Inductors store their energy in the form of a magnetic field that is created when a voltage is applied across the terminals of an inductor. The growth of the current flowing through the inductor is not instant but is determined by the inductors own selfinduced or back emf value. Then for an inductor coil, this back emf voltage V_{L} is proportional to the rate of change of the current flowing through it.
This current will continue to rise until it reaches its maximum steady state condition which is around five time constants when this selfinduced back emf has decayed to zero. At this point a steady state current is flowing through the coil, no more back emf is induced to oppose the current flow and therefore, the coil acts more like a short circuit allowing maximum current to flow through it.
However, in an alternating current circuit which contains an AC Inductance, the flow of current through an inductor behaves very differently to that of a steady state DC voltage. Now in an AC circuit, the opposition to the current flowing through the coils windings not only depends upon the inductance of the coil but also the frequency of the applied voltage waveform as it varies from its positive to negative values.
The actual opposition to the current flowing through a coil in an AC circuit is determined by the AC Resistance of the coil with this AC resistance being represented by a complex number. But to distinguish a DC resistance value from an AC resistance value, which is also known as Impedance, the term Reactance is used.
Like resistance, reactance is measured in Ohm’s but is given the symbol “X” to distinguish it from a purely resistive “R” value and as the component in question is an inductor, the reactance of an inductor is called Inductive Reactance, ( X_{L} ) and is measured in Ohms. Its value can be found from the formula.
Inductive Reactance
Where: X_{L} is the Inductive Reactance in Ohms, ƒ is the frequency in Hertz and L is the inductance of the coil in Henries.
We can also define inductive reactance in radians, where Omega, ω equals 2πƒ.
So whenever a sinusoidal voltage is applied to an inductive coil, the back emf opposes the rise and fall of the current flowing through the coil and in a purely inductive coil which has zero resistance or losses, this impedance (which can be a complex number) is equal to its inductive reactance. Also reactance is represented by a vector as it has both a magnitude and a direction (angle). Consider the circuit below.
AC Inductance with a Sinusoidal Supply
This simple circuit above consists of a pure inductance of L Henries ( H ), connected across a sinusoidal voltage given by the expression: V(t) = V_{max} sin ωt. When the switch is closed this sinusoidal voltage will cause a current to flow and rise from zero to its maximum value. This rise or change in the current will induce a magnetic field within the coil which in turn will oppose or restrict this change in the current.
But before the current has had time to reach its maximum value as it would in a DC circuit, the voltage changes polarity causing the current to change direction. This change in the other direction once again being delayed by the selfinduced back emf in the coil, and in a circuit containing a pure inductance only, the current is delayed by 90^{o}.
The applied voltage reaches its maximum positive value a quarter ( 1/4ƒ ) of a cycle earlier than the current reaches its maximum positive value, in other words, a voltage applied to a purely inductive circuit “LEADS” the current by a quarter of a cycle or 90^{o} as shown below.
Sinusoidal Waveforms for AC Inductance
This effect can also be represented by a phasor diagram were in a purely inductive circuit the voltage “LEADS” the current by 90^{o}. But by using the voltage as our reference, we can also say that the current “LAGS” the voltage by one quarter of a cycle or 90^{o} as shown in the vector diagram below.
Phasor Diagram for AC Inductance
So for a pure loss less inductor, V_{L} “leads” I_{L} by 90^{o}, or we can say that I_{L} “lags” V_{L} by 90^{o}.
There are many different ways to remember the phase relationship between the voltage and current flowing through a pure inductor circuit, but one very simple and easy to remember way is to use the mnemonic expression “ELI” (pronounced Ellie as in the girls name). ELI stands for Electromotive force first in an AC inductance, L before the current I. In other words, voltage before the current in an inductor, E, L, I equals “ELI”, and whichever phase angle the voltage starts at, this expression always holds true for a pure inductor circuit.
The Effect of Frequency on Inductive Reactance
When a 50Hz supply is connected across a suitable AC Inductance, the current will be delayed by 90^{o} as described previously and will obtain a peak value of I amps before the voltage reverses polarity at the end of each half cycle, i.e. the current rises up to its maximum value in “T secs“.
If we now apply a 100Hz supply of the same peak voltage to the coil, the current will still be delayed by 90^{o} but its maximum value will be lower than the 50Hz value because the time it requires to reach its maximum value has been reduced due to the increase in frequency because now it only has “1/2 T secs” to reach its peak value. Also, the rate of change of the flux within the coil has also increased due to the increase in frequency.
Then from the above equation for inductive reactance, it can be seen that if either the Frequency OR the Inductance is increased the overall inductive reactance value of the coil would also increase. As the frequency increases and approaches infinity, the inductors reactance and therefore its impedance would also increase towards infinity acting like an open circuit.
Likewise, as the frequency approaches zero or DC, the inductors reactance would also decrease to zero, acting like a short circuit. This means then that inductive reactance is “directly proportional to frequency” and has a small value at low frequencies and a high value at higher frequencies as shown.
Inductive Reactance against Frequency

The inductive reactance of an inductor increases as the frequency across it increases therefore inductive reactance is proportional to frequency ( X_{L} α ƒ ) as the back emf generated in the inductor is equal to its inductance multiplied by the rate of change of current in the inductor.
Also as the frequency increases the current flowing through the inductor also reduces in value.

We can present the effect of very low and very high frequencies on a the reactance of a pure AC Inductance as follows:
In an AC circuit containing pure inductance the following formula applies:
So how did we arrive at this equation. Well the self induced emf in the inductor is determined by Faraday’s Law that produces the effect of selfinduction in the inductor due to the rate of change of the current and the maximum value of the induced emf will correspond to the maximum rate of change. Then the voltage in the inductor coil is given as:
then the voltage across an AC inductance will be defined as:
Where: V_{L} = IωL which is the voltage amplitude and θ = + 90^{o} which is the phase difference or phase angle between the voltage and current.
In the Phasor Domain
In the phasor domain the voltage across the coil is given as:
and in Polar Form this would be written as: X_{L}∠90^{o} where:
AC through a Series R + L Circuit
We have seen above that the current flowing through a purely inductive coil lags the voltage by 90^{o} and when we say a purely inductive coil we mean one that has no ohmic resistance and therefore, no I^{2}R losses. But in the real world, it is impossible to have a purely AC Inductance only.
All electrical coils, relays, solenoids and transformers will have a certain amount of resistance no matter how small associated with the coil turns of wire being used. This is because copper wire has resistivity. Then we can consider our inductive coil as being one that has a resistance, R in series with an inductance, L producing what can be loosely called an “impure inductance”.
If the coil has some “internal” resistance then we need to represent the total impedance of the coil as a resistance in series with an inductance and in an AC circuit that contains both inductance, L and resistance, R the voltage, V across the combination will be the phasor sum of the two component voltages, V_{R} and V_{L}.
This means then that the current flowing through the coil will still lag the voltage, but by an amount less than 90^{o} depending upon the values of V_{R} and V_{L}, the phasor sum. The new angle between the voltage and the current waveforms gives us their Phase Difference which as we know is the phase angle of the circuit given the Greek symbol phi, Φ.
Consider the circuit below were a pure noninductive resistance, R is connected in series with a pure inductance, L.
Series ResistanceInductance Circuit
In the RL series circuit above, we can see that the current is common to both the resistance and the inductance while the voltage is made up of the two component voltages, V_{R} and V_{L}. The resulting voltage of these two components can be found either mathematically or by drawing a vector diagram. To be able to produce the vector diagram a reference or common component must be found and in a series AC circuit the current is the reference source as the same current flows through the resistance and the inductance. The individual vector diagrams for a pure resistance and a pure inductance are given as:
Vector Diagrams for the Two Pure Components
We can see from above and from our previous tutorial about AC Resistance that the voltage and current in a resistive circuit are both in phase and therefore vector V_{R} is drawn superimposed to scale onto the current vector. Also from above it is known that the current lags the voltage in an AC inductance (pure) circuit therefore vector V_{L} is drawn 90^{o} in front of the current and to the same scale as V_{R} as shown.
Vector Diagram of the Resultant Voltage
From the vector diagram above, we can see that line OB is the horizontal current reference and line OA is the voltage across the resistive component which is inphase with the current. Line OC shows the inductive voltage which is 90^{o} in front of the current therefore it can still be seen that the current lags the purely inductive voltage by 90^{o}. Line OD gives us the resulting supply voltage. Then:
 V equals the r.m.s value of the applied voltage.
 I equals the r.m.s. value of the series current.
 V_{R} equals the I.R voltage drop across the resistance which is inphase with the current.
 V_{L} equals the I.X_{L} voltage drop across the inductance which leads the current by 90^{o}.
As the current lags the voltage in a pure inductance by exactly 90^{o} the resultant phasor diagram drawn from the individual voltage drops V_{R} and V_{L} represents a right angled voltage triangle shown above as OAD. Then we can also use Pythagoras theorem to mathematically find the value of this resultant voltage across the resistor/inductor ( RL ) circuit.
As V_{R} = I.R and V_{L} = I.X_{L} the applied voltage will be the vector sum of the two as follows:
The quantity represents the impedance, Z of the circuit.
The Impedance of an AC Inductance
Impedance, Z is the “TOTAL” opposition to current flowing in an AC circuit that contains both Resistance, ( the real part ) and Reactance ( the imaginary part ). Impedance also has the units of Ohms, Ω‘s. Impedance depends upon the frequency, ω of the circuit as this affects the circuits reactive components and in a series circuit all the resistive and reactive impedance’s add together.
Impedance can also be represented by a complex number, Z = R + jX_{L} but it is not a phasor, it is the result of two or more phasors combined together. If we divide the sides of the voltage triangle above by I, another triangle is obtained whose sides represent the resistance, reactance and impedance of the circuit as shown below.
The RL Impedance Triangle
Then: ( Impedance )^{2} = ( Resistance )^{2} + ( j Reactance )^{2} where j represents the 90^{o} phase shift.
This means that the positive phase angle, θ between the voltage and current is given as.
Phase Angle
While our example above represents a simple nonpure AC inductance, if two or more inductive coils are connected together in series or a single coil is connected in series with many noninductive resistances, then the total resistance for the resistive elements would be equal to: R_{1} + R_{2} + R_{3} etc, giving a total resistive value for the circuit.
Likewise, the total reactance for the inductive elements would be equal to: X_{1} + X_{2} + X_{3} etc, giving a total reactance value for the circuit. This way a circuit containing many chokes, coils and resistors can be easily reduced down to an impedance value, Z comprising of a single resistance in series with a single reactance, Z^{2} = R^{2} + X^{2}.
AC Inductance Example No1
In the following circuit, the supply voltage is defined as: V_{(t)} = 230 sin( 314t  30^{o} ) and L = 2.2H. Determine the value of the current flowing through the coil and draw the resulting phasor diagram.
The voltage across the coil will be the same as the supply voltage. Converting this time domain value into polar form gives us: V_{L} = 230 ∠30^{o} (v). The inductive reactance of the coil is: X_{L} = ωL = 314 x 2.2 = 690Ω. Then the current flowing through the coil can be found using Ohms law as:
With the current lagging the voltage by 90^{o} the phasor diagram will be.
AC Inductance Example No2
A coil has a resistance of 30Ω and an inductance of 0.5H. If the current flowing through the coil is 4amps. What will be the value of the supply voltage if its frequency is 50Hz.
The impedance of the circuit will be:
Then the voltage drops across each component is calculated as:
The phase angle between the current and supply voltage is calculated as:
The phasor diagram will be.
In the next tutorial about AC Capacitance we will look at the Voltagecurrent relationship of a capacitor when a steady state sinusoidal AC waveform is applied to it along with its phasor diagram representation for both pure and nonpure capacitors.
As we can see For example No2 AC inductance,the VL lead Vs by 11 degree(9079).Thus what exactly mean?The system are good or not?
What it means is the example circuit is heavily inductive as it has a lagging power factor of 0.18 then power factor correction is required.
problems and multiple choice questions and answers
prof: hany frawillh
The magnitude of current is always shown larger than that of voltage in the phasor diagram. Please correct them!
Nice…
High frequency inductors and capacitors are commonly placed with
silver because
a. Reduce their DC resistances
b. Reduce their AC resistances
c. Increase their DC resistances
d. Increase their DC resistances
Please explain me about synchronous reactance in a 3 phase alternator.
Hello Venkat, An electrical machine consists of many turns or coils of wire producing a magnetic flux in the windings and as such has both resistance, R and inductive reactance, XL. If a synchronous generator is operated with noload attached it produces a terminal voltage because as the rotor spins, the rotating field cuts the armature conductors, inducing a voltage. As the load increases to a shortcircuit condition the generators terminal voltage decreases (voltage regulation) due to the resistive I^2.R losses but also due to the internal inductive and armature reactances of the windings because of the connected loads power factor.
Then the total voltage drop in a synchronous alternator under load is IZs = IRs + jIXs. As in most practical machines the resistance R is usually very small compared to the synchronous reactance, Xs so the synchronous impedance, Zs of the machine is equal to the synchronous reactance, and the synchronous reactance is proportional to the current.
Hi Scott. You talked about back emf in DC circuit lots of time. That’s what exactly cause the current cannot increase or decrease instantly. what about back emf in AC circuit.
Consider an ideal solenoid (no resistance, no leakage reactance etc.) connected across an AC supply. The back EMF induced in it will be exactly equal and in opposite direction to the source voltage (which means that when a certain terminal of the AC supply is positive, the side of solenoid connected with it would also be positive, and vice versa).
My question is, how will current flow at all when the EMFs of AC source and solenoid are cancelling each other out? It’s like having having a circuit with only two batteries and terminals of similar polarities shorted with each other.
Could you please explain it a little bit?
due to batteries of same emf by short circuiting no current will flow because there must be some potential difference tothe flow of current
As stated above, when an inductor is connected to an AC voltage, it creates a magnetic field. As the AC voltage changes polarity, it causes this magnetic field to expand and collapse in unison with the supply. This results in a backemf being produced as a result of the magnetic field developed around the coil. The amount of voltage induced in the coil depends on the rate of change of the magnetic field. The faster the magnetic field expands and collapses, the greater the induced voltage. The total effective voltage across the inductor is the difference between the applied voltage and the induced voltage. Then the induced voltage is always less than the applied voltage.
Also, inductance presents an opposition to the change in current flow in a coil. The voltage induced in an inductor does not depend on the amount of current but on the rate of change of the current and in a purely inductive circuit (containing no resistance), the voltage across an inductor will lead the inductor current by 90 deg. For an inductive coil, this opposition is known as inductive reactance (XL). Even though current is limited by inductive reactance, voltage is not. Inductive reactance also depends on frequency as XL = 2pifL, and the greater the frequency the greater its value and so the greater its current limiting action.
That answer is really in details. Thank you very much!!!
But I always can find some book writing VL=E=applied voltage if there is only a inductor in series with an AC supply.
So your KEY point is “actually back EMF is LESS than applied voltage in magnitude” which probably because not all the changing flux filed can fully induce EMF.
And since you said the total effective voltage is the difference between back emf and the applied voltage. How come we calculate the power is not the difference voltage times the current?
Thank you for this but I am not your private teacher and will not continue, please do some of your own research.
Sorry Storr. My bad! automatically changed by my computer!!
Thank you anyway, Scott. Your website is really great!
i like your teaching method.i will wel come of more article.
Very good explanation. Rwandan Engineer.
Excellent explanation,,,,,the sequence of explanation is excellent