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:
- $v_R$ is in phase with $i$;
- $v_L$ leads $i$ by $\pi/2$;
- $v_C$ lags $i$ by $\pi/2$.
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, $$v_m^2=v_{Rm}^2+(v_{Cm}-v_{Lm})^2$$ $$=i_m^2\left[R^2+(X_C-X_L)^2\right].$$
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.
- Find the reactances at the given frequency: $X_L=\omega L$, $X_C=1/\omega C$.
- Find $Z=\sqrt{R^2+(X_C-X_L)^2}$, then $I=V/Z$ (rms) or $i_m=v_m/Z$ (peak).
- Find the phase from $\tan\phi=(X_C-X_L)/R$ and say whether the current leads or lags.
- Element voltages: $V_R=IR$, $V_L=IX_L$, $V_C=IX_C$. Check: $\sqrt{V_R^2+(V_L-V_C)^2}=V$.
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.
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.
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 $\omega_2-\omega_1=R/L$.
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.
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 $V$.
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
- 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 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 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 $R=10\ \Omega$, $L=0.10$ 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.
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 $C=25\ \mu\text{F}$.
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)
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.
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.
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 $5$.
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.
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.
- 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 $\nu_0$.
- 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.
- For each $R$ find $Q$ and the bandwidth (from the graph and from $R/2\pi L$).
- 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).
🧪 Self-check quiz
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