Alternating Current

Lessons

Senior 6 · Physics

Alternating Current

4 lessons

Lesson 3 of 4

Series RL, RC, LC and RLC Circuits: Impedance and Resonance

25 min

Voltages that do not add up

A resistor and a capacitor are connected in series to a 12 V rms, 50 Hz supply. One voltmeter across the resistor reads 7.2 V; another across the capacitor reads 9.6 V. Their sum, 16.8 V, is more than the supply. Has Kirchhoff's voltage law failed? No. The law holds at every instant, but the two voltages reach their peaks at different instants, so their rms values cannot simply be added. In Lesson 1 you added phase-shifted voltages as phasors; in Lesson 2 you found the phase of each component. Put the two together and you can analyse any series AC circuit, including the tuned circuit that lets a radio pick out one station.

The method for a series circuit

Figure 3.1: A series RLC circuit with a voltmeter across each component.

In a series circuit the same current flows through every component, so the current is the natural reference phasor. Then, from Lesson 2:

$V_{R} = I R$ is drawn along $I$ ;
$V_{L} = I X_{L}$ is drawn 90° ahead of $I$ (up);
$V_{C} = I X_{C}$ is drawn 90° behind $I$ (down).

The supply voltage is the phasor sum. Because $V_{L}$ and $V_{C}$ point in opposite directions, they partly cancel:

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

Figure 3.2: (a) The voltage phasors of a series RLC circuit with $X_{L} > X_{C}$ . (b) Dividing every side by $I$ gives the impedance triangle.

The ratio $\frac{V}{I}$ is the circuit's impedance, and the angle by which the supply voltage leads the current is the phase angle $\phi$ :

Z = \sqrt{R^2 + (X_L - X_C)^2}, \qquad \tan\phi = \frac{X_L - X_C}{R}, \qquad \cos\phi = \frac{R}{Z}

If $X_{L} > X_{C}$ the circuit is inductive ( $\phi > 0$ , voltage leads current); if $X_{L} < X_{C}$ it is capacitive ( $\phi < 0$ , current leads voltage). The simpler circuits are special cases:

Circuit	Impedance $Z$	Phase of $V$ relative to $I$
series RL	$\sqrt{R^2 + X_L^2}$	leads by $\tan^{-1}\frac{X_L}{R}$ , between 0° and 90°
series RC	$\sqrt{R^2 + X_C^2}$	lags by $\tan^{-1}\frac{X_C}{R}$ , between 0° and 90°
series LC (no resistance)	$\left\|X_L - X_C\right\|$	leads by 90° if $X_{L} > X_{C}$ ; lags by 90° if $X_{L} < X_{C}$
series RLC	$\sqrt{R^2 + (X_L - X_C)^2}$	$\tan\phi = \frac{X_L - X_C}{R}$

Definition

The impedance $Z$ of an AC circuit is the ratio of the rms voltage across it to the rms current through it, $Z = \frac{V_{\text{rms}}}{I_{\text{rms}}}$ , measured in ohms. It combines resistance and reactance as phasors, not as a simple sum. The phase angle $\phi$ is the angle by which the voltage leads the current. Example: $R = 12\,\mathrm{\Omega}$ in series with $X_L = 5.0\,\mathrm{\Omega}$ gives $Z = 13\,\mathrm{\Omega}$ and $\phi = 22.6^\circ$ .

Worked example 1: the voltmeter puzzle

In the opening circuit, $R = 150\,\mathrm{\Omega}$ and $C = 15.9\,\mathrm{\mu F}$ , on 12 V rms, 50 Hz. Find the impedance, the current, both voltmeter readings and the phase angle.

Solution

$X_C = \frac{1}{2\pi \times 50 \times 15.9 \times 10^{-6}} = 200\,\mathrm{\Omega}$ . $Z = \sqrt{150^2 + 200^2} = 250\,\mathrm{\Omega}$ , so $I = \frac{12}{250} = 0.048\,\mathrm{A}$ .

$V_R = IR = 0.048 \times 150 = 7.2\,\mathrm{V}$ and $V_C = IX_C = 0.048 \times 200 = 9.6\,\mathrm{V}$ .

Figure 3.3: The phasors of Worked example 1, drawn to scale.

$\tan\phi = \frac{-X_C}{R} = -\frac{200}{150}$ , so $\phi = -53.1^\circ$ : the supply voltage lags the current by 53.1°. Check: $\sqrt{7.2^2 + 9.6^2} = \sqrt{51.84 + 92.16} = \sqrt{144} = 12\,\mathrm{V}$ . The rms voltages add as phasors, and Kirchhoff's law is safe.

Worked example 2: a real coil

A coil takes 2.0 A from a 12 V DC supply and 1.5 A from a 12 V rms, 50 Hz supply. Find its resistance, its inductance and the phase angle on AC.

Solution

On DC there is no reactance ( $f = 0$ ): $R = \frac{12}{2.0} = 6.0\,\mathrm{\Omega}$ .

On AC the coil is a series RL circuit: $Z = \frac{12}{1.5} = 8.0\,\mathrm{\Omega}$ , so $X_L = \sqrt{Z^2 - R^2} = \sqrt{64 - 36} = 5.29\,\mathrm{\Omega}$ .

$L = \frac{X_L}{2\pi f} = \frac{5.29}{2\pi \times 50} = 1.68 \times 10^{-2}\,\mathrm{H} = 16.8\,\mathrm{mH}$ .

$\tan\phi = \frac{5.29}{6.0}$ , so $\phi = 41.4^\circ$ : the voltage leads the current by 41.4°.

Worked example 3: a series RLC circuit

A 40 Ω resistor, a 0.30 H inductor of negligible resistance and a 20 µF capacitor are in series across a 24 V rms, 50 Hz supply. Find (a) the impedance and current, (b) the voltage across each component, (c) the phase angle.

Solution

(a) $X_L = 2\pi \times 50 \times 0.30 = 94.2\,\mathrm{\Omega}$ ; $X_C = \frac{1}{2\pi \times 50 \times 20 \times 10^{-6}} = 159.2\,\mathrm{\Omega}$ .

Z = \sqrt{40^2 + (94.2 - 159.2)^2} = \sqrt{1600 + 4213} = 76.2\,\mathrm{\Omega}, \qquad I = \frac{24}{76.2} = 0.315\,\mathrm{A}

(b) $V_R = 0.315 \times 40 = 12.6\,\mathrm{V}$ ; $V_L = 0.315 \times 94.2 = 29.7\,\mathrm{V}$ ; $V_C = 0.315 \times 159.2 = 50.1\,\mathrm{V}$ .

(c) $\tan\phi = \frac{94.2 - 159.2}{40} = -1.62$ , so $\phi = -58.4^\circ$ . The circuit is capacitive: the current leads the supply voltage by 58.4°.

Notice that $V_C = 50.1\,\mathrm{V}$ is about twice the supply voltage. The capacitor must be rated for it: $50.1\sqrt{2} = 71\,\mathrm{V}$ peak.

Resonance

As the supply frequency rises, $X_{L}$ grows and $X_{C}$ shrinks (Figure 2.3). At one frequency they are equal. Then $X_{L} - X_{C} = 0$ and:

the impedance is smallest, $Z = R$ , so the current is largest, $I = \frac{V}{R}$ ;
$\tan\phi = 0$ : the supply voltage and current are in phase, and the circuit behaves as a pure resistance;
$V_{L}$ and $V_{C}$ are equal and opposite and cancel, but each can be much larger than the supply voltage.

Setting $\omega_0 L = \frac{1}{\omega_0 C}$ gives $\omega_0^2 = \frac{1}{LC}$ , so the resonant frequency is

f_0 = \frac{1}{2\pi\sqrt{LC}}

It depends only on $L$ and $C$ ; the resistance decides how sharp the resonance is. The quality factor $Q$ measures this. It is the ratio of the voltage across the inductor (or capacitor) to the supply voltage at resonance:

Q = \frac{V_L}{V} = \frac{IX_L}{IR} = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}

A small $R$ gives a large $Q$ : a tall, narrow peak. That is what a radio needs to pick one station and reject its neighbours.

Figure 3.4: Resonance curves for the circuit of Worked example 3 with two values of R.

Definition

Resonance in a series RLC circuit is the condition $X_{L} = X_{C}$ , reached at the resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$ . At resonance $Z = R$ (a minimum), the current is a maximum, and the current is in phase with the supply voltage. The quality factor $Q = \frac{\omega_0 L}{R}$ is the voltage magnification at resonance. Example: $L = 10\,\mathrm{mH}$ and $C = 0.10\,\mathrm{\mu F}$ resonate at 5.03 kHz.

Worked example 4: tuning the circuit to resonance

For the circuit of Worked example 3 ( $R = 40\,\mathrm{\Omega}$ , $L = 0.30\,\mathrm{H}$ , $C = 20\,\mathrm{\mu F}$ , 24 V rms), find (a) the resonant frequency, (b) the current at resonance, (c) the voltage across the capacitor at resonance and (d) the Q factor.

Solution

(a) $f_0 = \frac{1}{2\pi\sqrt{0.30 \times 20 \times 10^{-6}}} = \frac{1}{2\pi \times 2.449 \times 10^{-3}} = 65.0\,\mathrm{Hz}$ .

(b) $Z = R = 40\,\mathrm{\Omega}$ , so $I = \frac{24}{40} = 0.60\,\mathrm{A}$ , nearly twice the 0.315 A at 50 Hz.

(c) At $f_{0}$ , $X_C = X_L = 2\pi \times 65.0 \times 0.30 = 122\,\mathrm{\Omega}$ , so $V_C = 0.60 \times 122 = 73\,\mathrm{V}$ .

(d) $Q = \frac{X_L}{R} = \frac{122}{40} = 3.1$ . Check: $\frac{V_C}{V} = \frac{73}{24} = 3.1$ . A 24 V supply puts 73 V rms (104 V peak) across the capacitor, so it must be rated for that.

Experiment 3.1: finding a resonance curve

Apparatus: a signal generator, a coil of about 10 mH, a 1.0 µF capacitor, a 10 Ω resistor, an oscilloscope (or a multimeter that reads AC correctly at a few kilohertz; many cheap meters do not).

Method:

Connect the generator, coil, capacitor and resistor in series. Set the generator to about 1 V rms.
Connect the oscilloscope across the 10 Ω resistor. The voltage across it is $V_{R} = I R$ , so it measures the current.
Starting at 200 Hz, record $V_{R}$ at steps of about 200 Hz up to 4 kHz. Take smaller steps near the peak. Keep the generator's output constant.
Plot $I = \frac{V_R}{10}$ against $f$ . Compare the peak with $f_0 = \frac{1}{2\pi\sqrt{LC}} \approx 1.6\,\mathrm{kHz}$ .
Connect the oscilloscope's second channel across the capacitor near the peak: it shows a voltage several times the supply.

Expect a curve like Figure 3.4. The peak is usually a little lower and broader than predicted, because the coil's own resistance adds to the 10 Ω.

Warning

Common mistakes. (1) Adding rms voltages directly: $V \neq V_R + V_L + V_C$ . (2) Writing $Z = R + X_{L} + X_{C}$ , or $Z = \sqrt{R^2 + X_L^2 + X_C^2}$ : $X_{L}$ and $X_{C}$ are opposite, so subtract them first. (3) Thinking the current is zero at resonance: it is at its MAXIMUM. (4) Forgetting the coil's resistance: a real inductor is an R and an L in series. (5) Assuming a component voltage can never exceed the supply: in an RLC circuit near resonance it can, by the factor $Q$ .

Tip

Draw the current along the reference line FIRST in every series problem. Then $V_{R}$ goes along it, $V_{L}$ up, $V_{C}$ down. Divide each by $I$ and the same diagram becomes the impedance triangle: $R$ along, $X_{L} - X_{C}$ up, $Z$ along the hypotenuse.

Resonance in AC Circuits Visualised (Decipher Physics). Watch the phasors of a series RLC circuit as the frequency is swept through resonance, and how the resistance sets the height and width of the peak.

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Practice

A 12 Ω resistor and an inductor of reactance 5.0 Ω are in series on a 26 V rms supply. Find the impedance, the current and the phase angle.
A 1.0 kΩ resistor and a 1.0 µF capacitor are in series on a 10 V rms, 50 Hz supply. Find the impedance, the current and the phase angle. Does the current lead or lag?
Find the resonant frequency of a 10 mH inductor in series with a 0.10 µF capacitor.
In a series RLC circuit, $R = 20\,\mathrm{\Omega}$ , $X_L = 50\,\mathrm{\Omega}$ and $X_C = 35\,\mathrm{\Omega}$ , on a 50 V rms supply. Find $Z$ , $I$ , the three component voltages and $\phi$ . Is the circuit inductive or capacitive?
A coil takes 0.50 A from 6.0 V DC and 0.30 A from 6.0 V rms at 50 Hz. Find its resistance and inductance.
Sketch on one set of axes how the current in a series RLC circuit varies with frequency for a small and a large resistance. Mark $f_{0}$ and explain why it is the same for both.
At resonance, the voltage across the capacitor of a series circuit on a 5.0 V supply is 60 V. Find Q. Explain how this is possible without breaking the conservation of energy.
Mugisha measures 7.0 V across the resistor and 9.0 V across the inductor of a series RL circuit and concludes that the supply is 16 V. Evaluate his conclusion, and find the actual supply voltage.

Show answersHide answers

$Z = \sqrt{144 + 25} = 13\,\mathrm{\Omega}$ ; $I = 2.0\,\mathrm{A}$ ; $\phi = \tan^{-1}\frac{5.0}{12} = 22.6^\circ$ (voltage leads).
$X_C = 3.18\,\mathrm{k\Omega}$ ; $Z = \sqrt{1000^2 + 3183^2} = 3.34\,\mathrm{k\Omega}$ ; $I = 3.0\,\mathrm{mA}$ ; $\phi = -72.6^\circ$ . The current leads the voltage by 72.6°.
$f_0 = \frac{1}{2\pi\sqrt{10^{-2} \times 10^{-7}}} = 5.03\,\mathrm{kHz}$ .
$Z = \sqrt{400 + 225} = 25\,\mathrm{\Omega}$ ; $I = 2.0\,\mathrm{A}$ ; $V_R = 40\,\mathrm{V}$ , $V_L = 100\,\mathrm{V}$ , $V_C = 70\,\mathrm{V}$ ; $\phi = \tan^{-1}\frac{15}{20} = 36.9^\circ$ . Inductive ( $X_{L} > X_{C}$ ); check $\sqrt{40^2 + 30^2} = 50\,\mathrm{V}$ .
$R = \frac{6.0}{0.50} = 12\,\mathrm{\Omega}$ ; $Z = \frac{6.0}{0.30} = 20\,\mathrm{\Omega}$ ; $X_L = \sqrt{400 - 144} = 16\,\mathrm{\Omega}$ ; $L = \frac{16}{2\pi \times 50} = 51\,\mathrm{mH}$ .
Both curves peak at the same $f_0 = \frac{1}{2\pi\sqrt{LC}}$ , which does not involve $R$ . The small-R curve is taller ( $I_{\max} = \frac{V}{R}$ ) and narrower; the large-R curve is lower and broader.
$Q = \frac{60}{5.0} = 12$ . Energy is not created: $V_{L}$ and $V_{C}$ are equal and opposite, so they cancel in the loop. Energy sloshes between the capacitor's electric field and the inductor's magnetic field each cycle; the supply only replaces what $R$ turns into heat.
Wrong: the rms voltages are 90° apart and add as phasors. $V = \sqrt{7.0^2 + 9.0^2} = \sqrt{130} = 11.4\,\mathrm{V}$ .

Key points

Series circuit: same current, so draw $I$ as reference; $V_{R}$ along, $V_{L}$ up, $V_{C}$ down.
$Z = \sqrt{R^2 + (X_L - X_C)^2}$ , $\tan\phi = \frac{X_L - X_C}{R}$ , $\cos\phi = \frac{R}{Z}$ .
Resonance: $X_{L} = X_{C}$ , $f_0 = \frac{1}{2\pi\sqrt{LC}}$ , $Z = R$ , current maximum and in phase with $V$ .
$Q = \frac{\omega_0 L}{R}$ : component voltages at resonance can be $Q$ times the supply.

Sources and credits

15.4: RLC Series Circuits with AC and 15.6: Resonance in an AC Circuit, from University Physics Volume 2 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Video: Resonance in AC Circuits Visualised | Series RLC Circuit, Resonant Frequency & Q Factor, Decipher Physics, embedded from YouTube.

The experiment, the four diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Voltages that do not add up

2The method for a series circuit

3Resonance

4Experiment 3.1: finding a resonance curve

Voltages that do not add up

The method for a series circuit

Resonance

Experiment 3.1: finding a resonance curve