Inductor is pretty much and electromagnet

[Farn]

Few real basic questions about inductors/inductance:

As I understand it an inductor is pretty much and electromagnet (coil of wire around a ferromagnetic material). Am I right?

I know inductors resist changes in current. The more inductance a circuit has the greater it resists AC. Does this mean that it actually lowers the peek current in an AC circuit, or does it merely delay the peak current (set it out of phase with voltage)?
For example I have a simple circuit without an inductor that's running off a wall outlet. It draws 1amp of current at its peek. Now, if I were to add an inductor, would the peek be lowered to let's say .5amps, or will the same peek of 1amp still be achieved but a little later?

 

[russ_watters]

You don't need the core to have an inductor - just the coil.

An inductor is a load so it will decrease the amount of current traveling through the circuit - but remember, the current is constant everywhere on a non-branching circuit.

 

[sheldon]

When you have an inductive load, current lag's voltage and when you have a capacitive load, current leads voltage. You’re right about the phase properties. This is important when trying to balance a load in a building or whatever, because you can create very large tank circuits that raise your power factor and KVARS. This creates noise and runaway neutrals. To answer your question on what an inductor is it is anything that can conduct for the most part. A single wire without any turns is still an inductor. It still has Henry's.

 

[Rockazella]

when you have a capacitive load, current leads voltage

How could the current ever lead the voltage? What causes the electrons to flow before they get a push?

 

[sheldon]

here you go.


http://users.erols.com/renau/impedance.html

 

[Ivan Seeking]

Originally posted by Farn
The more inductance a circuit has the greater it resists AC. Does this mean that it actually lowers the peek current in an AC circuit...

Any inductor will have at least a small amount of resistance. So in this way the peak current is always affected slightly by the introduction of another component. In a pure DC circuit, the peak current is otherwise unaffected by an inductor. In an AC circuit, this is true if the inductor becomes saturated - all of the magnetic dipoles of the iron core are fully aligned with the field. At this point any further increases in the current are virtually unaffected. Likewise, as the current decreases, the inductance is not seen until we drop back down below the saturation level. An air core inductor never saturates in practice but has a very small inductance value.

Most importantly, with the exception of the slight resistance added to the circuit, an inductor does not in principle limit the peak current. However, if the frequency is high enough, the peak current is never seen simply because the current does not have enough time to reach its maximum value.

 

[Farn]

Alright thanks fellows!

 

[Rockazella]

sheldon,
I looked at the site but I'm afraid it’s a bit over my head too many numbers too little logic for me. Would you mind explaining how current can lead voltage? Is it actually that there is current increasing before there is a voltage, or is it more just a mathematical way to express it?

 

[Ivan Seeking]

Originally posted by Rockazella
sheldon,
I looked at the site but I'm afraid it’s a bit over my head too many numbers too little logic for me. Would you mind explaining how current can lead voltage? Is it actually that there is current increasing before there is a voltage, or is it more just a mathematical way to express it?

Not on the first cycle. We see the effect on the next cycle as the current from the first lags behind the voltage. This makes it appear to lead the voltage from the next cycle. The current leading the voltage is a function of timing and the delay caused by inductors...and the fact that we don't know which cycle actually causes the current to flow when we measure it. The model is correct but a little misleading. If we apply a DC source, as the voltage is impressed upon the circuit, we will see the voltage first followed by the current rise.

 

[sheldon]

Originally posted by Ivan Seeking
Not on the first cycle. We see the effect on the next cycle as the current from the first lags behind the voltage. This makes it appear to lead the voltage from the next cycle. The current leading the voltage is a function of timing and the delay caused by inductors...and the fact that we don't know which cycle actually causes the current to flow when we measure it. The model is correct but a little misleading. If we apply a DC source, as the voltage is impressed upon the circuit, we will see the voltage first followed by the current rise.

Thanks Ivan, I wasn't sure how to explain it. I would like to add that you can have a lot of current with small voltage, the power remains the same. As the voltage catches up the current decreases making it appear to be leading the voltage. We are also talking about AC not DC so the effects of caps. and coils are different with reactance. When you apply AC to a coil, imagine when the ac is on the rising slope the magnetic field around the coil is building up until the cycle starts on its down slope. The reason the voltage is lagging is because the magnetic field starts to colapse around the coil and is delayed behind the voltage which is inducing the current cycle and causing the current to be behind the voltage. The opposite is true with a capacitor as the voltage rises through a capacitor it builds up charge instead of magnetic storage. When the cycle starts on its down slope, The charge in the capacitor holds the voltage until the current leaves and then it starts to discharge causing the voltage to lag the current. Like I said you can have current with low voltage and visa versa. You still have the same power put the reactance of the circuit will be your load, reactance is like resistance even is explained in ohms but is not actual resistance like in a DC circuit. Reactance is the opposition to the flow of electric current resulting from inductance and capacitance (rather than resistance). I hope this helps and enjoy your questions

 

[mmwave]

Originally posted by Rockazella
How could the current ever lead the voltage? What causes the electrons to flow before they get a push?

I didn't see a direct answer to this question so...

When they say current leads the voltage (or the reverse) they are talking about the voltage drop across the device and the current into the positive terminal.

You are right in thinking that there must be a voltage difference for current to flow at all. If you had an ac voltage source and two resistors in series the voltage & current reach peaks at the same time. Replace one resistor with an inductor and the voltage drop across the inductor will peak before the current peaks. The source voltage and the inductor voltage will peak at the same time however.

 

[Ivan Seeking]

Originally posted by mmwave
I didn't see a direct answer to this question so...

When they say current leads the voltage (or the reverse) they are talking about the voltage drop across the device and the current into the positive terminal.

You are right in thinking that there must be a voltage difference for current to flow at all. If you had an ac voltage source and two resistors in series the voltage & current reach peaks at the same time. Replace one resistor with an inductor and the voltage drop across the inductor will peak before the current peaks. The source voltage and the inductor voltage will peak at the same time however.

Again this is really a function of timing in the circuit. In a capacitive circuit, the voltage across the CAPACITOR lags the current. In a simple circuit, this means that the voltage lags the current across the circuit. This does not mean that the voltage from the source lags the current from the source. It means that as the source voltage rises, current flows into the capacitor. Since the voltage across a cap is given by Q = CV, where Q is charge and C is the capacitance, we see that as charge Q flows into the cap, the voltage across the cap starts to rise; eventually nearly equaling the source voltage if given enough time T. So at first we see a virtual short circuit Q = 0 -> V = 0 across the circuit, and then the voltage across the circuit rises to approach the peak magnitude of the source voltage. So the current flows, and then the voltage rises across the capacitor.

In an inductive circuit, the current lags the voltage due to the inductor’s resistance to changes in current flow. We apply the potential, and then the current rises at a rate determined by the inductance.


I should add that if we consider the internal resistance of the voltage source, this gets a bit more complicated.