You should possibly consider looking at the transformer in two ways; one without a load and one with a load on the secondary.

Without a load on the secondary, the transformer is just an inductor and if you have components (such as L1 and R1) in series with the primary, the voltage developed on the primary will not be the full AC amount from your generator. It's a simple case of calculating the impedances and volt-drops. This is with the secondary unconnected remember.

The primary has inductance like any other coil but, for a transformer to more effective, it is desirable for the primary's self inductance to be high in power applications. If you looked at how much current flowed into the primary (secondary open circuit) you would find that the current was small compared to when driving a load on the secondary and it may have an inductance of several henries.

With 10 henries inductance, at 50Hz the impedance is 3142 ohms and from 230VAC would take a current of 73mA - that current through R1 (10 ohm) hardly drops any voltage.

It's a different matter when there is a load on the secondary. If the turns ratio is 1:1 and you have 100 ohms on the secondary, it is reasonable to argue that the impedance presented to the primary circuit is also 100 ohms. This assumes power out is close to power in. In fact the impedance relationship between primary and secondary is related to turns ratio squared. For instance if it is a 10:1 step-down transformer with a load of 100 ohms, the equivalent impedance at the primary is 10k ohms i.e. 10 x 10 x 100.

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 100 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: -

Here is a good document from Elliott Sound Products and please note what it says about maximum flux density therefore saturation:

**Why doesn't the core saturate more under load conditions?** Imagine two coils sharing the same magnetic core. Ignore magnetization currents and losses. The primary is 100 turns and the secondary is 10 turns. If the secondary load current is 10A, the primary current must be 1A and the ampere-turns is therefore the same on both coils. Are these ampere-turns additive or subtractive? They are subtractive and this can easily be seen with dot notation....

If current is flowing into the dot on the primary, current is flowing out of the dot on the secondary and this produces opposing fluxes in the magnetic material. When you think about this you have to be consistent and use the right-hand rule to see that the two fluxes oppose and cancel.

Because the dots are at the top on both coils, they are wound in the same direction and the currents are flowing in (primary) and out (secondary) therefore due to the RH rule the fluxes (due to ampere-turns) are cancelled.

Let's look at the formula and equivalent circuit for a transmission line.

(1) Impedance rather than reactance.

Reactance refers to the opposition to the change in current (of an inductor) or voltage (for a capacitor) - single components. The transmission line has \$R,L\$ and \$C\$ components - impedance is the ratio of voltage phasor to current phasor.

(2) It is \$50\Omega\$ because the ratio of inductance to capacitance per unit length produces that value. As \$R << j\omega L\$ and \$G \to 0\$, these values can be ignored and so the expression reduces to \$\sqrt{L/C}\$ (frequency independent).

(3) Nope, but it's generally a good idea to keep things as standard as possible. You may find it difficult to find a suitable connector for your \$167\Omega\$ transmission line. There's also a lot of information available for designing **standard** transmission lines on PCBs, etc. The magic number in my book is 376.73031... the impedance of free space. Now without that one we'd live in a different universe.

(4) Going back to the formula. At low frequencies \$R\$ may be significant as the reactance of the inductor will be small). At very high frequencies the dielectric losses may become significant.

## Best Answer

Diagram!

This is for a complex impedance:

Resistance \$R\$ is in phase with the applied voltage, so the vector points in the same X direction. The impedance of a capacitor is almost completely reactive, i.e. its resistive part is much smaller than the \$\dfrac{1}{j \omega C} \$. The \$j\$ causes a \$\theta\$ = 90° rotation, and since the \$j\$ (= \$\sqrt{-1}\$) is in the denominator the angle is negative \$\left( \dfrac{1}{j} = -j \right) \$.

To calculate the current \$ I = \dfrac{U}{Z} \$, we note that when dividing by an impedance with angle \$\theta\$ we subtract the angle from our reference, so that the angle's sign is inverted.

The result shows how for a capacitive load the current leads the voltage by an angle \$ \theta\$, where \$ 0 \le \theta \le 90°\$.

For inductive loads a similar diagram can be drawn, only \$ j \omega L\$ points in the opposite direction of \$\dfrac{1}{j \omega C} \$, and the current will trail the voltage.

edit(after your edit of the question)So, resistance will cause the current to be in phase with the voltage. If there's an imaginary term (the \$j\$) then that term represents the reactance, either capacitive or inductive, and

In an ideal world, if you don't have capacitors or coils you wouldn't have reactance either. But a circuit may have parasitic impedance: the length of a PCB trace will cause an inductive reactance (it behaves as a coil), and two adjacent traces will have a capacitive reactance (they behave as a capacitor). Parasitic impedances are unintentional, and most of the time a nuisance, though sometimes a designer can make good use of them.

You can measure components impedances with an

RLC-meter, which will give you resistance in series or parallel with a reactance (inductive or capacitive).Reactance will show as a phase shift in voltage or current. This phase shift can be shown on an oscilloscope in X-Y mode; a zero phase shift will show a straight line, a 90° phase shift will show a circle, anything in between will give you an ellipse.