Understanding processors

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By hubpagesrui

Understanding processors

 The great technical advantage of the alternating current in comparison with the current lies in the possibility of obtaining from the first, any voltage you want, almost without loss, by means of transformers. Ordinarily, the place of use, requires low voltages that are dangerous to the human body (it's common to use voltage of 117 volts and 220 volts).
Furthermore, the transport of electricity from the place of their generation to their use, they should be performed under higher voltages possible (220 000 V or 380 000 V). But for the most economical operation of the machines that produce electricity, it is an average voltage of a few thousand volts. Therefore, throughout the distribution system there is always the need to transform the voltage.

A transformer consists of a core closed upon itself, formed by layers of iron (to reduce losses due to eddy currents), sweet (because they are easily magnetized and desmagnetizável), which is a primary winding (A) and a secondary (B), as illustrated.

 We will consider to start, that the winding (B) is open, ie that he did some stream flows.
Suppose that the terminals of the winding (A) is applied an alternating voltage U1 = Uo.senw.t that produces the same electric current im (current pickup). This current excites the iron core of a magnetic flux F = (FMM) m / Rm = n1.im/Rm, n1 being the number of turns of the primary winding and the reluctance of the Romans iron core. (FMM) m = n1.im is a magnetomotive force.
But as im changing over time, so does the flow F and the flow variable determining each of the n1 turns of primary winding an emf - dF / dt, and a total electromotive force E = - n1.dF / dt. If R1 is the ohmic resistance of the primary winding, we U1 + E = R1.im.
 

However, R1 is always small and we can consider R1 = 0, obtaining then U1 + E = 0, ie,

Uo.senw.t = n1.dF/dt or dF / dt = (Uo/n1). Senw.t

Obtained by integration:

F = - (Uo/w.n1). Cosw.t = (Uo/w.n1). Sen (w.t - p / 2)

The magnetic flux in the iron core has, therefore, the same way that the sinusoidal primary voltage U1, but has its late phase of p / 2 with respect to voltage. The above relationship between flow and F primary voltage U1 is always the primary winding, even when there is current in the secondary winding, since it is a necessary consequence of the relationship always valid U1 + E = 0.

  Now consider the current pickup im producing the flow F. Since F = n1.im/Rm clear that im is fixed at all times the value of F. However, the reluctance Rm depends on the magnetic resistivity m or the magnetic permeability m = 1 / h of iron core, which is undergoing a cyclical pickup permanently.
The permeability, in turn, depends (although not unequivocally) the magnetic excitation, ie, the im ultimately, so that Rm is a function of important (in fact, in a way very similar to Ohm's Law ).
Therefore, the relationship between im and F is very complicated, and as F has a purely sinusoidal development, is not the same with taxes. But, F is the only one that matters. Note that im has a finite value, even if put U1 + E = 0, this is due to the hypothesis R1 = 0. This is a case quite similar to the current induction in a superconducting (R = 0).

The flow F runs through the iron core and thus also passes through n2 of turns of secondary winding; record that started the discussion considering an open circuit in the secondary. The flow F, which varies with time, that induces a secondary winding electromotive force E2 = - n2.dF/dt and therefore dF / dt = E2 / (-n2), which led to the expression Uo.senw.t = n1. dF / dt provide:

             Uo.senw.t = n1.E2 / (-n2) or E2 = - (n2/n1). Uo.senw.t = - (n2/n1). U1 = (n2/n1). Uo sin (wt - p)

Thus, the electromotive force induced in the secondary winding (E2), which in the case of a winding open fully corresponds to the voltage that appears between its terminals (U2) is greater or less than the primary voltage U1 in n2/n1 ratio (ratio of processing), and has a development purely sinusoidal, phase has a delay of ap with respect to primary voltage.

When the secondary winding is being crossed by current i2 (think initially a purely resistive load), it will produce an additional flow F2 in the iron core, which also goes through the primary winding, thereby result disturbed the equilibrium condition in the primary winding, expressed by the equation U1 + E = 0. However, this is restored instantly, because besides the current pickup im, comes from an additional current i1 at the expense of the generator that feeds the primary winding, which is exactly the intensity required to produce the flow F1 that exactly cancels the flow F2 i2. With load, the current in the primary increases!
Thus, regardless of consumption in the iron core is always the flow F determined solely by the primary voltage and the existence of the current im.

The stream valley F2 = F2 and F1 and n2.i2/Rm = n1.i1/Rm, because it is F1 + F2 = 0 we can by n1i1 = - n2i2, ie i1 = - (n2/n1). i2.

At any given moment, the power of the secondary current worth E2.i2 = P2 = - (n2/n1). U1.i2. The power of current i1 (a part of the primary current, because there is still a part im) is P1 = U1.i1 = - (n2/n1). U1.i2. There is, therefore, P1 = P2.
The power supplied by the current i1 of the primary winding is collected in full in the secondary circuit. If you dispense with the current pickup im, a transformer converts a given strain in a different one without loss of energy.
If the iron core magnetic hysteresis not present, would not have to experience a cyclical pickup in the course of a period of alternating current, and the current im would be a pure reactance, so that the average expense for a period would be zero. For its part, a cyclical pickup requires work proportional to the area of the hysteresis loop, which must be made by the current i. However, in practice, im always small in comparison with i1 flowing through the primary winding, which means that the power im involved with is only a small fraction of that total expenditure involved in the primary.

Other causes of energy loss are: the slight dispersion of magnetic lines in the air, the angles of the nucleus, where a few field lines are closed flows through the air outside the primary and secondary windings and the fact that the ohmic resistance the two windings can not be considered strictly zero. However, these losses of energy (and some others not quoted) are very small and a good transformer will operate with a yield of 95%.

We have seen that the secondary electromotive force E2 is due solely to the flow variable over time produced by the current im, which is independent of the load. Because it is not possible to manufacture a transformer that will work economically and is constructed without iron (core), although the two windings are available together as possible. The flow is proportional to the permeability me, so im will be greater the lower the m If we replace the rail, air, im become one m times, ie, hundreds of times more intense and longer represent a tiny fraction of total current, the power is now considerable im2.R1 --- the processor will not result economic.

The advantage presented by the high voltages for the transmission of electric power can be seen from the following considerations: Let R be the resistance of long distance drivers eia intensity of current flowing through them. Driving consumes a power i2.R = DP. Where E is the electromotive force responsible for driving, the total power is P = Ei Therefore, in conducting the fraction disappears DP / P = Ri / E that results unusable. However, the higher E, for a given power P = Ei, the lower the intensity i. Therefore, the relative loss (DP / P) decreases as E increases.

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