AC Excitation of Magnetic Circuits

i Learning Objectives

By the end of this chapter you should be able to:

Explain what excitation of a magnetic circuit means and why most machines are excited from AC.
Contrast DC and AC excitation: which quantity fixes the flux and which adjusts itself.
Derive the EMF equation \(E_{rms} = 4.44\,fN\phi_m\) and state the phase relation between flux and EMF.
Use \(V \approx 4.44\,fN\phi_m\) to find the working flux and flux density of a core.

Demonstrative Video

Section 1

What Excitation Means

The magnetic circuits of transformers, AC machines, relays and contactors must be magnetised before they can do any work. Supplying the current that sets up this working flux is called excitation. A magnetic circuit can be excited from either DC or AC, and the two cases behave in fundamentally different ways.

Magnetic core with an exciting coil — A magnetic core excited by an N-turn coil.

Section 2

DC Excitation: the Current Decides

With a steady DC supply, the coil current is limited only by the winding resistance:

\[I = \frac{V}{R}\]

The current fixes the MMF \(NI\), and the flux then adjusts itself to whatever value the B–H curve of the core allows for that MMF. The coil inductance plays a part only during transients, when the current is building up at switch-on or decaying at switch-off. In steady state, \(L\,di/dt = 0\).

DC excitation is used for electromagnets and for the field windings of DC and synchronous machines.

Section 3

AC Excitation: the Voltage Decides

With AC, the inductance acts all the time, not only during switching. The applied voltage is balanced almost entirely by the EMF induced by the changing flux. The winding resistance drop is only a few per cent of \(V\), so:

\[V \approx E = N\frac{d\phi}{dt}\]

The reversal. In most AC magnetic circuits, the flux is fixed by the applied voltage and frequency. The magnetising current then adjusts itself to whatever value the B–H curve demands for that flux.

For economic reasons the working flux density is usually set just beyond the linear part of the magnetisation curve. The magnetising current is then not exactly sinusoidal, and the inductance is not strictly constant. For most practical calculations, however, the circuit parameters are treated as constant.

DC excitation

The current is set by \(V/R\).
The flux follows from the B–H curve.
Inductance matters only in transients.
No core loss in steady state.

AC excitation

The flux is set by \(V/(4.44fN)\).
The current follows from the B–H curve.
Inductance matters all the time.
Hysteresis and eddy-current losses occur in the core.

Section 4

The EMF Equation

Let the core flux vary sinusoidally:

\[\phi = \phi_m \sin \omega t = \phi_m \sin 2\pi f t \qquad (1)\]

The EMF induced in the \(N\)-turn coil is

\[\begin{aligned} e &= N\frac{d\phi}{dt} = N\frac{d}{dt}\left(\phi_m \sin 2\pi f t\right)\\ &= 2\pi f N\phi_m \cos 2\pi f t = 2\pi f N \phi_m \sin\!\left(2\pi f t + \frac{\pi}{2}\right) \qquad (2) \end{aligned}\]

The peak and RMS values are therefore

\[E_m = 2\pi f N \phi_m, \qquad E_{rms} = \frac{E_m}{\sqrt{2}} = \frac{2\pi}{\sqrt 2}\,fN\phi_m = 4.44\,fN\phi_m\]

≈

AC Excitation Equation

\(V \approx E_{rms} = 4.44\,f N \phi_m = 4.44\,f N B_m A\)

Comparing (1) and (2), the induced EMF leads the flux by 90°, and therefore also leads the magnetising current that produces the flux. The induced EMF and the resistance drop together oppose the applied voltage. Since the resistance drop is small, \(E\) and \(V\) may be taken as equal in magnitude.

At a fixed frequency, \(\phi_m \propto V\). If the frequency changes, \(\phi_m \propto V/f\). This is why variable-frequency drives keep the ratio \(V/f\) constant.

1 Worked Example — Working flux of a coil

Problem. A 500-turn coil on a core of cross-section 20 cm² is connected to 230 V, 50 Hz. Neglecting resistance, find the peak flux and the peak flux density.

\[\phi_m = \frac{V}{4.44 f N} = \frac{230}{4.44\times 50\times 500} = 2.07\ \text{mWb}\]

\[B_m = \frac{\phi_m}{A} = \frac{2.07\times10^{-3}}{20\times10^{-4}} = 1.04\ \text{T}\]

The flux density depends only on V, f, N and A. The core material affects the current needed, not the flux.

2 Worked Example — Wrong supply

Problem. The coil of Example 1 has a resistance of 2 Ω. What happens if it is connected (a) to 230 V, 60 Hz, and (b) to 230 V DC?

(a) The flux falls in proportion to \(1/f\):

\[\phi_m = 2.07 \times \frac{50}{60} = 1.73\ \text{mWb}, \qquad B_m = 0.86\ \text{T}\]

(b) On DC there is no induced EMF in steady state, so only the resistance limits the current:

\[I = \frac{230}{2} = 115\ \text{A}\]

This is far beyond the coil rating. An AC coil connected to the same DC voltage will overheat and burn out.

Section 5

Summary and Key Formulas

Excitation means setting up the working flux of a magnetic circuit.
DC: the current is set by resistance, and the flux follows. AC: the flux is set by voltage and frequency, and the current follows.
The induced EMF leads the flux by 90°, and \(E_{rms} = 4.44fN\phi_m\).
Core flux is proportional to \(V/f\).

Quantity	Formula
Instantaneous EMF	\(e = N\,d\phi/dt\)
Peak EMF	\(E_m = 2\pi f N\phi_m\)
RMS EMF	\(E_{rms} = 4.44 fN\phi_m\)
Working flux	\(\phi_m \approx V/(4.44fN)\)

Section 6

Common Mistakes

Thinking a better core material lowers the flux. Under AC, a higher-permeability core only lowers the magnetising current. The flux is still \(V/(4.44fN)\).
Using the peak flux with the RMS factor wrongly. The factor 4.44 converts peak flux to RMS EMF.
Connecting an AC coil to DC. With no induced EMF to limit it, the current is only \(V/R\).

Section 7

Chapter Review

Practice Problems

P1 A 1000-turn coil is connected to 400 V, 50 Hz. Find the peak core flux.

Show answer
\[\phi_m = \frac{400}{4.44\times50\times1000} = 1.80\ \text{mWb}\]
P2 What peak flux density is allowed if a core of 25 cm² must carry the flux of P1?

Show answer
\(B_m = 1.80\times10^{-3}/25\times10^{-4} = 0.72\ \text{T}\).
P3 A coil designed for 230 V, 50 Hz is to be used at 25 Hz without exceeding its rated flux. What voltage should be applied?

Show answer
Keep \(V/f\) constant: \(V = 230\times 25/50 = 115\ \text{V}\).

Multiple-Choice Questions

MCQ 1. With AC excitation and negligible resistance, the core flux is determined by
(a) core permeability (b) applied voltage and frequency (c) winding resistance (d) magnetising current only

Show answer
(b), since \(\phi_m = V/(4.44fN)\).
MCQ 2. The induced EMF in an AC-excited coil
(a) lags the flux by 90° (b) leads the flux by 90° (c) is in phase with the flux (d) is in anti-phase with the flux

Show answer
(b) leads by 90°, because \(e \propto d\phi/dt\).
MCQ 3. If the supply frequency is doubled at constant voltage, the core flux
(a) doubles (b) halves (c) is unchanged (d) quadruples

Show answer
(b) halves.

Conceptual Questions

Why does inductance matter only during transients with DC excitation, but all the time with AC?
Why is the magnetising current non-sinusoidal when the core works beyond the knee of the B–H curve?
Explain why an AC contactor coil burns out when connected to a DC supply of the same voltage.