Option A: Alternating Currents

What you'll learn

How a rotating coil in a magnetic field produces a sinusoidal induced emf.
What peak, period, frequency and rms values mean for a.c. waveforms.
How inductors and capacitors affect phase and oppose a.c. using reactance.
How phasors, impedance, resonance frequency and Q factor describe series RCL circuits.

1. Rotating coils and electromagnetic induction

An alternating current, or a.c., is a current whose direction changes periodically. In this option, you mostly deal with sinusoidal a.c., meaning it varies like a sine or cosine wave.

A rotating coil in a magnetic field is the basic model of an a.c. generator: mechanical rotation changes the magnetic flux linkage, so an emf is induced.

Definition

Faraday's law

Faraday’s law says that the induced emf is equal to the negative rate of change of flux linkage:

-\frac{\Delta(N\Phi)}{\Delta t}

For a smoothly rotating coil, this becomes a calculus idea: the induced emf depends on how fast the flux linkage is changing.

Flux linkage in a rotating coil

Magnetic flux, $Φ\Phi$ , through a flat coil depends on the component of magnetic field perpendicular to the coil:

Φ=BAcos⁡θ\Phi = BA\cos\theta

where $B$ is magnetic flux density in tesla (T), $A$ is area in square metres (m²), and $θ\theta$ is the angle between the coil’s normal and the field.

For a coil with $N$ turns, the flux linkage is:

NΦ=BANcos⁡θN\Phi = BAN\cos\theta

If the coil rotates at angular speed $ω\omega$ , then:

θ=ωt\theta = \omega t

so the flux linkage becomes:

NΦ=BANcos⁡ωtN\Phi = BAN\cos \omega t

The induced emf is therefore:

-\frac{d}{dt}(BAN\cos \omega t) = \omega BAN\sin\omega t

The peak emf is:

V0=ωBANV_0 = \omega BAN

Rotating coil in a magnetic field showing flux linkage as a cosine wave and induced emf as a sine wave

Key Idea

Generator waveform

The flux linkage is greatest when the coil normal is parallel to the magnetic field, but the induced emf is greatest when the flux linkage is changing fastest.

Example

Finding the rms emf and power from a rotating coil

A coil has 200 turns, area $4.0 \times 10^{-3}\ \text{m}^2$ , and rotates at 50 Hz in a uniform magnetic field of 0.20 T. It is connected to a 40 Ω resistor. Find the peak emf, rms emf and mean power dissipated.

Convert frequency to angular speed:

s−1\omega = 2\pi f = 2\pi(50\ \text{Hz}) = 3.14 \times 10^2\ \text{rad}\ \text{s}^{-1}

Use the peak emf equation:

V0=ωBANV_0 = \omega BAN

VV_0 = (3.14 \times 10^2)(0.20)(4.0 \times 10^{-3})(200) = 50\ \text{V}

Convert peak emf to rms emf:

VV_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{50\ \text{V}}{\sqrt{2}} = 35\ \text{V}

Use rms values for mean power in the resistor:

\frac{V^2}{R} = \frac{(35\ \text{V})^2}{40\ \Omega} = 31\ \text{W}

2. Describing a.c. waveforms

Definition

Waveform terms

For an alternating potential difference or current:

Frequency, $f$ , is the number of complete cycles per second, measured in hertz (Hz).
Period, $T$ , is the time for one complete cycle, measured in seconds (s), with $T = 1/ f$ .
Peak value, such as $V_0$ or $I_0$ , is the maximum magnitude from zero.
Rms value means “root mean square”: the equivalent d.c. value that gives the same mean power dissipation in a resistor.

For a sinusoidal current or potential difference:

\frac{I_0}{\sqrt{2}}

\frac{V_0}{\sqrt{2}}

Here, $I$ and $V$ are rms values unless otherwise stated. For the rotating coil:

Vrms=ωBAN2V_{\text{rms}} = \frac{\omega BAN}{\sqrt{2}}

Mean power in a resistor

For a resistor in an a.c. circuit, use rms values in the usual power equations:

I^2R = \frac{V^2}{R}

Common Mistake

Using peak values in power equations

If the question asks for mean power in a resistor, use rms values. Using $V_0$ instead of $VrmsV_{\text{rms}}$ makes the power twice too large for a sinusoidal waveform.

3. Using an oscilloscope

A CRO or PC-based oscilloscope displays potential difference on the vertical axis and time on the horizontal axis.

To measure:

a.c. potential difference: measure peak-to-peak height, then convert to peak or rms if sinusoidal
d.c. potential difference: measure the vertical displacement from the zero line
frequency: measure the period from the time base, then use $f = 1/ T$
current: measure the potential difference across a known resistor and use $I = V / R$

Common Mistake

Oscilloscopes and mains electricity

Do not connect an oscilloscope directly to mains circuits unless the equipment and probes are rated and the setup is safely isolated. UK mains is about 230 V rms, which has a peak value of about 325 V.

Example

Reading a sinusoidal waveform from an oscilloscope

A sine wave is displayed with 5.6 divisions peak-to-peak. The vertical scale is 0.50 V per division. One cycle takes 4.0 divisions, and the time base is 2.0 ms per division. The waveform is across a 10.0 Ω resistor.

Find the peak-to-peak potential difference:

VV_{\text{pp}} = (5.6\ \text{div})(0.50\ \text{V div}^{-1}) = 2.8\ \text{V}

Convert peak-to-peak to peak, then to rms:

VV_0 = \frac{2.8\ \text{V}}{2} = 1.4\ \text{V}

VV_{\text{rms}} = \frac{1.4\ \text{V}}{\sqrt{2}} = 0.99\ \text{V}

Find the period:

(4.0\ \text{div})(2.0\ \text{ms div}^{-1}) = 8.0\ \text{ms} = 8.0 \times 10^{-3}\ \text{s}

Calculate frequency and rms current:

\frac{1}{T} = \frac{1}{8.0 \times 10^{-3}\ \text{s}} = 125\ \text{Hz}

AI_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{0.99\ \text{V}}{10.0\ \Omega} = 0.099\ \text{A}

4. Inductors and capacitors in a.c.

A resistor dissipates energy as heat. An ideal inductor or capacitor stores energy temporarily and returns it to the circuit, so its mean power dissipation is zero.

Definition

Reactance

Reactance is the opposition of an inductor or capacitor to alternating current. It is measured in ohms (Ω) and is defined using rms values:

\frac{V_{\text{rms}}}{I_{\text{rms}}}

Inductor: current lags potential difference

For an ideal inductor, the current lags behind the potential difference by 90°.

Its reactance is:

XL=ωLX_L = \omega L

where $L$ is inductance in henries (H). Increasing frequency increases inductive reactance.

Capacitor: current leads potential difference

For an ideal capacitor, the current leads the potential difference by 90°.

Its reactance is:

XC=1ωCX_C = \frac{1}{\omega C}

where $C$ is capacitance in farads (F). Increasing frequency decreases capacitive reactance.

Example

Comparing inductive and capacitive reactance

A 12.0 V rms, 50 Hz supply is connected separately to a 0.25 H inductor and a 10 µF capacitor. Find the rms current in each ideal component and state the phase relationship.

Calculate angular frequency:

s−1\omega = 2\pi f = 2\pi(50\ \text{Hz}) = 314\ \text{rad}\ \text{s}^{-1}

For the inductor:

ΩX_L = \omega L = (314)(0.25\ \text{H}) = 78.5\ \Omega

AI_{\text{rms}} = \frac{12.0\ \text{V}}{78.5\ \Omega} = 0.153\ \text{A}

The current lags the potential difference by 90°.

For the capacitor:

ΩX_C = \frac{1}{\omega C} = \frac{1}{(314)(10 \times 10^{-6}\ \text{F})} = 318\ \Omega

AI_{\text{rms}} = \frac{12.0\ \text{V}}{318\ \Omega} = 0.0377\ \text{A}

The current leads the potential difference by 90°.

5. Phasors, phase angle and impedance

A phasor is a rotating vector used to represent a sinusoidal quantity. In series a.c. circuits, the current is the same through every component, so it is usually taken as the horizontal reference phasor.

Potential differences must be added as phasors, not just as ordinary scalar numbers, because they may not be in phase.

Phasor diagrams for series RL, RC and RCL circuits with impedance triangle

Definition

Impedance and phase angle

Impedance, $Z$ , is the total opposition of a circuit to a.c.:

\frac{V_{\text{rms}}}{I_{\text{rms}}}

The phase angle, $ϕ\phi$ , is the angle between the supply potential difference and the current.

For a series RCL circuit:

\sqrt{R^2 + (X_L - X_C)^2}

and

tan⁡ϕ=XL−XCR\tan\phi = \frac{X_L - X_C}{R}

If $X_L > X_C$ , the circuit is overall inductive, so the current lags the supply potential difference. If $X_C > X_L$ , it is overall capacitive, so the current leads.

Example

Calculating impedance and phase angle

A series RCL circuit has $60\ \Omega$ , $0.20\ \text{H}$ and $50\ \mu\text{F}$ . It is connected to a 12.0 V rms supply at 100 Hz. Find the current and phase angle.

Calculate the reactances:

ΩX_L = \omega L = 2\pi(100)(0.20) = 126\ \Omega

ΩX_C = \frac{1}{\omega C} = \frac{1}{2\pi(100)(50 \times 10^{-6})} = 31.8\ \Omega

Find the net reactance and impedance:

ΩX_L - X_C = 126\ \Omega - 31.8\ \Omega = 94.2\ \Omega

\sqrt{(60\ \Omega)^2 + (94.2\ \Omega)^2} = 112\ \Omega

Use $Z = V / I$ :

AI_{\text{rms}} = \frac{12.0\ \text{V}}{112\ \Omega} = 0.107\ \text{A}

Calculate the phase angle:

ϕ=tan⁡−1(94.260)=57.5∘\phi = \tan^{-1}\left(\frac{94.2}{60}\right) = 57.5^\circ

Since $X_L > X_C$ , the circuit is inductive: the supply potential difference leads the current by 57.5°.

6. Resonance in a series RCL circuit

In a series RCL circuit, resonance occurs when the inductive and capacitive reactances are equal:

X_L = X_C

Using $XL=ωLX_L = \omega L$ and $XC=1/(ωC)X_C = 1/(\omega C)$ :

ωL=1ωC\omega L = \frac{1}{\omega C}

ω2=1LC\omega^2 = \frac{1}{LC}

So the resonant angular frequency is:

ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}

and the resonant frequency is:

f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}

At resonance, the net reactance is zero, so:

Z = R

The current is maximum, and the current is in phase with the supply potential difference.

Series RCL resonance curve showing high-Q and low-Q resonance

Q factor

At resonance, the potential differences across the inductor and capacitor are equal in magnitude and opposite in phase. They can be much larger than the supply potential difference, but they cancel in the phasor sum.

Definition

Q factor

For a series RCL circuit at resonance, the Q factor is:

\frac{V_L}{V_R} = \frac{V_C}{V_R}

A larger Q factor means a sharper, narrower resonance curve.

Example

Finding resonance frequency and Q factor

A series RCL circuit has $0.10\ \text{H}$ , $4.0\ \mu\text{F}$ and total resistance $20\ \Omega$ . The rms supply potential difference is 5.0 V. Find the resonance frequency and Q factor.

Use the resonance formula:

f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}

Hzf_0 = \frac{1}{2\pi\sqrt{(0.10)(4.0 \times 10^{-6})}} = 252\ \text{Hz}

Calculate the resonant angular frequency:

s−1\omega_0 = 2\pi f_0 = 2\pi(252\ \text{Hz}) = 1.58 \times 10^3\ \text{rad}\ \text{s}^{-1}

At resonance, find the inductive reactance:

ΩX_L = \omega_0 L = (1.58 \times 10^3)(0.10\ \text{H}) = 158\ \Omega

Use $Q = V_L/V_R = X_L/R$ at resonance:

\frac{158\ \Omega}{20\ \Omega} = 7.9

Since $V_R$ equals the supply potential difference at resonance:

VV_L = QV_R = (7.9)(5.0\ \text{V}) = 40\ \text{V}

The inductor and capacitor each have about 40 V rms across them, even though the supply is only 5.0 V rms.

Tip

Spotting resonance experimentally

To investigate resonance, keep the supply amplitude constant, vary the frequency, measure the rms current, and plot current against frequency. The peak gives $f_0$ , and a narrower peak means a larger Q factor.

Exam technique

In the exam

Check whether a value is peak, peak-to-peak or rms before using it in a power or impedance equation.
For series RCL circuits, calculate $X_L$ and $X_C$ first, then decide whether the circuit is inductive or capacitive.
Use phasor addition for potential differences; do not add $V_R$ , $V_L$ and $V_C$ as simple scalar magnitudes unless they are in phase.

Self review

Check yourself

Why is the induced emf in a rotating coil zero when the flux linkage is maximum?
In a series RCL circuit with $X_C > X_L$ , does the current lead or lag the supply potential difference?
Why can the potential difference across the inductor be larger than the supply potential difference at resonance?

What you'll learn

How a rotating coil in a magnetic field produces a sinusoidal induced emf.
What peak, period, frequency and rms values mean for a.c. waveforms.
How inductors and capacitors affect phase and oppose a.c. using reactance.
How phasors, impedance, resonance frequency and Q factor describe series RCL circuits.

1. Rotating coils and electromagnetic induction

An alternating current, or a.c., is a current whose direction changes periodically. In this option, you mostly deal with sinusoidal a.c., meaning it varies like a sine or cosine wave.

A rotating coil in a magnetic field is the basic model of an a.c. generator: mechanical rotation changes the magnetic flux linkage, so an emf is induced.

Definition

Faraday's law

Faraday’s law says that the induced emf is equal to the negative rate of change of flux linkage:

-\frac{\Delta(N\Phi)}{\Delta t}

For a smoothly rotating coil, this becomes a calculus idea: the induced emf depends on how fast the flux linkage is changing.