Chapter 7 · Alternating Current

7.2 Phasors, the Series LCR Circuit and Resonance

In a series LCR circuit the voltages across $R$, $L$ and $C$ are out of step, so they add as phasors: the current is $V/Z$ with $Z=\sqrt{R^2+(X_L-X_C)^2}$. When $X_L=X_C$ the circuit resonates — the principle behind every radio tuner.

Turn the tuning knob of an old radio and you are turning a variable capacitor. Hundreds of stations broadcast at once, and the aerial picks up all of them; yet the speaker plays just one. The tuner is a coil and a capacitor in series with the small resistance of the wires, and it responds strongly to one frequency only — the frequency at which the reactance of the coil equals that of the capacitor. Move the capacitor plates and that frequency moves, and with it the station you hear. The same resonance selects the channel in a phone, a Wi-Fi card and a wireless charger, and must be avoided in power networks, where it can build up dangerously large voltages.

In Section 7.1 each component was alone. Now we join a resistor, an inductor and a capacitor in series to one source. The same current flows through all three, but their voltages are out of step with one another, so they cannot simply be added as numbers. Phasors solve the problem: we add the voltages as arrows, find the impedance $Z$ and the phase angle, and then study resonance and how sharp it is — measured by the bandwidth and the quality factor $Q$.

The Series LCR Circuit and Phasor Addition

A source $v=v_m\sin\omega t$ drives a resistor $R$, an inductor $L$ and a capacitor $C$ in series. There is only one path, so the same current $i=i_m\sin(\omega t+\phi)$ flows through every element; $\phi$ is the phase of the current relative to the source voltage, still to be found. From Section 7.1, the peak voltages are $v_{Rm}=i_mR$, $v_{Lm}=i_mX_L$, $v_{Cm}=i_mX_C$, and relative to the current:

Kirchhoff's loop rule holds at every instant: $v=v_R+v_L+v_C$. Instantaneous values are vertical projections of phasors, and the projection of a sum of arrows is the sum of their projections, so the phasor of $v$ is the arrow sum of the three voltage phasors. Draw the current phasor $\mathbf I$ as reference. $\mathbf V_R$ lies along it; $\mathbf V_L$ points $90^\circ$ ahead and $\mathbf V_C$ $90^\circ$ behind, so these two are exactly opposite and combine into one arrow of length $|v_{Lm}-v_{Cm}|$ perpendicular to $\mathbf V_R$. By Pythagoras, $$v_m^2=v_{Rm}^2+(v_{Cm}-v_{Lm})^2$$ $$=i_m^2\left[R^2+(X_C-X_L)^2\right].$$

📘 Definition 7.6 — Impedance
The impedance of a series LCR circuit is $$Z=\sqrt{R^2+(X_C-X_L)^2},$$ measured in ohms. It is the total opposition of the circuit to AC: $i_m=v_m/Z$ and $I=V/Z$. $R$, $|X_C-X_L|$ and $Z$ form a right-angled triangle, the impedance triangle, similar to the triangle of voltage phasors.
📐 Result 7.5 — Current and phase in a series LCR circuit

For $v=v_m\sin\omega t$ applied to $R$, $L$ and $C$ in series, the current is $i=i_m\sin(\omega t+\phi)$ with $$i_m=\frac{v_m}{Z},\qquad \tan\phi=\frac{X_C-X_L}{R}.$$ If $X_L\gt X_C$ then $\phi\lt0$: the current lags the voltage and the circuit is inductive. If $X_C\gt X_L$ then $\phi\gt0$: the current leads and the circuit is capacitive. If $X_L=X_C$, then $\phi=0$ and $Z=R$. The magnitude of $\phi$ lies between $0$ and $\pi/2$.

The same results hold for any two of the three elements: put $X_C=0$ for an RL circuit (no capacitor) and $X_L=0$ for an RC circuit (no inductor). Note that the rms voltages across the parts do not add up to the supply voltage — they add as arrows.

🧭 Problem-solving strategy — a series LCR circuit
  1. Find the reactances at the given frequency: $X_L=\omega L$, $X_C=1/\omega C$.
  2. Find $Z=\sqrt{R^2+(X_C-X_L)^2}$, then $I=V/Z$ (rms) or $i_m=v_m/Z$ (peak).
  3. Find the phase from $\tan\phi=(X_C-X_L)/R$ and say whether the current leads or lags.
  4. Element voltages: $V_R=IR$, $V_L=IX_L$, $V_C=IX_C$. Check: $\sqrt{V_R^2+(V_L-V_C)^2}=V$.
✏️ Example 1 — Impedance, current and phase
A $30\ \Omega$ resistor, a coil and a capacitor are in series across a $200$ V AC supply. At the supply frequency the coil's reactance is $80\ \Omega$ and the capacitor's is $40\ \Omega$. Find (a) the impedance, (b) the current, (c) the phase angle, (d) the rms voltage across each element.
(a) Impedance $Z=\sqrt{30^2+(40-80)^2}$ $=\sqrt{900+1600}=50\ \Omega$.
(b) Current $I=\dfrac{V}{Z}=\dfrac{200}{50}=4.0$ A.
(c) Phase $\tan\phi=\dfrac{40-80}{30}=-\dfrac43$, so $\phi=-53.1^\circ$: the current lags the voltage by $53^\circ$ (the circuit is inductive, $X_L\gt X_C$).
(d) Voltages $V_R=(4.0)(30)=120$ V, $V_L=(4.0)(80)=320$ V, $V_C=(4.0)(40)=160$ V.
Check $\sqrt{120^2+(320-160)^2}$ $=\sqrt{14400+25600}$ $=200$ V ✓. The coil has $320$ V across it — more than the supply. The arithmetic sum $120+320+160=600$ V means nothing.
✏️ Example 2 — From voltmeter readings
In a series LCR circuit an AC voltmeter reads $60$ V across the resistor, $150$ V across the inductor and $70$ V across the capacitor. (a) What is the supply voltage? (b) Does the current lead or lag, and by how much? (c) If $R=20\ \Omega$, find the current, the impedance and the two reactances.
(a) Supply $V=\sqrt{V_R^2+(V_L-V_C)^2}$ $=\sqrt{60^2+80^2}=100$ V.
(b) Phase $V_L\gt V_C$, so the circuit is inductive and the current lags. $\tan|\phi|=\dfrac{80}{60}$, $|\phi|=53.1^\circ$.
(c) Values $I=\dfrac{V_R}{R}=\dfrac{60}{20}=3.0$ A; $Z=\dfrac{100}{3.0}\approx33.3\ \Omega$; $X_L=\dfrac{150}{3.0}=50\ \Omega$; $X_C=\dfrac{70}{3.0}\approx23.3\ \Omega$.
Check $\sqrt{20^2+(50-23.3)^2}$ $=\sqrt{400+711}$ $\approx33.3\ \Omega$ ✓.
Interactive 1 — Phasor Diagram of a Series LCR Circuit

A $100$ V (rms) source drives $R$, $L$ and $C$ in series. On the left the voltage phasors are drawn head to tail: $\mathbf V_R$ (green) along the current phasor $\mathbf I$ (blue), $\mathbf V_L$ (purple) $90^\circ$ ahead, $\mathbf V_C$ (amber) $90^\circ$ behind; their sum is the source phasor $\mathbf V$ (red). On the right, $v(t)$ and $i(t)$ (shapes scaled to the same height) show the phase difference. Change $\nu$ and watch the triangle flip from inductive to capacitive.

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$X_L$, $X_C$ (Ω)—
$Z$ (Ω)—
current $I$ (rms)—
phase $\phi$ of $i$—
$V_R$, $V_L$, $V_C$ (rms, V)—

Resonance

As the frequency rises, $X_L=\omega L$ grows and $X_C=1/\omega C$ falls. At low frequency the capacitor dominates (capacitive circuit, current leading); at high frequency the coil dominates (inductive, current lagging). In between there is one frequency at which the two reactances are equal and cancel: the impedance is then as small as it can be, $Z=R$, and the current is at its largest.

📘 Definition 7.7 — Resonance in a series LCR circuit
A series LCR circuit is in resonance when $X_L=X_C$, that is $\omega_0L=\dfrac{1}{\omega_0C}$. The resonant angular frequency and resonant frequency are $$\omega_0=\frac{1}{\sqrt{LC}},\qquad \nu_0=\frac{1}{2\pi\sqrt{LC}}.$$
📐 Result 7.6 — Properties of series resonance

At $\omega=\omega_0$: (i) $Z=R$, its minimum value; (ii) the current has its maximum value $i_m=v_m/R$ (rms $V/R$); (iii) $\phi=0$ — the current is in phase with the source voltage; (iv) the voltages across $L$ and $C$ are equal in size and opposite in phase, so together they are zero at every instant, and each equals $I\omega_0L$, which can be much larger than the supply voltage. The resonant frequency depends only on $L$ and $C$, not on $R$. A circuit needs both an inductor and a capacitor to resonate.

Resonance is useful because the current is large only near $\omega_0$. A tuning circuit with an adjustable capacitor responds strongly to the station whose frequency equals its $\omega_0$ and weakly to the others. It can also be dangerous: in a power circuit near resonance the voltage across a capacitor or a coil can be many times the supply voltage and break down the insulation.

✏️ Example 3 — A circuit at resonance
A series circuit has $R=20\ \Omega$, $L=0.10$ H and $C=10\ \mu\text{F}$ and is connected to a $100$ V (rms) variable-frequency source. Find (a) the resonant angular frequency and frequency, (b) the current at resonance, (c) the rms voltages across $L$ and across $C$ at resonance, (d) the current at $50$ Hz.
(a) Resonance $\omega_0=\dfrac{1}{\sqrt{(0.10)(10\times10^{-6})}}$ $=\dfrac{1}{10^{-3}}=1000$ rad/s; $\ \nu_0=\dfrac{1000}{2\pi}\approx159$ Hz.
(b) Current $Z=R$, so $I_0=\dfrac{100}{20}=5.0$ A.
(c) L and C $X_L=X_C=\omega_0L=100\ \Omega$, so $V_L=V_C=(5.0)(100)=500$ V — five times the supply voltage, cancelling each other exactly.
(d) At 50 Hz $X_L=31.4\ \Omega$, $X_C=318.3\ \Omega$: $Z=\sqrt{20^2+(318.3-31.4)^2}$ $\approx287.6\ \Omega$, so $I\approx0.35$ A — the circuit is far from resonance and strongly capacitive.
Check Interactive 2 opens on this circuit: at $159$ Hz it reads $I=5.00$ A and $\phi\approx0$.
✏️ Example 4 — Tuning a medium-wave radio
The tuning circuit of a radio has a coil of inductance $200\ \mu\text{H}$ and a variable capacitor. (a) What capacitance tunes it to a station at $1000$ kHz? (b) The medium-wave band runs from about $540$ kHz to $1600$ kHz. Over what range must the capacitance vary?
Formula $\nu_0=\dfrac{1}{2\pi\sqrt{LC}}$ gives $C=\dfrac{1}{4\pi^2\nu_0^2L}$.
(a) 1000 kHz $C=\dfrac{1}{4\pi^2(10^6)^2(200\times10^{-6})}$ $\approx1.27\times10^{-10}$ F $=127$ pF.
(b) Range $C\propto1/\nu_0^2$. At $540$ kHz: $C=127\times\left(\dfrac{1000}{540}\right)^2\approx434$ pF; at $1600$ kHz: $C=127\times\left(\dfrac{1000}{1600}\right)^2\approx49.5$ pF. So $C$ must vary from about $50$ pF to $430$ pF.
Lesson A $3$-to-$1$ range of frequency needs a $9$-to-$1$ range of capacitance, because $\nu_0\propto1/\sqrt C$.

Sharpness of Resonance: Bandwidth and Quality Factor

How selective is a resonant circuit? Away from $\omega_0$ the current falls, and it falls faster when $R$ is small. A convenient measure uses the half-power frequencies $\omega_1\lt\omega_0\lt\omega_2$ at which the current amplitude has fallen to $1/\sqrt2$ of its peak value. Since the power in $R$ is proportional to $I^2$, the power there is half its maximum. At these frequencies $Z=\sqrt2R$, so $|X_L-X_C|=R$: $$\omega L-\frac{1}{\omega C}=\pm R\quad\Longrightarrow\quad L\omega^2\mp R\omega-\frac1C=0 .$$ Taking the positive root of each quadratic, $$\omega_{2,1}=\frac{\pm R+\sqrt{R^2+4L/C}}{2L},$$ and the difference is exactly $\omega_2-\omega_1=R/L$.

📘 Definition 7.8 — Bandwidth and quality factor
The bandwidth of a series LCR circuit is $\omega_2-\omega_1=2\Delta\omega$, the width of the resonance curve between its half-power points. The quality factor is $$Q=\frac{\omega_0}{2\Delta\omega}=\frac{\omega_0L}{R}.$$ A large $Q$ means a narrow, sharp resonance.
📐 Result 7.7 — Sharpness of resonance

For a series LCR circuit: $$2\Delta\omega=\frac{R}{L},$$ $$Q=\frac{\omega_0L}{R}=\frac{1}{\omega_0CR}=\frac1R\sqrt{\frac LC}.$$ In hertz the bandwidth is $\nu_2-\nu_1=\dfrac{R}{2\pi L}$. At resonance the voltage across $L$ (or across $C$) is $Q$ times the supply voltage, so $Q$ is also the "voltage magnification". Reducing $R$ leaves $\omega_0$ unchanged but raises the peak current and narrows the curve.

✏️ Example 5 — Half-power points of Example 3
For the circuit of Example 3 ($R=20\ \Omega$, $L=0.10$ H, $C=10\ \mu\text{F}$, $100$ V), find (a) the half-power angular frequencies, (b) the bandwidth and the quality factor, (c) the current and phase at the upper half-power point, (d) $Q$ if $R$ were reduced to $5.0\ \Omega$.
(a) Roots $\sqrt{R^2+4L/C}=\sqrt{400+40\,000}\approx201.0$; $\ \omega_2=\dfrac{20+201.0}{0.20}\approx1105$ rad/s; $\ \omega_1=\dfrac{-20+201.0}{0.20}\approx905$ rad/s.
(b) Width and Q $\omega_2-\omega_1=200$ rad/s $=R/L$ ✓ (about $31.8$ Hz). $\ Q=\dfrac{\omega_0L}{R}=\dfrac{(1000)(0.10)}{20}=5.0$, and $\dfrac{\omega_0}{2\Delta\omega}=\dfrac{1000}{200}=5.0$ ✓.
(c) At ω₂ $Z=\sqrt2R\approx28.3\ \Omega$, $I=\dfrac{100}{28.3}\approx3.54$ A $=\dfrac{5.0}{\sqrt2}$; $\ X_L-X_C=R$, so $\tan\phi=-1$: the current lags by $45^\circ$. (At $\omega_1$ it leads by $45^\circ$.)
(d) Smaller R $Q=\dfrac{100}{5.0}=20$; the bandwidth shrinks to $50$ rad/s and the peak current rises to $20$ A.
Note $\omega_1\omega_2=905\times1105\approx1.0\times10^6=\omega_0^2$: the resonant frequency is the geometric mean of the half-power frequencies.
✏️ Example 6 — Same resonant frequency, different sharpness
Circuit A has $L=0.10$ H, $C=10\ \mu\text{F}$; circuit B has $L=1.0$ H, $C=1.0\ \mu\text{F}$. Each has $R=20\ \Omega$. Compare their resonant frequencies, quality factors and bandwidths. Which is better for picking out one signal from its neighbours?
Resonance Both have $LC=1.0\times10^{-6}\ \text{s}^2$, so both resonate at $\omega_0=1000$ rad/s ($159$ Hz).
Q A: $Q=\dfrac1{20}\sqrt{\dfrac{0.10}{10^{-5}}}=\dfrac{100}{20}=5.0$. B: $Q=\dfrac1{20}\sqrt{\dfrac{1.0}{10^{-6}}}=\dfrac{1000}{20}=50$.
Bandwidth A: $R/L=200$ rad/s. B: $R/L=20$ rad/s.
Verdict B is ten times sharper and far more selective (though both carry the same peak current, $V/R$). Interactive 2 lets you switch between A and B.
⚠️ Common mistakes with series LCR circuits

1. Adding rms voltages as numbers. $V\ne V_R+V_L+V_C$. Use $V=\sqrt{V_R^2+(V_L-V_C)^2}$.

2. Losing the sign of the phase. $X_L\gt X_C$: inductive, current lags. $X_C\gt X_L$: capacitive, current leads. Decide from the reactances, not from memory.

3. Thinking $R$ changes the resonant frequency. $\omega_0=1/\sqrt{LC}$ only; $R$ sets the height and width of the peak.

4. Mixing rad/s and Hz. $\omega_0=1/\sqrt{LC}$ is in rad/s; divide by $2\pi$ for hertz. The bandwidth $R/L$ is in rad/s; $R/2\pi L$ in Hz.

5. Assuming no element can have more than the supply voltage. Near resonance $V_L$ and $V_C$ are each about $Q$ times $V$.

Interactive 2 — The Resonance Explorer

A $100$ V (rms) source of variable frequency drives a series LCR circuit. The thick curve is the rms current against frequency for your $R$; the faint curves are for $R/2$ and $2R$. The dashed line marks $I_0/\sqrt2$ and the bracket the bandwidth between the half-power points. Switch the graph to the phase angle, or choose circuit B (same $\omega_0$, ten times the $Q$). Play sweeps the frequency through resonance.

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$\nu_0$ and $Q$—
half-power $\nu_1$, $\nu_2$—
bandwidth $\nu_2-\nu_1$—
at $\nu$: $Z$ and $I$—
at $\nu$: phase $\phi$—
🚀 Beyond the board syllabus (JEE Advanced) — LC oscillations

In JEE Advanced 2026 only (it lists LC circuits); not in the CBSE, JEE Main or NEET 2026 syllabi, which list only the series LCR circuit and resonance.

Charge a capacitor $C$ to $q_m$ and connect it across an inductor $L$ with no resistance and no source. The capacitor starts to discharge through the coil; the coil's back emf slows the growth of the current, and once the capacitor is empty the coil keeps the current going and charges the capacitor the other way. The charge oscillates. With $q$ the charge and $i=dq/dt$, the loop rule $\dfrac qC+L\dfrac{di}{dt}=0$ gives $$\frac{d^2q}{dt^2}=-\frac{1}{LC}\,q,$$ the equation of SHM (Class 11 Physics, Chapter 13) with $\omega_0=1/\sqrt{LC}$ — the same frequency as series resonance. So $q=q_m\cos\omega_0t$ and $i=-q_m\omega_0\sin\omega_0t$, with peak current $i_m=q_m\omega_0$.

Energy. The energy passes back and forth between the electric field of the capacitor and the magnetic field of the coil, and the total stays constant: $$\frac{q^2}{2C}+\frac12Li^2=\frac{q_m^2}{2C}.$$ It is the electrical twin of a mass on a spring: $q\leftrightarrow x$, $L\leftrightarrow m$ (inertia), $1/C\leftrightarrow k$ (stiffness). Any real circuit has some resistance, which turns energy into heat and makes the oscillations die away. Example: $C=10\ \mu\text{F}$ charged to $10$ V ($q_m=100\ \mu\text{C}$) and $L=0.10$ H oscillate at $\omega_0=1000$ rad/s ($159$ Hz) with $i_m=0.10$ A; the energy $\tfrac12CV^2=0.50$ mJ equals $\tfrac12Li_m^2$ ✓.

Exercises

I
Impedance and phase.
  • 1 $R=40\ \Omega$, $X_L=70\ \Omega$ and $X_C=40\ \Omega$ are in series across $250$ V. Find $Z$, $I$ and the phase angle; does the current lead or lag?
  • 2 A $6.0\ \Omega$ resistor and an inductor of reactance $8.0\ \Omega$ are in series across $10$ V. Find the current and its phase.
  • 3 A $100\ \Omega$ resistor and a $20\ \mu\text{F}$ capacitor are in series across a $220$ V, $50$ Hz supply. Find the impedance, the current and the phase angle.
  • 4 In a series LCR circuit the rms voltages across $R$, $L$ and $C$ are $80$ V, $100$ V and $40$ V. Find the supply voltage.
  • 5 Explain how the rms voltage across the capacitor of a series LCR circuit can be larger than the rms supply voltage.
  • 6 A coil draws $4.0$ A from a $12$ V battery and $2.4$ A from a $12$ V (rms), $50$ Hz supply. Find its resistance, its reactance and its inductance.
  • 7 Is a series circuit with $L=0.10$ H and $C=10\ \mu\text{F}$ inductive or capacitive at $50$ Hz? Does the current lead or lag?
  • 8 Find the impedance of $R=20\ \Omega$, $L=0.10$ H and $C=10\ \mu\text{F}$ in series at $\omega=2000$ rad/s.
1 $Z=\sqrt{40^2+30^2}=50\ \Omega$; $I=5.0$ A; $\tan\phi=\dfrac{40-70}{40}=-0.75$, $\phi=-36.9^\circ$: the current lags by $37^\circ$.
2 $Z=\sqrt{36+64}=10\ \Omega$; $I=1.0$ A, lagging by $\tan^{-1}\dfrac{8}{6}=53.1^\circ$.
3 $X_C=\dfrac{1}{2\pi(50)(20\times10^{-6})}\approx159\ \Omega$; $Z=\sqrt{100^2+159^2}\approx188\ \Omega$; $I\approx1.17$ A, leading by $\tan^{-1}1.59\approx57.9^\circ$.
4 $V=\sqrt{80^2+(100-40)^2}=100$ V.
5 $V_C=IX_C$ and $V=IZ$. When $X_L$ nearly cancels $X_C$, $Z$ can be much smaller than $X_C$, so $V_C\gt V$. The large voltages across $L$ and $C$ are opposite in phase and almost cancel.
6 D.c.: $R=\dfrac{12}{4.0}=3.0\ \Omega$. A.c.: $Z=\dfrac{12}{2.4}=5.0\ \Omega$, so $X_L=\sqrt{25-9}=4.0\ \Omega$ and $L=\dfrac{4.0}{100\pi}\approx12.7$ mH.
7 $X_L=31.4\ \Omega\lt X_C=318\ \Omega$: capacitive, the current leads.
8 $X_L=200\ \Omega$, $X_C=\dfrac{1}{(2000)(10^{-5})}=50\ \Omega$; $Z=\sqrt{20^2+150^2}\approx151\ \Omega$.
II
Resonance.
  • 1 Find the resonant angular frequency and frequency of a series circuit with $L=0.50$ H and $C=20\ \mu\text{F}$.
  • 2 The circuit of part 1 has $R=40\ \Omega$ and is connected to a $230$ V source at its resonant frequency. Find the current and the rms voltage across the inductor.
  • 3 What capacitance must be put in series with a $0.20$ H inductor for resonance at $50$ Hz?
  • 4 The resistance of a series LCR circuit is doubled. What happens to (a) the resonant frequency, (b) the current at resonance?
  • 5 The inductance of a resonant circuit is doubled. How must the capacitance change to keep the same resonant frequency?
  • 6 What is the phase angle between the current and the source voltage at resonance? What is the impedance?
  • 7 Below its resonant frequency, is a series LCR circuit inductive or capacitive? Does the current lead or lag?
  • 8 An FM receiver uses a $0.25\ \mu\text{H}$ coil. What capacitance tunes it to $100$ MHz?
1 $\omega_0=\dfrac{1}{\sqrt{(0.50)(20\times10^{-6})}}$ $=\dfrac{1}{\sqrt{10^{-5}}}$ $\approx316$ rad/s; $\nu_0\approx50.3$ Hz.
2 $I=\dfrac{230}{40}=5.75$ A; $X_L=\omega_0L\approx158\ \Omega$; $V_L=(5.75)(158.1)\approx909$ V.
3 $C=\dfrac{1}{4\pi^2\nu^2L}$ $=\dfrac{1}{4\pi^2(50)^2(0.20)}$ $\approx5.07\times10^{-5}$ F $\approx50.7\ \mu\text{F}$.
4 (a) Unchanged ($1/\sqrt{LC}$). (b) Halved ($V/R$).
5 $LC$ must stay the same, so $C$ must be halved.
6 Zero — the current is in phase with the voltage; $Z=R$.
7 Below $\omega_0$, $X_C\gt X_L$: capacitive, the current leads.
8 $C=\dfrac{1}{4\pi^2(10^8)^2(0.25\times10^{-6})}$ $\approx1.0\times10^{-11}$ F $=10$ pF.
III
Bandwidth and quality factor.
  • 1 Find $Q$ for $L=0.50$ H, $C=20\ \mu\text{F}$, $R=10\ \Omega$.
  • 2 Find the bandwidth of the circuit in part 1 in rad/s and in Hz.
  • 3 Find the half-power angular frequencies of $L=0.10$ H, $C=10\ \mu\text{F}$, $R=50\ \Omega$, and check that their difference is $R/L$.
  • 4 A circuit with $Q=50$ is driven at resonance by a $2.0$ V (rms) source. What is the rms voltage across its capacitor?
  • 5 Two tuning circuits have the same resonant frequency, with $Q=10$ and $Q=100$. Which is more selective? Which has the wider bandwidth, and by what factor?
  • 6 Show that $Q=\dfrac{\omega_0L}{R}$ can be written as $\dfrac1R\sqrt{\dfrac LC}$.
  • 7 A series circuit resonates at $1000$ rad/s with a bandwidth of $50$ rad/s. Find $Q$; if $L=0.10$ H, find $R$ and $C$.
1 $\omega_0\approx316.2$ rad/s; $Q=\dfrac{(316.2)(0.50)}{10}\approx15.8$.
2 $R/L=20$ rad/s; $\dfrac{20}{2\pi}\approx3.2$ Hz.
3 $\sqrt{R^2+4L/C}$ $=\sqrt{2500+40\,000}$ $\approx206.2$; $\omega_2=\dfrac{50+206.2}{0.20}\approx1281$ rad/s, $\omega_1=\dfrac{-50+206.2}{0.20}\approx781$ rad/s; difference $500$ rad/s $=\dfrac{50}{0.10}$ ✓.
4 $V_C=QV=(50)(2.0)=100$ V.
5 $Q=100$ is more selective. $2\Delta\omega=\omega_0/Q$, so the $Q=10$ circuit has $10$ times the bandwidth.
6 $\omega_0L=\dfrac{L}{\sqrt{LC}}=\sqrt{\dfrac LC}$, so $Q=\dfrac{\omega_0L}{R}=\dfrac1R\sqrt{\dfrac LC}$.
7 $Q=\dfrac{1000}{50}=20$. $R=L\times(\text{bandwidth})$ $=(0.10)(50)=5.0\ \Omega$; $C=\dfrac{1}{\omega_0^2L}$ $=\dfrac{1}{(10^6)(0.10)}$ $=10\ \mu\text{F}$.
IV
Harder problems.
  • 1 $R=10\ \Omega$, $L=0.10$ H and $C=100\ \mu\text{F}$ are in series across a $100$ V (rms) source with $\omega=500$ rad/s. Find the current and its phase. Is $\omega$ above or below resonance?
  • 2 For $R=20\ \Omega$, $L=0.10$ H, $C=10\ \mu\text{F}$, at what angular frequency does the current lag the voltage by $45^\circ$?
  • 3 In a series LCR circuit across a $220$ V supply, a voltmeter reads $220$ V across the resistor alone. What does it read across the inductor and capacitor taken together? What can you say about the frequency?
  • 4 Show that the two frequencies at which a series LCR circuit carries the same current satisfy $\omega_1\omega_2=\omega_0^2$. Check with Example 5.
  • 5 A $100\ \Omega$ resistor, an inductor and a capacitor are in series across $200$ V. With the capacitor short-circuited, the current lags the voltage by $60^\circ$; with the inductor short-circuited instead, the current leads by $60^\circ$. Find the current in each of these cases and in the full circuit.
  • 6 In Example 3 at resonance, find the largest energy stored in the inductor and the largest energy stored in the capacitor. Comment.
1 $X_L=50\ \Omega$, $X_C=\dfrac{1}{(500)(10^{-4})}=20\ \Omega$; $Z=\sqrt{100+900}\approx31.6\ \Omega$; $I\approx3.16$ A, lagging by $\tan^{-1}3\approx71.6^\circ$. $\omega_0=\dfrac{1}{\sqrt{10^{-5}}}\approx316$ rad/s, so $500$ rad/s is above resonance (inductive) ✓.
2 A lag of $45^\circ$ needs $X_L-X_C=R$ — the upper half-power point of Example 5: $\omega\approx1105$ rad/s ($176$ Hz).
3 $V^2=V_R^2+(V_L-V_C)^2$ with $V_R=V$ gives $V_L-V_C=0$: the voltmeter reads $0$ across $L$ and $C$ together. The circuit is at resonance.
4 Equal currents mean equal $|X_L-X_C|$ with opposite signs: $\omega_2L-\dfrac{1}{\omega_2C}=\dfrac{1}{\omega_1C}-\omega_1L$, so $L(\omega_1+\omega_2)=\dfrac{\omega_1+\omega_2}{\omega_1\omega_2C}$, giving $\omega_1\omega_2=\dfrac1{LC}=\omega_0^2$. Example 5: $(905)(1105)\approx1.0\times10^6$ ✓.
5 $\tan60^\circ=\dfrac{X_L}{R}=\dfrac{X_C}{R}$, so $X_L=X_C=173\ \Omega$. Either case: $Z=\dfrac{R}{\cos60^\circ}=200\ \Omega$, $I=1.0$ A. Full circuit: $X_L=X_C$ — resonance — so $Z=R$ and $I=2.0$ A.
6 $i_m=5.0\sqrt2\approx7.07$ A: $U_L=\tfrac12Li_m^2$ $=\tfrac12(0.10)(50)=2.5$ J. $v_{Cm}=500\sqrt2\approx707$ V: $U_C=\tfrac12Cv_{Cm}^2$ $=\tfrac12(10^{-5})(5.0\times10^5)$ $=2.5$ J. They are equal: at resonance energy sloshes back and forth between the coil and the capacitor, and the source only makes up the loss in $R$.
🎯 Exam corner — series LCR and resonance BoardJEE MainJEE AdvancedNEET

E1. Board · 5 marks A source $v=v_m\sin\omega t$ is connected to $R$, $L$ and $C$ in series. Using a phasor diagram, derive the impedance of the circuit and the phase angle between current and voltage. State the condition for resonance, and find the resonant frequency for $L=40$ mH and $C=25\ \mu\text{F}$.

Phasors Take the current phasor $\mathbf I$ as reference. $\mathbf V_R$ (length $i_mR$) is along $\mathbf I$; $\mathbf V_L$ ($i_mX_L$) is $90^\circ$ ahead; $\mathbf V_C$ ($i_mX_C$) is $90^\circ$ behind. $\mathbf V_L+\mathbf V_C$ has length $i_m|X_C-X_L|$ perpendicular to $\mathbf V_R$.
Impedance $\mathbf V=\mathbf V_R+\mathbf V_L+\mathbf V_C$: $v_m^2=i_m^2[R^2+(X_C-X_L)^2]$, so $i_m=\dfrac{v_m}{Z}$ with $Z=\sqrt{R^2+(X_C-X_L)^2}$.
Phase From the triangle, $\tan\phi=\dfrac{X_C-X_L}{R}$, with $i=i_m\sin(\omega t+\phi)$.
Resonance $X_L=X_C$, i.e. $\omega_0=\dfrac1{\sqrt{LC}}$; $Z=R$, current maximum and in phase. Here $LC=(0.040)(25\times10^{-6})$ $=1.0\times10^{-6}\ \text{s}^2$, so $\omega_0=1000$ rad/s and $\nu_0=\dfrac{1000}{2\pi}\approx159$ Hz. Exam tip: label all four phasors and the angle $\phi$ — the diagram carries marks.

E2. Board · Case study The tuning circuit of a medium-wave receiver has a coil of inductance $250\ \mu\text{H}$ and resistance $15\ \Omega$ in series with a variable capacitor. (i) What capacitance tunes it to a station at $900$ kHz? (2 marks) (ii) What is the quality factor at this frequency? (1 mark) (iii) Find the bandwidth in kHz and explain why a smaller coil resistance would help. (1 mark)

(i) $C=\dfrac{1}{4\pi^2\nu_0^2L}$ $=\dfrac{1}{4\pi^2(9.0\times10^5)^2(2.5\times10^{-4})}$ $\approx1.25\times10^{-10}$ F $=125$ pF.
(ii) $\omega_0L=2\pi(9.0\times10^5)(2.5\times10^{-4})$ $\approx1414\ \Omega$, so $Q=\dfrac{\omega_0L}{R}$ $=\dfrac{1414}{15}\approx94$.
(iii) $\nu_2-\nu_1=\dfrac{R}{2\pi L}$ $=\dfrac{15}{2\pi(2.5\times10^{-4})}$ $\approx9.5\times10^3$ Hz $\approx9.5$ kHz $\left(=\dfrac{\nu_0}{Q}\right)$. A smaller $R$ raises $Q$ and narrows the band, so neighbouring stations are rejected better. Exam tip: in case studies keep $\omega$ and $\nu$ apart.

E3. JEE Main In a series LCR circuit the rms voltages across $R$, $L$ and $C$ are $40$ V, $80$ V and $50$ V. The rms voltage of the source is (A) $170$ V (B) $50$ V (C) $70$ V (D) $10$ V.

Phasors $V=\sqrt{V_R^2+(V_L-V_C)^2}$ $=\sqrt{40^2+30^2}=50$ V.
Answer (B). Exam tip: (A) is the arithmetic sum — always wrong for out-of-phase voltages.

E4. NEET A series LCR circuit has $L=\dfrac1\pi$ H, $C=\dfrac{1}{\pi}\ \mu\text{F}$ and $R=100\ \Omega$. Its resonant frequency is (A) $50$ Hz (B) $250$ Hz (C) $500$ Hz (D) $1000$ Hz.

Formula $\sqrt{LC}=\sqrt{\dfrac{10^{-6}}{\pi^2}}=\dfrac{10^{-3}}{\pi}$, so $\nu_0=\dfrac{1}{2\pi\sqrt{LC}}=\dfrac{\pi}{2\pi\times10^{-3}}=500$ Hz.
Answer (C). Exam tip: $R$ is a distractor — it does not affect $\nu_0$.

E5. JEE Advanced (One or more correct.) A series circuit with $L=0.10$ H, $C=10\ \mu\text{F}$ and $R=10\ \Omega$ is connected to a $50$ V (rms) source of variable frequency. Which statements are correct? (A) The resonant angular frequency is $1000$ rad/s. (B) At resonance the rms voltage across the capacitor is $500$ V. (C) The bandwidth is $100$ rad/s. (D) The quality factor is $5$.

(A) $\omega_0=1/\sqrt{10^{-6}}=1000$ rad/s — true.
(B) $I_0=\dfrac{50}{10}=5.0$ A, $X_C=100\ \Omega$, $V_C=500$ V — true.
(C), (D) $R/L=100$ rad/s — (C) true. $Q=\dfrac{\omega_0L}{R}=\dfrac{100}{10}=10$ — (D) false (and $V_C=QV=500$ V agrees with (B)).
Answer (A), (B), (C). Exam tip: $Q=\omega_0/(2\Delta\omega)=1000/100=10$ is a quick consistency check.

E6. JEE Advanced A $4.0\ \mu\text{F}$ capacitor charged to $50$ V is connected across a $10$ mH inductor of negligible resistance. The maximum current in the circuit is (A) $0.50$ A (B) $1.0$ A (C) $2.0$ A (D) $0.10$ A.

Concept LC oscillations are beyond the NCERT syllabus (see the 🚀 box). Energy is conserved: $\tfrac12CV_0^2=\tfrac12Li_m^2$.
Numbers $i_m=V_0\sqrt{\dfrac CL}=50\sqrt{\dfrac{4.0\times10^{-6}}{1.0\times10^{-2}}}$ $=50(0.020)=1.0$ A. (Equivalently $q_m\omega_0=(200\ \mu\text{C})(5000\ \text{rad/s})$.)
Answer (B). Exam tip: the energy route avoids having to find $\omega_0$ at all.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. A wireless phone charger transfers energy between two coils using a resonant circuit. How sharp should that resonance be — and what happens to the delivered current if the frequency or the component values drift slightly?

Why it matters. Wireless charging pads, electric-toothbrush chargers, implanted medical devices (such as heart pumps and cochlear implants) and experimental road charging for electric buses all use resonant coils. A high quality factor gives a large current from a small drive, but a very sharp resonance is easily lost if the coil warms up, a phone is placed slightly off-centre, or the capacitor ages. Designers in Korea, Germany, the USA and India balance these effects.

What to do.

  1. Use these illustrative values for the transmitter circuit: $L=10\ \mu\text{H}$, $C=129$ nF, driven by a $5.0$ V (rms) source. Find $\nu_0$.
  2. In a spreadsheet, compute $Z$ and $I$ from $100$ kHz to $180$ kHz in steps of $1$ kHz for $R=0.20\ \Omega$, $0.50\ \Omega$ and $1.0\ \Omega$. Plot the three resonance curves.
  3. For each $R$ find $Q$ and the bandwidth (from the graph and from $R/2\pi L$).
  4. Suppose the capacitance drifts by $2\%$. By what fraction does the current at the original drive frequency fall for each $R$?

Tools: Spreadsheet; Interactive 2 for checking the shape of the curves.

Hand in: Your three curves, a table of $Q$ and bandwidth, the effect of the $2\%$ drift, and a one-paragraph recommendation: high $Q$ or moderate $Q$?

Going further. Real chargers adjust their frequency automatically to stay at resonance. Find out how a circuit can "tell" that it is at resonance using the phase between $v$ and $i$ (Result 7.5).

Instructor note: $\nu_0=\dfrac{1}{2\pi\sqrt{(10^{-5})(1.29\times10^{-7})}}\approx140$ kHz; $I_0=V/R=25$ A, $10$ A, $5.0$ A; $Q\approx44$, $18$, $8.8$; bandwidth $R/2\pi L\approx3.2$, $8.0$, $16$ kHz. A $2\%$ rise in $C$ moves $\nu_0$ down by about $1\%$ ($1.4$ kHz); at the original frequency the reactances then differ by about $2\%$ of $X_L$ (about $0.17\ \Omega$), and the current falls to roughly $76\%$, $95\%$ and $99\%$ of its peak for the three resistances. A high-$Q$ circuit therefore needs active frequency tracking.

🧪 Self-check quiz

0/5 correct

Q1. In a series circuit $R=3\ \Omega$, $X_L=8\ \Omega$ and $X_C=4\ \Omega$. The impedance is

Q2. At resonance in a series LCR circuit,

Q3. The resistance of a series LCR circuit is increased. The resonance curve

Q4. A series LCR circuit is driven below its resonant frequency. Then

Q5. The quality factor of a series circuit with $L=0.10$ H, $C=10\ \mu\text{F}$ and $R=10\ \Omega$ is