Inductance impedance relationship problems

Impedance vs frequency (video) | Khan Academy

inductance impedance relationship problems

Resistance is a value which measure how much the component “resist” the passage of . The formula to calculate inductance: . A number of them are rife with spelling problems and I to find it very bothersome to tell the truth. Expressed mathematically, the relationship between the voltage dropped across the inductor and rate of current change through the inductor is as such. Calculate current and/or voltage in simple inductive, capacitive, and resistive circuits. Entering the frequency and inductance into Equation {X}_{L}=2\pi \text{ fL} . Construct a problem in which you calculate the relative reduction in voltage of.

So, for example, if we have a cosine wave, if we have a cosine as a function of time, omega t, we can express that in terms of complex exponentials like this: And what I'm gonna do now is I'm gonna say, let's look at what happens when we use this as an input signal.

This is not a real input signal, it's an imaginary, rotating vector. But if I have two of these, I can reassemble them into a cosine wave. And we like to use these exponentials, because they go through the differential equations of a circuit really easily. These are the inputs that we know how to solve when we do differential equations. So what I'm gonna do is develop the i-v equations for the resistor, inductor, and capacitor in terms of this kind of an input, when the voltage or the current looks like this, what do those equations look like?

So we're gonna start with, we'll start with the resistor. And we know from that that just Ohm's law is v equals i times R. And just for the moment I'm gonna assume that the current, let's assume that i equals e to the plus j omega t.

Electrical impedance - Wikipedia

So if this is i, what is v for our resistor? Well, we just plug i in here and we get v equals R times e to the plus j omega t. Now I'm gonna do something that looks like it's a little too simple, but it's gonna get interesting soon.

I wanna look at the ratio of voltage to current. In this situation where we're driving with this complex exponential. So voltage turned out to be R e to the plus j omega t and that's the voltage I get if I put current though the resistor of e to the plus j omega t. And what does that equal to? Well, these two are the same so they cancel, and I get the ratio of voltage to current is R for a resistor. So for our resistor we just proved that v over i equals R.

So this isn't news, we didn't make a discovery here, this is just Ohm's law. And for a resistor, the voltage over the current is always equal to the resistance.

Impedance vs frequency

This is gonna get more interesting now as we go do inductors and capcitors. So let's do an inductor. It has a value of L henries. And for an inductor we know that voltage equals L di dt. All right, and let's do the same thing again.

Let i equal e to the plus j omega t. So it's a complex exponential current that we're forcing through our inductor. And let's go ahead and work out what v is. So v equals L times d dt of this value here, e to the plus j omega t. Or v equals, now we take the derivative and the j omega term comes down to multiply L, so we get j omega L times what?

Times the same thing, e to the plus j omega t, this is the beautiful thing about exponentials, they give us back themselves. All right, now let's do this. Let's take, once more, what's the ratio of voltage to current?

And that equals, here's the voltage, j omega L times e to the plus j omega t, and let's divide that by i which is i is right here, i is e to the plus j omega t.

inductance impedance relationship problems

And we get v over i equals j omega L. So now we have an equation for v over i for an inductor. And this is interesting, this time we have the inductance value which we expected, and there's also this omega, j omega term that comes in.

So this tells us this is frequency, omega is frequency. So this tells us that the ratio of v to i for an inductor is dependent on frequency. Now we'll do the same thing for a capacitor. So here's a capacitor. And there's it's capacitance value in farads. And for the capacitor, we know that i equals C times dv dt. And this time let's let v of t, let's let v equal e to the plus j omega t.

  • Resistance, Capacitance, Inductance, Impedance and Reactance
  • AC Inductor Circuits
  • Electrical impedance

So this time we're gonna force a voltage across our capacitor that is this imaginary, this complex exponential. This assumes power out is close to power in. In fact the impedance relationship between primary and secondary is related to turns ratio squared.

inductance impedance relationship problems

For instance if it is a In summary, for a power transformer, you'd like the primary inductance to be infinite but that is impractical so you live with something that doesn't take too much current when the secondary is open circuit.

The off-load current that flows is real current taken from the AC power and if everyone had low-impedance transformers the electricity companies would be supplying a load of current that doesn't get them revenue.

This is a slight exaggeration but not far off the truth. On industrial sites power factor correction is used to minimize this effect but that's a whole new story! And if your transformer primary was ohm impedance you'd be seeing something less than half your AC voltage applied.

If R1 was zero then you'd see exactly half. As regards saturation I've shown the equivalent circuit of a transformer below. Note that saturation is caused by the current flowing through the magnetizing inductor which is nothing to do with load current: Imagine two coils sharing the same magnetic core.

Ignore magnetization currents and losses. The primary is turns and the secondary is 10 turns.