Resonance in AC circuits

Turn the tuning knob of an old radio and, for most positions, nothing happens — a hiss, maybe a whisper of a distant station bleeding through. Then, at one exact setting, a station snaps into focus, loud and clear, as if the circuit had suddenly decided to listen only to that one frequency. Nothing about the radio's antenna or speaker changed in that instant. What changed is that the receiver's internal LCR circuit was tuned so that its resonant frequency matched the frequency of the broadcast — and at resonance, a series LCR circuit stops fighting the applied signal and lets current flow through it almost as freely as if the coil and capacitor were not there at all. This is the punchline of the entire Alternating Current chapter's treatment of the series LCR circuit. Two earlier ideas — the inductive reactance $X_L=\omega L$ that grows with frequency, and the capacitive reactance $X_C=1/\omega C$ that shrinks with frequency — pull the circuit's impedance in opposite directions as frequency changes. At exactly one frequency their opposing effects become numerically equal and, because they act with opposite phase, they cancel completely. The circuit's total opposition to current collapses to bare resistance $R$ , current surges to its largest possible value, and the current swings in perfect step with the driving voltage. That is resonance. This article rebuilds that single idea from the phasor picture already familiar from reactance and impedance, derives the resonant angular frequency $\omega_0=1/\sqrt{LC}$ cleanly, and then works two numbers all the way through — a gentle "find the resonant frequency" problem and a fuller "find current, voltage, and power at resonance" problem — using the exact same $L$ , $C$ , $R$ values a CBSE board paper would use. Along the way we will meet the resonance curve, see why a smaller resistance makes tuning sharper, and clear up the single most common exam-losing confusion: resonance does not mean zero impedance.

Key takeaways

Resonance in a series LCR circuit happens at one specific angular frequency $\omega_0$ where $X_L=X_C$ , so the reactive parts cancel and only $R$ remains to oppose the current.
The resonant angular frequency is $\omega_0=\dfrac{1}{\sqrt{LC}}$ , independent of the resistance $R$ ; the resonant ordinary frequency is $f_0=\omega_0/2\pi$ .
At resonance the impedance is at its minimum, $Z_{min}=R$ (never zero for a real circuit), so the current amplitude is at its maximum, $I_{0,max}=V_0/R$ .
At resonance the phase angle $\phi=0$ , current and voltage are exactly in phase, and the power factor $\cos\phi=1$ — so average power $P_{avg}=V_{rms}I_{rms}$ is also maximised.
A smaller series resistance $R$ gives a taller, sharper resonance peak (more selective tuning); a larger $R$ gives a shorter, broader peak.
This sharp selectivity is exactly what lets a radio or TV tuner pick one broadcast frequency out of many crowding the airwaves.
Key idea: resonance is the frequency $\omega_0=1/\sqrt{LC}$ at which $X_L$ and $X_C$ exactly cancel, driving impedance down to $R$ and current up to its maximum, in-phase value.

The core insight

In a series LCR circuit, resonance is the single angular frequency $\omega_0=1/\sqrt{LC}$ at which inductive and capacitive reactance exactly cancel, so impedance falls to its minimum value $R$ , current rises to its maximum in-phase value $V_0/R$ , and power factor becomes $1$ .

Where it fits in the syllabus

Resonance is the natural conclusion of the Alternating Current chapter (a Unit VI topic in the CBSE 2025-26 syllabus), sitting immediately after reactance/impedance and the series LCR circuit, and immediately before power in AC circuits and the transformer. The prescribed scope names the resonance condition explicitly — $\omega_0=1/\sqrt{LC}$ — to be reached and used through the phasor/impedance picture already built up, with no new mathematics introduced. The formal differential-equation route to the LCR circuit and any quantitative $Q$ -factor/bandwidth mathematics are outside the prescribed scope; this article stays within phasors and algebra, exactly as the board exam does.

The physical picture

In a series LCR circuit driven by a sinusoidal source of peak voltage $V_0$ and angular frequency $\omega$ , the resistor, inductor, and capacitor each oppose the current differently, and their combined opposition — the impedance — is

Z=\sqrt{R^2+(X_L-X_C)^2},\qquad X_L=\omega L,\quad X_C=\frac{1}{\omega C}

where: $R$ is resistance (ohm, $\Omega$ ), $X_L$ is inductive reactance and $X_C$ is capacitive reactance (both in ohm), and $Z$ is the net impedance (ohm) that the peak current must divide into the peak voltage, $I_0=V_0/Z$ . Physically, $X_L$ rises linearly with frequency (a coil fights change in current harder as the current wiggles faster) while $X_C$ falls with frequency (a capacitor's charging/discharging opposition weakens as the voltage reverses more often) — the two reactances are pulling the circuit's total opposition in opposite directions as $\omega$ is swept. As $\omega$ is increased from a low value, $X_L$ climbs and $X_C$ drops; at one particular angular frequency, the resonant frequency $\omega_0$ , the two become exactly equal, $X_L=X_C$ , so the bracket $(X_L-X_C)$ inside the square root vanishes and the impedance falls to its smallest possible value:

Z_{min}=R

where: the reactive contribution has disappeared entirely, leaving only the resistor's plain opposition — the circuit behaves, at this one frequency, exactly like a purely resistive circuit even though the coil and capacitor are still very much present and carrying current. Because current amplitude is $I_0=V_0/Z$ , minimum impedance means maximum current:

I_{0,max}=\frac{V_0}{R}

and the phase angle, given generally by $\tan\phi=(X_L-X_C)/R$ , becomes $\tan\phi=0$ , so $\phi=0$ — current and voltage rise and fall together, exactly in step, and the power factor $\cos\phi=1$ takes its largest possible value.

Figure

A graph of current amplitude against angular frequency for a series LCR circuit, showing a low broad curve for large resistance and a tall sharp peaked curve for small resistance, both peaking at the same resonant angular frequency omega naught

Qualitatively, the current-versus-frequency curve rises from a small value at low $\omega$ (where $X_C$ dominates and starves the circuit of current), climbs to a peak exactly at $\omega=\omega_0$ , and falls again at high $\omega$ (where $X_L$ dominates). A smaller resistance $R$ produces a taller, sharper peak — the circuit responds strongly only within a narrow band of frequencies near $\omega_0$ , which is precisely the selectivity that lets a radio tuner reject a neighbouring station a few kilohertz away. A larger $R$ flattens and broadens the same peak. The formal quality-factor mathematics that describes exactly how sharp the peak is lies beyond this syllabus; the qualitative shape and its dependence on $R$ is what is examined.

Derivation AFinding the resonant angular frequency of a series LCR circuit

The goal is to find the one angular frequency at which a series LCR circuit's impedance is minimised and its current is maximised, starting from nothing more than the definitions of inductive and capacitive reactance and the impedance formula already established for the series LCR circuit via the phasor method. No differential equations are used — only algebra on the phasor-derived impedance expression, exactly as the CBSE syllabus prescribes.

Step 1. Start from the impedance of a series LCR circuit, obtained earlier from the phasor diagram of the three voltage phasors $V_R$ , $V_L$ , $V_C$ : $Z=\sqrt{R^2+(X_L-X_C)^2}$ , with $X_L=\omega L$ and $X_C=1/(\omega C)$ . This single formula already encodes the whole frequency dependence of the circuit, since $R$ is fixed but $X_L$ and $X_C$ both change with $\omega$ .

Step 2. Ask when $Z$ is smallest. Since $R^2$ is a fixed positive number, $Z$ is smallest exactly when the term $(X_L-X_C)^2$ is smallest — and a square is smallest, equal to zero, precisely when the quantity being squared is zero. So the condition for minimum impedance is simply $X_L-X_C=0$ , that is, $X_L=X_C$ .

Step 3. Substitute the definitions of the two reactances into this condition. $X_L=X_C$ becomes $\omega L=\dfrac{1}{\omega C}$ . Multiply both sides by $\omega$ to clear the fraction: $\omega^2 L=\dfrac{1}{C}$ , and hence $\omega^2=\dfrac{1}{LC}$ .

Step 4. Take the square root of both sides, keeping only the physically meaningful positive root (angular frequency cannot be negative), and name this special frequency $\omega_0$ : $\omega_0=\dfrac{1}{\sqrt{LC}}$ . The corresponding ordinary frequency, obtained by dividing by $2\pi$ , is $f_0=\dfrac{\omega_0}{2\pi}=\dfrac{1}{2\pi\sqrt{LC}}$ . Notice that $\omega_0$ depends only on $L$ and $C$ — the resistance $R$ never entered the calculation, so changing $R$ shifts nothing about where the peak sits, only how tall and sharp it is.

Step 5. Confirm the consequence. At $\omega=\omega_0$ , substitute $X_L=X_C$ back into $Z=\sqrt{R^2+(X_L-X_C)^2}$ to get $Z=\sqrt{R^2+0}=R$ , the smallest value impedance can take. Then $I_{0,max}=V_0/R$ is the largest current amplitude the circuit can deliver at any frequency for that supply, and $\tan\phi=(X_L-X_C)/R=0$ gives $\phi=0$ , so current and voltage are in phase and $\cos\phi=1$ — maximum power factor alongside maximum current.

Worked examples

Board level — resonant frequency of a given LCR combination

A series LCR circuit has $L=1\ \text{H}$ and $C=10\ \mu\text{F}$ . Find the resonant angular frequency $\omega_0$ and the resonant frequency $f_0$ .

Step 1. Convert $C$ into SI units (farads): $C=10\ \mu\text{F}=10\times10^{-6}\ \text{F}=1\times10^{-5}\ \text{F}$ .

Step 2. Compute the product $LC$ : $LC=1\times(1\times10^{-5})=1\times10^{-5}\ \text{s}^2$ .

Step 3. Apply the resonance formula: $\omega_0=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{1\times10^{-5}}}=\dfrac{1}{3.162\times10^{-3}}\approx 316.2\ \text{rad/s}$ .

Step 4. Convert to ordinary frequency: $f_0=\dfrac{\omega_0}{2\pi}=\dfrac{316.2}{6.283}\approx 50.3\ \text{Hz}$ .

The resonant angular frequency is $\omega_0\approx 316.2\ \text{rad/s}$ and the resonant frequency is $f_0\approx 50.3\ \text{Hz}$ — strikingly close to India's standard 50 Hz mains supply, meaning this particular $L$ - $C$ pair would draw an unusually large current if connected straight to the mains at exactly this frequency.

Competitive level — current, voltage, and power at resonance

A series LCR circuit with $R=10\ \Omega$ , $L=0.4\ \text{H}$ , and $C=25\ \mu\text{F}$ is connected to a $100\ \text{V}$ (RMS) AC source whose frequency is tuned exactly to the circuit's resonant frequency. Find (a) the resonant angular frequency, (b) the current at resonance, and (c) the average power delivered at resonance.

Step 1. Compute $LC$ : $LC=0.4\times(25\times10^{-6})=1\times10^{-5}\ \text{s}^2$ , the same product as the board-level example, so $\omega_0=\dfrac{1}{\sqrt{1\times10^{-5}}}\approx 316.2\ \text{rad/s}$ (and $f_0\approx 50.3\ \text{Hz}$ , again near mains frequency).

Step 2. Find the current at resonance. Since $Z=R$ at resonance, Ohm's law for RMS quantities gives $I_{rms}=\dfrac{V_{rms}}{R}=\dfrac{100}{10}=10\ \text{A}$ — a large current for a $100\ \text{V}$ source, precisely because impedance has collapsed to just $10\ \Omega$ .

Step 3. Find the average power at resonance. Since $\cos\phi=1$ at resonance, $P_{avg}=V_{rms}I_{rms}\cos\phi=100\times10\times1=1000\ \text{W}$ .

Step 4. Check the individual reactive voltages, since this is where competitive papers often add a twist. $X_L=\omega_0 L=316.2\times0.4\approx126.5\ \Omega$ , so $V_L=I_{rms}X_L=10\times126.5\approx1265\ \text{V}$ — over twelve times the $100\ \text{V}$ source voltage. Similarly $X_C=1/(\omega_0 C)=1/(316.2\times25\times10^{-6})\approx126.5\ \Omega$ , giving $V_C\approx1265\ \text{V}$ as well, equal to $V_L$ as resonance requires. These huge individual voltages exist across $L$ and $C$ separately but cancel each other in the net phasor sum, leaving only $V_R=I_{rms}R=10\times10=100\ \text{V}$ , which correctly equals the full source voltage.

At resonance this circuit draws $I_{rms}=10\ \text{A}$ and delivers the maximum possible average power, $P_{avg}=1000\ \text{W}$ , even though the individual inductor and capacitor voltages momentarily reach roughly $1265\ \text{V}$ each — a genuine safety consideration in real resonant circuits. Timing note: in a competitive exam, recognise the shortcut immediately — "at resonance" always means $Z=R$ , $\cos\phi=1$ , and $P_{avg}=V_{rms}^2/R$ directly, without needing to compute $Z$ from the general formula at all; save the individual $V_L$ , $V_C$ check only if the question explicitly asks about voltage across the reactive elements.

Common traps

Confusing resonance with "resistance being zero" — at resonance $Z=R$ , the impedance is minimised down to the resistance value, not down to zero; a real circuit always has $R>0$ and hence a finite, though maximised, current.
Forgetting that resonance is a frequency-specific phenomenon — a given $L$ - $C$ pair resonates only at the one angular frequency $\omega_0=1/\sqrt{LC}$ , though the current stays reasonably large in a band of frequencies close to it, more so when $R$ is small.
Mixing up $\omega_0$ (angular frequency, rad/s) with $f_0$ (ordinary frequency, Hz) — always divide or multiply by $2\pi$ correctly when converting between them; a factor-of- $2\pi$ slip is one of the most common numerical errors in this topic.
Assuming that because $\phi=0$ and $\cos\phi=1$ at resonance, the inductor and capacitor "have no effect" at that frequency — they still carry large individual voltages ( $V_L=I_0X_L$ and $V_C=I_0X_C$ , each of which can exceed the source voltage many times over, as the worked example above shows), they simply cancel each other's net phase contribution, not their individual physical presence.
Believing that changing $R$ shifts the resonant frequency $\omega_0$ — it does not, since $\omega_0=1/\sqrt{LC}$ contains no $R$ at all; changing $R$ only reshapes how tall and sharp the resonance peak is, not where it sits on the frequency axis.

You should be able to…

Derive the resonance condition $X_L=X_C$ and the resonant angular frequency $\omega_0=1/\sqrt{LC}$ for a series LCR circuit, starting from the impedance formula.
Explain why impedance is minimised to $Z=R$ (never zero) and current is maximised to $I_{0,max}=V_0/R$ at resonance.
State that current and voltage are exactly in phase at resonance, with phase angle $\phi=0$ and power factor $\cos\phi=1$ .
Compute the resonant frequency, current, and average power for a given $L$ , $C$ , $R$ , and source voltage, as in the two worked examples.
Sketch and describe qualitatively how the current-versus-frequency resonance curve's height and sharpness depend on the value of $R$ , while its peak location depends only on $L$ and $C$ .
Connect the resonance phenomenon to a real application such as tuning a radio or television receiver to a single broadcast frequency.

Ask the tutor: if you would like to see how changing $R$ reshapes the resonance curve, or want a numerical walkthrough with your own choice of $L$ , $C$ , and $R$ , ask the AI tutor to work through it with you.

If Z equals R at resonance, does that mean the inductor and capacitor stop doing anything?

No — they are working just as hard as ever, individually. What cancels at resonance is only their combined phase effect on the total circuit, because the voltage phasor across the inductor and the voltage phasor across the capacitor point in exactly opposite directions and are equal in size, so they cancel each other in the vector sum. Each one, on its own, can carry a voltage far larger than the source voltage — the worked example above finds around 1265 V across each of L and C from just a 100 V source. So L and C are very active at resonance; they simply neutralise each other's net contribution to the total impedance, leaving only R to determine the current.

Why does resonant frequency not depend on resistance R at all?

Because the resonance condition comes purely from setting the two reactances equal, $X_L=X_C$ , and neither $X_L=\omega L$ nor $X_C=1/(\omega C)$ contains $R$ anywhere in their formulas — resistance never enters that equation, so it cannot appear in the answer $\omega_0=1/\sqrt{LC}$ . What R does control is a separate question: how sharply the current peaks around that fixed frequency. A small R lets the current rise very high and fall off quickly on either side of $\omega_0$ (a tall, narrow peak, useful for precise tuning); a large R damps the peak into a shorter, broader hump, but the peak's location on the frequency axis never moves.

How does resonance actually let a radio pick one station out of many?

A radio tuner is, at its heart, a series LCR circuit connected to the antenna, and every broadcast station arriving at the antenna is a different frequency signal all mixed together. Turning the tuning dial changes C (or sometimes L) in the tuner circuit, which shifts the resonant frequency $\omega_0=1/\sqrt{LC}$ . Only the incoming signal whose frequency matches the current $\omega_0$ meets the low-impedance, high-current condition and gets amplified strongly by the circuit; every other station's frequency sees a much higher impedance and is suppressed. A tuner built with small R gives a sharp resonance peak, meaning it responds to a very narrow band of frequencies — which is exactly what is needed to separate two stations broadcasting only a little apart in frequency.

Is the formal derivation of the LCR circuit's differential equation something I need to know for boards?

No. CBSE's 2025-26 syllabus explicitly treats the series LCR circuit through the phasor (vector-diagram) method only — you are expected to build up $Z=\sqrt{R^2+(X_L-X_C)^2}$ and the resonance condition from the phasor picture, exactly as done in this article, and not from solving the underlying differential equation of the circuit. If you see that differential-equation approach in an older reference book or online video, it is genuinely correct physics, but it sits above this syllabus's ceiling and is not required or expected in a CBSE answer.