Why Does Current Lag Voltage in an Inductive Circuit or Coil?

Can any one tell me why current lags the voltage in any inductive circuit???

i m not sure but i think it is due to lanz's law. as innitianly the magnetic field opposes the flow of currnt which produces it.

When an emf is generated by a change in magnetic flux, the polarity of the induced emf is such that it produces a current whose magnetic field opposes the change which produces it. This opposition means that the "generated" voltage appears and then "opposes" the current flow thus causing the current to lag the voltage.

In a DC circuit, when a voltage source is connected across an inductor, a current will flow through it. This current generates a magnetic field which increases in strength from zero to its maximum value. As this magnetic field is being created it induces a back emf into the inductor coil due to the effects of electromagnetic induction which opposes the main source current flowing through the coil and is given as -L.di/dt

This effect of a current carrying inductor, or coil of inducing a voltage within itself is known as self-inductance. Once the circuit current flowing through the inductor reaches its maximum value a steady state exists between the current and the self generated magnetic field resulting in no induced voltage or current to oppose the main current flow and the inductive coil acts as a short circuit.

However, when an AC voltage is applied to a purely inductive coil or inductor, the same back emf is produced due to the self-inductance of the coil, but as the voltage is sinusoidal, this back emf opposes the rise and fall of the current through the coil during each and every cycle. As a sinusoidal voltage is represented by the expression of: v(t) = Vm.sin(wt) the instantaneous current flowing through the inductor due to its inductive reatance will therefore become i(t) = Im.sin(wt - pi/2).

Then you can see that the current lags behind the applied voltage by a quarter cycle as indicated by the pi/2 part of the current equations. In other words the phase difference between the two is 90 degs with voltage leading.

This site has more information about inductive circuits and "(inductive reactance)":http://www.electronics-tutorials.ws/inductor/AC-inductors.html

The result of all of this is:

Because an inductor is really just a length of wire, it behaves like that after a sufficient amount of time, but at the instant a voltage is applied across an inductor, it behaves like an open circuit (because of _back EMF_ as described by my forummates)! Thus at the moment voltage is applied across an inductor, there is no current, then as time passes, more and more current flows, thus the current lags the voltage.

The opposite is true for a capacitor, which is, essentially, and open circuit, but at time zero, behaves like a conductor! Thus, in a capacitor, the current _leads_ the voltage.

Dear as you said above inductor behaves like an open circuit when voltage applied across it?????
then why all inductive machines have inrush current???
why capacitor is behave like open circuit when we apply voltage across it???

The inrush current is high just to overcome the back emf.

An ordinary electric motor is inductive, because it works by magnetic forces produced by wire coils.

The motor's commutator repeatedly cuts off and re-connects the current, resulting in the current in the coils being intermittent.
Each time it re-connects one of the coils, the current has to build up over a short period of time.

This repeated cut-off and re-connection happens much faster as the motor increases speed, therefore each time it re-connects a coil, the current through that coil has less time to build up before it is disconnected.

Therefore the total average current through the motor becomes less as its speed increases.

When the shaft of the motor is driving a load, the motor is slowed down, so the current increases, taking more power in order to drive that load.
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A capacitor acts like a battery. As it charges up, it's own voltage is acting against (in the opposite direction to) the voltage that is charging it.

When it is uncharged, its voltage is zero, so there is nothing opposing the current flowing into it (except the resistance of the source charging it and the resistance of the circuit wires).

(The internal resistance of the capacitor is usually tiny, just being the resistance of the metal its plates are made of.)

As the capacitor charges, its voltage becomes progressively higher, opposing the voltage that is charging it, thus reducing the current.

When its voltage becomes equal and opposite to the voltage charging it, no more more current can flow.

Oops, forgot to say "ideal inductor/capacitor" The real world has such annoyances as "core saturation" and "stalled current". Also, I'm talking about the "transient response" [i.e. when a constant voltage is applied], NOT, for instance, an AC sinusoidal voltage.

Inrush current has causes in cases where an inductor is not behaving in an Ideal fashion, i.e. not following the formula:



http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indtra.html
http://www.learnabout-electronics.org/ac_theory/dc_ccts45.php

Also, when power is first applied to a [DC] motor at rest, because the motor isn't turning yet, the current that initially flows is the "stall current". But, if you were to look at the "inrush" current graph at startup, you would still see the characteristic logarithmic curve i.e. the current doesn't instantaneously rise to the stall current level--thus, the inductance of the winding behaves like an open circuit at the instant voltage is applied and then rises logarithmically to the current that is limited by only the resistance of the wire. When the shaft begins to turn, magnetics are involved, and it's a whole different beast.

incorrect