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Why doesn't an inductor act as a closed circuit to AC?

by Vishera
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Vishera
#1
Apr2-14, 02:43 PM
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I understand that an inductor acts as a closed circuit to DC because it's just a coiled wire but why doesn't it act the same way for AC? What does it act as in AC?
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NateTG
#2
Apr2-14, 02:51 PM
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You can think of the inductor as working a bit like inertia in stream of fluid: when the rate of flow changes, it 'resists' the change. It creates a voltage difference proportional to the rate of change in current - so when the current is constant then there is no voltage.
Vishera
#3
Apr2-14, 03:09 PM
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Quote Quote by NateTG View Post
You can think of the inductor as working a bit like inertia in stream of fluid: when the rate of flow changes, it 'resists' the change. It creates a voltage difference proportional to the rate of change in current - so when the current is constant then there is no voltage.
Wow. That was elegantly explained.

psparky
#4
Apr3-14, 10:03 AM
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Why doesn't an inductor act as a closed circuit to AC?

Couple different ways to look at this.
The impedance of a INDUCTOR is JWL

w=omega....or radians per second of your input signal.....such as the 377 in a wall outlet: 170sin377t volts
So clearly, as the frequency changes, so does the impedance.

What's the frequency of DC? Zero right? When you plug zero into JWL, you get zero resistance, or a "closed" circuit as you say above.

Then there is this formula:
L*di/d(t)=v(t)

In simple terms, You could say that any change in current makes a voltage accross the inductor.
In DC, there is no change in current, just a flat line. No change in current, no voltage....hence your closed or short circuit once again.

Works the same but opposite for Capacitors.
Impedance is 1/(jwc)

Impedance changes with frequency again.
Again, if w=0 you will have a infinite resistance....or open circuit.

C*dv/dt=it

Again, change in voltage induces current. In DC, no change in voltage, no current, open circuit......

Also important to note that the "j" in the impedance, shifts the current out of phase with voltage. If it is purely inductive circuit, the current will shift 90 degrees with a vector pointing down. If it is purely capacitive, it will be a current vector pointing straight up. Ussually tho, there is a combination between real and imaginary current....leading to a current angle somewhere between 0 and 90 degrees. The two derivatives above will accomplish the same thing.

One more, the AC voltage in a receptacle has a frequency of 60 hz. So the current changes directions 60 times per second. So when it hits that coil of wire, an electrical field is induced, but then it changes direction inducing the field the other way....over and over. Same for the capacitor, it starts to charge in one direction, then current changes direction inducing the charge the other way...over and over.
jim hardy
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Apr3-14, 02:15 PM
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The impedance of a capacitor inductor is JWL

i do that all the time too.
psparky
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Apr3-14, 02:22 PM
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Quote Quote by jim hardy View Post
The impedance of a capacitor inductor is JWL

i do that all the time too.
True that!
sophiecentaur
#7
Apr4-14, 05:14 AM
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Another alternative way to look at it:
The essence of an Inductor is that, when you pass a current through it, it has a magnetic field around it (this is true even for a 'wire' only it is more for many turns of wire and particularly if there's a magnetic 'core' inside it). Building up the field involves Energy and this energy is 'stored' in a dynamic way, by the Inductor. This Magnetic energy is analogous to Kinetic Energy of a moving mass. Along with this Energy, there is an electromagnetic equivalent to equivalent to the Mass. Applying an AC voltage across an inductor is the equivalent to trying to 'wobble' a massive object. The faster the wobble (higher the AC frequency) and the bigger the mass (Inductance), the smaller the displacements of the wobble (Currents flowing in the Inductor for a given applied voltage).


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