Alternating Current

Lessons

Senior 6 · Physics

Alternating Current

4 lessons

Lesson 2 of 4

Resistors, Inductors and Capacitors in AC Circuits

25 min

A coil that dims a lamp

Connect a lamp in series with a large coil to a 12 V DC supply and the lamp shines almost normally: the coil's copper wire has only a few ohms of resistance. Change to a 12 V AC supply of the same rms value and the lamp becomes dim. Push an iron core into the coil and the lamp nearly goes out. The coil's resistance has not changed, so something else must be opposing the alternating current. A capacitor does the opposite trick: in series with a lamp it blocks DC completely, yet the lamp glows on AC. This lesson explains both, using the phasors of Lesson 1.

Experiment 2.1: the coil and the capacitor on AC and DC

Apparatus: a power pack giving 12 V DC and 12 V AC (50 Hz), a 12 V, 0.1 A lamp, a switch, a coil of about 1000 turns with a removable laminated iron core, a non-polarised capacitor of about 100 µF rated at least 25 V, connecting leads.

Figure 2.1: The circuit for Experiment 2.1.

Method:

Connect the circuit of Figure 2.1 to the DC outlet. Note the lamp's brightness with and without the iron core in the coil.
Move to the AC outlet and repeat.
Replace the coil with the capacitor and repeat on DC and on AC.

Typical results:

In series with the lamp	12 V DC	12 V AC, 50 Hz
coil, no core	bright	slightly dimmer
coil with iron core	bright	very dim or out
capacitor	out (after a brief flash)	glows

Conclusions: a coil opposes AC more than DC, and more when its inductance is larger. A capacitor passes no steady current, but alternating current flows "through" it as the plates charge and discharge in turn.

Warning

Safety. Use a NON-polarised capacitor. An electrolytic (polarised) capacitor connected to AC is reverse-biased every half-cycle; it heats up and can burst. Use only the 12 V outlets of the power pack, and switch off before changing components: a coil with an iron core gives a sharp kick of voltage when the current is broken (Unit 3, Lesson 4).

A resistor: voltage and current in phase

Apply $v = V_0\sin\omega t$ across a resistance $R$ . Ohm's law holds at every instant, so $i = \frac{v}{R} = \frac{V_0}{R}\sin\omega t$ . The current reaches its peak $I_0 = \frac{V_0}{R}$ at the same moment as the voltage: they are in phase, and their phasors point the same way. Dividing by $\sqrt{2}$ gives $V_{\text{rms}} = I_{\text{rms}}R$ . The resistance does not depend on frequency.

An inductor: the voltage leads by 90°

Let a current $i = I_0\sin\omega t$ flow in a pure inductor (inductance $L$ , no resistance). The changing current induces a back emf $-L\frac{di}{dt}$ . With no resistance, the applied voltage must exactly balance it at every instant (Kirchhoff's loop rule):

v = L\frac{di}{dt} = L\frac{d}{dt}\left(I_0\sin\omega t\right) = \omega L I_0\cos\omega t = \omega L I_0\sin\left(\omega t + \frac{\pi}{2}\right)

Two results follow. The peak voltage is $V_0 = \omega L I_0$ , so the ratio $\frac{V_0}{I_0}$ , which acts like a resistance, is

X_L = \omega L = 2\pi f L

And the voltage leads the current by $\frac{\pi}{2}$ (90°). Physically: the back emf is greatest when the current is changing fastest, which is when $i$ passes through zero.

A capacitor: the current leads by 90°

Apply $v = V_0\sin\omega t$ across a capacitor $C$ . The charge on its plates is $q = C v$ , so the current is

i = \frac{dq}{dt} = C\frac{dv}{dt} = \omega C V_0\cos\omega t = \omega C V_0\sin\left(\omega t + \frac{\pi}{2}\right)

The peak current is $I_0 = \omega C V_0$ , giving

X_C = \frac{V_0}{I_0} = \frac{1}{\omega C} = \frac{1}{2\pi f C}

and the current leads the voltage by $\frac{\pi}{2}$ . Physically: the most charge flows onto the plates when the voltage is changing fastest, as $v$ passes through zero; when $v$ is at its peak the plates are full and $i = 0$ .

Figure 2.2: Phasors (top) and waveforms (bottom) for a resistor, an inductor and a capacitor. In each phasor diagram the current is drawn along the reference line.

Definition

Reactance $X$ is the opposition of an inductor or a capacitor to alternating current: the ratio of rms (or peak) voltage to rms (or peak) current, $X = \frac{V_{\text{rms}}}{I_{\text{rms}}}$ . Its unit is the ohm. Inductive reactance $X_L = 2\pi fL$ rises with frequency; capacitive reactance $X_C = \frac{1}{2\pi fC}$ falls with frequency. Example: a 0.10 H inductor at 50 Hz has $X_L = 31.4\,\mathrm{\Omega}$ .

Worked example 1: an inductor at two frequencies

A 0.20 H inductor of negligible resistance is connected to a 12 V rms supply. Find the rms current (a) at 50 Hz and (b) at 500 Hz. (c) Taking $v = V_0\sin\omega t$ at 50 Hz, write an equation for the current.

Solution

(a) $X_L = 2\pi fL = 2\pi \times 50 \times 0.20 = 62.8\,\mathrm{\Omega}$ ; $I_{\text{rms}} = \frac{12}{62.8} = 0.191\,\mathrm{A}$ .

(b) Ten times the frequency gives ten times the reactance: $X_L = 628\,\mathrm{\Omega}$ and $I_{\text{rms}} = 0.0191\,\mathrm{A} = 19.1\,\mathrm{mA}$ .

(c) $I_0 = \sqrt{2} \times 0.191 = 0.270\,\mathrm{A}$ , and the current lags the voltage by $\frac{\pi}{2}$ : $i = 0.270\sin\left(100\pi t - \frac{\pi}{2}\right)\,\mathrm{A}$ .

Worked example 2: a capacitor on the mains

A ceiling-fan motor uses a 2.5 µF capacitor (an illustrative value). Find (a) its reactance on the 230 V, 50 Hz mains and (b) the rms current through it. (c) What would the current be on a 230 V, 60 Hz supply?

Solution

(a) $X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50 \times 2.5 \times 10^{-6}} = 1.27 \times 10^{3}\,\mathrm{\Omega}$ .

(b) $I_{\text{rms}} = \frac{230}{1273} = 0.181\,\mathrm{A}$ .

(c) $X_C \propto \frac{1}{f}$ , so at 60 Hz the reactance falls by the factor $\frac{50}{60}$ and the current rises by $\frac{60}{50}$ : $I_{\text{rms}} = 0.181 \times 1.2 = 0.217\,\mathrm{A}$ .

How reactance depends on frequency

Figure 2.3: $X_{L}$ , $X_{C}$ and $R$ against frequency, for $L = 0.20\,\mathrm{H}$ , $C = 10\,\mathrm{\mu F}$ and $R = 100\,\mathrm{\Omega}$ .

At very low frequency (DC is $f = 0$ ) an inductor is just a piece of wire, $X_L \to 0$ , while a capacitor is a break in the circuit, $X_C \to \infty$ . At very high frequency the roles swap. This is why a capacitor is used to block DC but pass an audio signal between amplifier stages, and why a coil called a choke is used to stop high-frequency interference entering a power supply.

Why reactance costs no energy

In a pure inductor, $i = I_0\sin\omega t$ and $v = V_0\cos\omega t$ , so the power is

p = vi = V_0I_0\sin\omega t\cos\omega t = \frac{1}{2}V_0I_0\sin 2\omega t

This is positive for a quarter-cycle (energy flows INTO the magnetic field) and negative for the next (the same energy flows back to the supply). Its mean over a cycle is zero. The same is true for a capacitor, whose energy is stored in its electric field. An ideal inductor or capacitor takes no average power. Only resistance turns electrical energy into heat.

Component	Opposition	Depends on $f$ ?	Phase of $V$ relative to $I$	Average power
resistor	$R$	no	in phase	$I_{\text{rms}}^2R$
inductor	$X_L = \omega L$	$X_L \propto f$	leads by 90°	0
capacitor	$X_C = \frac{1}{\omega C}$	$X_C \propto \frac{1}{f}$	lags by 90°	0

Worked example 3: when the reactances are equal

At what frequency does the reactance of a 0.20 H inductor equal that of a 10 µF capacitor? What is the common reactance?

Solution

Set $2\pi fL = \frac{1}{2\pi fC}$ . Then $f^2 = \frac{1}{4\pi^2LC}$ , so

f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.20 \times 10 \times 10^{-6}}} = \frac{1}{2\pi \times 1.414 \times 10^{-3}} = 113\,\mathrm{Hz}

and $X_L = X_C = 2\pi \times 112.5 \times 0.20 = 141\,\mathrm{\Omega}$ . This is the crossing point in Figure 2.3. You will meet this frequency again in Lesson 3: it is the resonant frequency.

Warning

Common mistakes. (1) Using $f$ where $\omega$ is needed: $X_L = 2\pi fL$ , not $f L$ . (2) Forgetting powers of ten: $\mathrm{\mu F} = 10^{-6}\,\mathrm{F}$ , $\mathrm{mH} = 10^{-3}\,\mathrm{H}$ . (3) Saying a capacitor "lets current through its dielectric": charge flows on and off the plates, never across the gap. (4) Adding reactance to resistance directly, $R + X$ : they are 90° apart, so they add as phasors (Lesson 3).

Tip

CIVIL: in a Capacitor, I comes before V; V comes before I in an inductor (L). And the frequency rule: L for "large at high f", C for "cut at high f".

RLC Circuits (5 of 19): Inductive Reactance, Phase Shift and Phasor Diagrams (Step by Step Science). Watch how the voltage phasor across an inductor is drawn 90° ahead of the current phasor, and how the reactance grows with frequency.

✦

Practice

Find the reactance of a 50 mH inductor at 50 Hz and at 5.0 kHz.
Find the reactance of a 0.47 µF capacitor at 50 Hz and at 5.0 kHz.
A pure inductor takes 0.40 A rms from a 24 V, 50 Hz supply. Find its inductance.
The frequency of the supply to a capacitor is doubled while the rms voltage stays the same. What happens to the rms current? Explain.
A voltage $v = 10\sin(200\pi t)$ V is applied to a 100 µF capacitor. Find the peak current and write an equation for $i$ .
Show that the product $\omega L$ has the unit of the ohm.
Draw the phasor diagram, with the current as reference, for (a) a pure inductor and (b) a pure capacitor. Explain from the physics of each component why the voltage leads in one and lags in the other.
Keza says: "An inductor in an AC circuit wastes energy, because it opposes the current." Evaluate her statement.

Show answersHide answers

$X_L = 2\pi \times 50 \times 0.050 = 15.7\,\mathrm{\Omega}$ ; at 5.0 kHz, $1.57\,\mathrm{k\Omega}$ .
$X_C = \frac{1}{2\pi \times 50 \times 0.47 \times 10^{-6}} = 6.77\,\mathrm{k\Omega}$ ; at 5.0 kHz, $67.7\,\mathrm{\Omega}$ .
$X_L = \frac{24}{0.40} = 60\,\mathrm{\Omega}$ ; $L = \frac{60}{2\pi \times 50} = 0.19\,\mathrm{H}$ .
It doubles: $X_C = \frac{1}{2\pi fC}$ halves, so $I = \frac{V}{X_C}$ doubles. The plates charge and discharge twice as often, so twice the charge flows per second.
$X_C = \frac{1}{200\pi \times 10^{-4}} = 15.9\,\mathrm{\Omega}$ ; $I_0 = \frac{10}{15.9} = 0.628\,\mathrm{A}$ ; $i = 0.628\sin(200\pi t + \frac{\pi}{2}) = 0.628\cos(200\pi t)$ A.
$[\omega L] = \mathrm{s^{-1}} \times \mathrm{H}$ , and from $\varepsilon = L\frac{dI}{dt}$ , $1\,\mathrm{H} = 1\,\mathrm{V\,s\,A^{-1}}$ . So $\mathrm{s^{-1} \cdot V\,s\,A^{-1}} = \mathrm{V\,A^{-1}} = \mathrm{\Omega}$ .
(a) $V$ drawn 90° anticlockwise from $I$ (up); (b) $V$ drawn 90° clockwise from $I$ (down). In an inductor the back emf, and so the applied voltage, is largest when $\frac{di}{dt}$ is largest, a quarter-cycle BEFORE the current peaks. In a capacitor the current is largest when $\frac{dv}{dt}$ is largest, a quarter-cycle BEFORE the voltage peaks, so the voltage lags.
Wrong for an ideal inductor. It opposes the current by a back emf, not by friction-like resistance. It stores energy in its magnetic field for a quarter-cycle and returns it in the next, so the average power is zero. A real coil does waste some energy, but only through its resistance (and core losses), not through its reactance.

Key points

Resistor: $V$ and $I$ in phase; $R$ is independent of $f$ .
Inductor: $V$ leads $I$ by 90°; $X_L = 2\pi fL$ .
Capacitor: $I$ leads $V$ by 90°; $X_C = \frac{1}{2\pi fC}$ .
An ideal inductor or capacitor stores and returns energy each cycle and takes no average power.

Sources and credits

15.3: Simple AC Circuits, 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: RLC Circuits (5 of 19) Inductive Reactance; Phase Shift, Phasor Diagrams, Frequency, An Explanation, Step by Step Science, embedded from YouTube.

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

1A coil that dims a lamp

2Experiment 2.1: the coil and the capacitor on AC and DC

3A resistor: voltage and current in phase

4An inductor: the voltage leads by 90°

5A capacitor: the current leads by 90°

6How reactance depends on frequency

7Why reactance costs no energy

A coil that dims a lamp

Experiment 2.1: the coil and the capacitor on AC and DC

A resistor: voltage and current in phase

An inductor: the voltage leads by 90°

A capacitor: the current leads by 90°

How reactance depends on frequency

Why reactance costs no energy