So far we have looked at the behaviour of inductors connected to DC supplies and hopefully by now we know that when a DC voltage is applied across an inductor, the growth of the current through it is not instant but is determined by the inductors self-induced or back emf value.

Also we saw that the inductors current continues to rise until it reaches its maximum steady state condition after five time constants. The maximum current flowing through an inductive coil is limited only by the resistive part of the coils windings in Ohms, and as we know from Ohms law, this is determined by the ratio of voltage over current, V/R.

When an alternating or AC voltage is applied across an inductor the flow of current through it behaves very differently to that of an applied DC voltage. The effect of a sinusoidal supply produces a phase difference between the voltage and the current waveforms. Now in an AC circuit, the opposition to current flow through the coils windings not only depends upon the inductance of the coil but also the frequency of the AC waveform.

The opposition to current flowing through a coil in an AC circuit is determined by the AC resistance, more commonly known as **Impedance** (Z), of the circuit. But resistance is always associated with DC circuits so to distinguish DC resistance from AC resistance the term **Reactance** is generally used.

Just like resistance, the value of reactance is also measured in Ohm’s but is given the symbol X, (uppercase letter “X”), to distinguish it from a purely resistive value.

As the component we are interested in is an inductor, the reactance of an inductor is therefore called “Inductive Reactance”. In other words, an inductors electrical resistance when used in an AC circuit is called **Inductive Reactance**.

**Inductive Reactance** which is given the symbol X_{L}, is the property in an AC circuit which opposes the change in the current. In our tutorials about Capacitors in AC Circuits, we saw that in a purely capacitive circuit, the current I_{C} “LEADS” the voltage by 90^{o}. In a purely inductive AC circuit the exact opposite is true, the current I_{L} “LAGS” the applied voltage by 90^{o}, or (π/2 rads).

In the purely inductive circuit above, the inductor is connected directly across the AC supply voltage. As the supply voltage increases and decreases with the frequency, the self-induced back emf also increases and decreases in the coil with respect to this change.

We know that this self-induced emf is directly proportional to the rate of change of the current through the coil and is at its greatest as the supply voltage crosses over from its positive half cycle to its negative half cycle or vice versa at points, 0^{o} and 180^{o} along the sine wave.

Consequently, the minimum rate of change of the voltage occurs when the AC sine wave crosses over at its maximum or minimum peak voltage level. At these positions in the cycle the maximum or minimum currents are flowing through the inductor circuit and this is shown below.

These voltage and current waveforms show that for a purely inductive circuit the current lags the voltage by 90^{o}. Likewise, we can also say that the voltage leads the current by 90^{o}. Either way the general expression is that the current lags as shown in the vector diagram. Here the current vector and the voltage vector are shown displaced by 90^{o}. *The current lags the voltage*.

We can also write this statement as, V_{L} = 0^{o} and I_{L} = -90^{o} with respect to the voltage, V_{L}. If the voltage waveform is classed as a sine wave then the current, I_{L} can be classed as a negative cosine and we can define the value of the current at any point in time as being:

Where: ω is in radians per second and t is in seconds.

Since the current always lags the voltage by 90^{o} in a purely inductive circuit, we can find the phase of the current by knowing the phase of the voltage or vice versa. So if we know the value of V_{L}, then I_{L} must lag by 90^{o}. Likewise, if we know the value of I_{L} then V_{L} must therefore lead by 90^{o}. Then this ratio of voltage to current in an inductive circuit will produce an equation that defines the **Inductive Reactance**, **X _{L}** of the coil.

We can rewrite the above equation for inductive reactance into a more familiar form that uses the ordinary frequency of the supply instead of the angular frequency in radians, ω and this is given as:

Where: ƒ is the Frequency and L is the Inductance of the Coil and 2πƒ = ω.

From the above equation for inductive reactance, it can be seen that if either of the **Frequency** or **Inductance** was increased the overall inductive reactance value would also increase. As the frequency approaches infinity the inductors reactance would also increase to infinity acting like an open circuit.

However, as the frequency approaches zero or DC, the inductors reactance would decrease to zero, acting like a short circuit. This means then that inductive reactance is “proportional” to frequency. In other words, inductive reactance increases with frequency resulting in X_{L} being small at low frequencies and X_{L} being high at high frequencies and this demonstrated in the following graph:

The slope shows that the “Inductive Reactance” of an inductor increases as the supply frequency across it increases. Therefore |

Then we can see that at DC an inductor has zero reactance (short-circuit), at high frequencies an inductor has infinite reactance (open-circuit).

A coil of inductance 150mH and zero resistance is connected across a 100V, 50Hz supply. Calculate the inductive reactance of the coil and the current flowing through it.

So far we have considered a purely inductive coil, but it is impossible to have a pure inductance as all coils, relays or solenoids will have a certain amount of resistance no matter how small associated with the coils turns of wire being used. Then we can consider our simple coil as being a resistance in series with an inductance.

In an AC circuit that contains both inductance, L and resistance, R the voltage, V 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 new phase angle between the voltage and the current is known as the phase angle of the circuit and is given the Greek symbol phi, Φ.

To be able to produce a vector diagram of the relationship between the voltage and the current, a reference or common component must be found. In a series connected R-L circuit the current is common as the same current flows through each component. The vector of this reference quantity is generally drawn horizontally from left to right.

From our tutorials about resistors and capacitors we know that the current and voltage in a resistive AC circuit are both “in-phase” and therefore vector, V_{R} is drawn superimposed to scale on the current or reference line.

We also know from above, that the current “lags” the voltage in a purely inductive circuit and therefore vector, V_{L} is drawn 90^{o} in front of the current reference and to the same scale as V_{R} and this is shown below.

In the vector diagram above it can be seen that line OB represents the current reference line, line OA is the voltage of the resistive component and which is in-phase with the current. Line OC shows the inductive voltage which is 90^{o} in front of the current, therefore it can be seen that the current lags the voltage by 90^{o}. Line OD gives us the resultant or supply voltage across the circuit. The voltage triangle is derived from Pythagoras theorem and is given as:

In a DC circuit, the ratio of voltage to current is called resistance. However, in an AC circuit this ratio is known as **Impedance**, **Z** with units again in Ohms. Impedance is the total resistance to current flow in an “AC circuit” containing both resistance and inductive reactance. If we divide the sides of the voltage triangle above by the current, another triangle is obtained whose sides represent the resistance, reactance and impedance of the coil. This new triangle is called an “Impedance Triangle”

A solenoid coil has a resistance of 30 Ohms and an inductance of 0.5H. If the current flowing through the coil is 4 amps. Calculate,

a) The voltage of the supply if the frequency is 50Hz.

b) The phase angle between the voltage and the current.

There is one other type of triangle configuration that we can use for an inductive circuit and that is of the “Power Triangle”. The power in an inductive circuit is known as **Reactive Power** or **volt-amps reactive**, symbol **Var** which is measured in volt-amps. In a RL series AC circuit, the current lags the supply voltage by an angle of Φ^{o}.

In a purely inductive AC circuit the current will be out-of-phase by a full 90^{o} to the supply voltage. As such, the total reactive power consumed by the coil will be equal to zero as any consumed power is cancelled out by the generated self-induced emf power. In other words, the net power in watts consumed by a pure inductor at the end of one complete cycle is zero, as energy is both taken from the supply and returned to it.

The Reactive Power, ( Q ) of a coil can be given as: I^{2} x X_{L} (similar to I^{2}R in a DC circuit). Then the three sides of a power triangle in an AC circuit are represented by apparent power, ( S ), real power, ( P ) and the reactive power, ( Q ) as shown.

Note that an actual inductor or coil will consume power in watts due to the resistance of the windings creating an impedance, Z.

Error! Please fill all fields.

I have a choke (inductor) rated at 16.5 hy and a dc measurement of 100 ohms. I need to know how much current it will safely carry. Is it possible to determine this? Any help..Thanks. Geo.

I(dc) = V/R

I(ac) = V/Z

Z^2 = R^2 + XL^2

XL = 2pi.f.L

sir i need answer to this question…..

the f.f at inductive reatance is called what……….?

a) p.f b) active p.f c) reactive p.f d) inductive p.f

Whatever, f.f (form factor?) is, I guess C as its reactive

I just need to know when we are calculating phase angle , how we select it is a tan, cos or sin. Please guide.

thanks

It depends which combinations of resistance, reactance or impedance you are using to calculate it.

What type of power can be used if when a 3 phase system is being used and induction occurs when E and I cross. What power is derived when the E and I cross? Even if minimal or what is derived?

This is an odd question, it depends were they cross. There is more information here about Power.

Why we r taking negative sign for inductive reactance xL in admittance method plzzz tel me

Admittance is the inverse of impedance, then the angle for a given admittance is the negative of the same angle for the corresponding impedance. Consequently, an admittance angle for the branch currents is positive for a capacitive susceptance, (+jB) and negative for an inductive susceptance, (-jB).

I want know how draw a graph between frequency and Inductive reactance

As shown in this tutorial, Inductive Reactance is given as: 2pi.f.L If you know the value of the Inductance, L then calculate XL for each frequency value with Frequency along the X-axis and Inductive Reactance along the Y-axis.

I wanted to know how to make an inductor and further using it in my PCB board circuit plz suggest me a complete video.

How do we make a graph xl

There are a number of ways to do this but the easiest is to use a fixed voltage of choice with a variable frequency. Connect the test inductor in series with a fixed known value of resistance across the supply. Measure the voltage and current across the components at various frequencies and calculate to plot a graph of XL against frequency.

OK.I UNDERSTAND THE TOPICS ABOUT INDUCTIVE REACTANCE!!!!!!!!!!

Need help….if v=Vmaxsin100(omega*t)….inductance is 28 milli Henry….how the hell am i gonna find max current???

|Z|=Veff/Ieff -> Ieff = Veff / |Z| = Veff / (L*omega)

then (assuming sinusoidal voltage and current)

Imax = Vmax / (L*omega)