance ( $X_L$ or $X_C$ ) does not dissipate energy at all — it just stores and releases it every cycle — and it changes with frequency, growing for $X_L$ and shrinking for $X_C$ as frequency rises. Impedance $Z=\sqrt{R^2+(X_L-X_C)^2}$ is the single number that combines both effects for the purpose of relating $V_{rms}$ and $I_{rms}$ , but unlike plain resistance it is not a fixed property of the circuit alone — change the source frequency and $Z$ changes too.

Do I need to memorise the derivation with the phasor triangle for the board exam?

You should be comfortable enough to redraw the triangle from scratch and read off both $Z$ and $\tan\phi$ , because CBSE frequently asks for exactly this diagram, either as a standalone question or as the first step of a numerical. You do not need to memorise a rigid sequence of steps by rote — understanding why $V_L$ and $V_C$ are collinear and opposite, and why $V_R$ is perpendicular to their difference, lets you reconstruct the whole triangle and both formulas ( $Z$ and $\tan\phi$ ) confidently under exam pressure, which is safer than memorising the final formulas alone and forgetting where they came from.

What happens to Z if I keep increasing the frequency without limit?

As frequency $\omega$ rises, $X_L=\omega L$ grows without bound while $X_C=1/(\omega C)$ shrinks toward zero, so $(X_L-X_C)$ eventually becomes large and positive, and $Z\approx X_L=\omega L$ at very high frequency — the circuit becomes essentially a pure inductor, and the capacitor's effect becomes negligible. At very low frequency the opposite happens: $X_C$ dominates and $Z\approx X_C=1/(\omega C)$ , so the circuit behaves like a pure capacitor. Somewhere in between, at $\omega_0=1/\sqrt{LC}$ , the two reactances are exactly equal and cancel, giving the minimum possible impedance $Z=R$ — that special frequency is resonance, covered in the next article in this chapter.