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Any piece of wire of length that conducts current is an inductor. A piece of wire stretched across the room will be an inductor with one turn and an air core. Its inductance will depend on its length and thickness. This is because any current will form a magnetic field capable of storing energy. If this wire is now wound around several times, the inductance increases and the component becomes smaller. The inductance can also be increased by winding the coils around magnetic materials such as iron, iron powder, ferrite (a ceramic material containing iron oxide), nickel, cobalt or oxides of some other metals.

There is no such thing as a perfect inductor. The most significant effect will be the resistance of the wire and capacitive effects between any windings.

Perfect inductors do not dissipate power. This includes a straight wire but this is far from a perfect inductor. The inductance will be so small as will the resistance of the wire so, if there is enough current available, the current through the resistive part of the impedance will be enough to melt the wire.

Inductors do not dissipate power because they store energy in a magnetic field on one half a cycle and return that energy to the circuit on the other half.


When a DC voltage is applied to a coil, a magnetic field will build up until the current through the coil is limited by the resistance or the wire. This magnetic field will take some time to build up and so the current will not immediately be at its maximum.

This magnetic field will remain in place until the voltage is removed. At this time, the magnetic field will collapse but the current will continue until it does. The voltage at each end of the coil will become whatever it needs to be to maintain or remain consistent with this current as it collapses.

When an AC voltage is applied, the coil will charge in both directions as the voltage swings in both directions. Because the resistance of a perfect inductor is zero, the current will continue to build up so long as the voltage is positive. Consider the following diagram.

Current v Voltage in an inductor.

NOTE: This is the continuation of previous wave forms. The initial current is negative because the coil has been charged in the previous cycle.

At point A, the absolute value of the voltage is at a maximum. The coil is therefore charging up and the current is moving upwards as rapidly as it can but hasn't had time to fully charge but has had time to reach zero.

As the voltage between point A and point B falls, so does the RATE at which the coil charges and so at point B, where the voltage is 0, the coil neither continues to charge nor discharge. In a perfect coil, current will continue to increase while there is a positive voltage because there is zero resistance so it will continue to charge right up to point B. After point B, the voltage is negative so the coil current will start to charge in the opposite diection. Coil current is still positive because it hasn't had time to discharge to zero yet.

Rule:- In an inductive circuit, the applied voltage leads the current. For a perfect inductor this will be by and angle of Π/2 (90o).


The inductance of a coil represents its opposition to a change in current. This is inpdependant of frequency and is measured in Henries. There are many things which will affect a coil's inductance. Increasing the number of turns, increasing the diameter or decreasing the length will all increase the inductance. The magnetic properties of the surrounding material will also affect the inductance.

A couple of calculators for inductors with air cores are at INDUCTOR CALCULATOR 1 and INDUCTOR CALCULATOR 2.

As descibed above, an inductor will resist a change in current. The higher the frequency, the faster the voltage moves from one level to another making it more difficult for the current to change. The amount of current is reduced with increasing frequency and therefore the impedance of the inductor has increased. The impedance of a single perfect inductor is called it's reactance given by the formula

XL = + ω x L

XL is the inductive reactance,
ω is the anuglar frequency (2Πf) and
L is the inductance.
This magical formula simply comes from the definitions of each term. The plus sign differentiates Inductive and Capacitive reactance.

NOTE:- Reactance and impedance are two different things. Reactance only has magnitude (including sign) and can be used to calculate the current through a coil, capacitor or combination provided the phase angle is always ±Π/2. Impedance is a complex quantity including both reactive and resistive components even though one component might be 0. In the case of a purely reactive circuit, Impedance = Reactance. In the case of a purely resistive circuit Impedance = Resistance. In any other case |Impedance| = √ ( reactance2 + resistance2 ) and the phase angle of resulting impedance is arctan(reactance/resistance).


In parallel, there will be current through both or all coils. Since the phase of this current is the same for all, the current can simply be added. If two coils are in parallel, the current through each is given by:-

IL1 = V / XL1 ... WHILE ... IL2 = V / XL2 ... GIVING ... I = V / XL1 + V / XL2

I is the total current,
IL1 is the current through coil 1,
IL2 is the current through coil 2,
V is the applied voltage,
XL1 is the reactance of coil 1 and
XL2 is the reactance of coil 2.

The total reactance, XL,will depend on the current and voltage thus:-

XL = V / I

Substituting I for V / XL1 + V / XL2 we get:-

XL = V / ( V / XL1 + V / XL2 ) ... OR ... XL = V / V x ( 1/XL1 + 1/XL2 )

Finally cancelling out the voltage we get the formula for combined reactance of two inductors in parallel:-

XL = 1 / ( 1/XL1 + 1/XL2 )

Because the phase angle is the same and because XL = ω x L, substituting XL for ω x L, XL1 for ω x L1 and XL2 for ω x L2 respectively, after cancelling out the we get simply:-

ω x L1 = 1 / ( 1 / ( ω x L1 ) + ( 1 / ω x L2 ) ) ... OR ... L = 1 / ( 1 / L1 + 1 / L2 )

INDUCTORS IN SERIESInductors in series

Two inductors in series will both impede the current. Because the phase angle is the same, the respective reactances can be simply added thus:-

XL = XL1 + XL2

Since inductive reactance XL is ω x L, substituting XL for ω x L, XL1 for ω x L1 and XL2 for ω x L2 respectively we get:-

ω x L = ω x L1 + ω x L2 ... OR ... L = L1 + L2

NOTE:- This formula is for two separate inductors. If both are wound on the same former, mutual inductance will take effect and the impedance will multiply by 4 for double the turns.

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