# Complex forms of electrical laws

1. Jun 16, 2015

### ahmed markhoos

Hi,

I'm trying this summer to finish my mathematical methods book. I'm investigating right now the chapter of complex numbers, the end of the chapter has some applications in electricity and how can complex numbers make the work easier.

The problem is that I didn't found it easier nor reasonable.

I have two main questions. First the book says, instead of writing $$I = I_0 sin(wt)$$ we can write it as $$I = I_0 e^{iwt}$$ but we know that $$e^{iwt} = cos(wt)+isin(wt)$$, so how the imaginary number and cosine disappear?

Second, how that can be helpful for solving problems? any examples?

2. Jun 16, 2015

### Hesch

Complex numbers are mainly used for two purposes:

1) The grid has three phases with the phase voltages: 230/ 0°, 230/ 120°, 230/ 240°. Thus the currents in a 3-phase circuit connected to the grid also have complex values.
If the phase voltage is 230V ( from phase to neutral ), why is the voltage between two phases not 460V or 0V? Try to calculate this voltage by complex numbers.

2) Components in a circuit as capacitors or inductances has complex impedances: ZC = 1/(jωC) or 1/(sC), ZL = jωL or sL. So when you calculate currents and voltages in circuits containing such components, it's much easier to calculate them by means of complex numbers. Your calculator will do the work for you. You don't have to start a fight with Pythagoras, triangle calculations, etc.

3. Jun 16, 2015

### ahmed markhoos

Thanks,

still cannot find reasonable explanation for why does $$I = I_0 e^{iwt} = I_0 sin(wt)$$ it did appeared in various occasions like the some of light waves, instead of using $$Σsin(t+nδ)$$ we use $$Σe^{i(t+nδ)}$$

4. Jun 16, 2015

### Hesch

You are the one, that is right here: ( error in your textbook ).

I0*eiwt = I0(cos(wt) + jsin(wt) )

5. Jun 16, 2015

### ahmed markhoos

They used it many times, they ignored the cosine term and the i's. I still don't know way.

Mathematical Methods In the physical sciences, by Mary Boas 3rd edition

6. Jun 16, 2015

### Hesch

You could call "I0*sin(wt)" the reactive current

"I0*cos(wt)" is the active current.

7. Jun 17, 2015

In my opinion, it is only a semantic issue. Actually the real a.c. sinusoidal current as time function it is i=sqrt(2)*I*sin(wt+f) where i=instantaneous current value I=effective current value [the r.m.s-root mean square], and f it is the angle at start of record when t=0.

If f=90 then you may present the instantaneous current as i=sqrt(2)*I*cos(wt).

The complex presentation introduces an imaginary part in order to facilitate calculation in

sinusoidal current form [harmonic 1 or fundamental] and forgetting the sqrt(2).

Conventional the real part is connected to cosines and the [artificial] imaginary part to sinus.

8. Jun 17, 2015

### Svein

The imaginary part of the expression $I=I_{0}e^{i\omega t}$ specifies the reactive current, which must be handled in a different way than the real part - the active current. The reactive current cannot do any work, but is part of the total current across a power grid. That is why you have different values for active power (measured in Watt) and apparent power (measured in VA).

9. Jun 17, 2015

I agree with you Svein,in principle.
However, in my opinion ,first of all, if one represents a phasor [what is currently termed as “vector”] in real and imaginary space it is not always connected with “reactive property” of the phasor. We may, for instance represent phase A voltage as V+0j then the phase B voltage as V[cos (240o)+sin(240o)j] and C as V[cos(120o)+sin(120o)j] and still no reactive power it is involved.
Second-much noise about nothing-we are speaking about an imaginary power since reactive power does not exist physically but only mathematically.
We could consider as reactive part of the current only if we consider all voltage as “phase zero” and then all current-active as phase zero and reactive as 90o-lagging or leading.
At the beginning was only D.C. current. The power was calculated simply as V*I.
Compared with A.C. [On D.C.] here we have only active power, the reactive is always zero and the apparent is the same as active. We have also magnetic field around the conductor when current flow through it and also we have electric field between capacitor plates if a potential difference is between them. But these fields do not require power to exist.
The power which heat a resistance is I^2*R. The instant current value is the same with the average-or r.m.s. and we may use upper case in any case.
In A.C. we may define a value of a D.C. which may heat the resistance with the same power-the r.m.s and it is written with upper case letter as D.C..
Then we may consider i=sqrt(2)*I*sin(wt+f) or i=sqrt(2)*I*cos(wt) as I said.
Since the instant power is v*i [v-instant; i=instant] the total power in a cycle will be integral(v.i)=V*I*cos(f).Here f is the angular difference between v and i at the same time t.
In a.c. system here the field requires power to exist-variable. However it is only a loan for a half of a cycle. In the next half of a cycle the field refunds it. So no "reactive" power requires more active [ mechanical or thermal] power to be transferred here.
However the current level rises with the field intensity and produces losses in all resistive elements through which the current flows. Nothing it is new here I think.

10. Jun 17, 2015

### debelino

There is no reactive current, its just a mathematical trick to get the solution easier. Phasors active and reactive power, phasors.... its just an abstraction in order to simplify calculations
watch this video

11. Jun 17, 2015

### Hesch

I disagree in that: No imaginary power → no energy.
The power is there physically, but integrated as for a period, it will become zero. During the period this imaginary power is integrated to electrical/magnetic energy as for a capacitor/inductance. The current is phase shifted ( 90° ) which leads to that power is going in one direction in one half of the period, then delivered back the next half of a period.

The energy exists ( try to put 10A ac through an ideal inductance = 1Henry, switch of the supply at that instance where the (imagenary) current has its maximum, see what happens ). To obtain energy, (imaginary) power is required.

12. Jun 18, 2015

Thank you Hesch. Still I see it is a semantic issue. However I agree with you, of course.

Thank you debelino. It is very interesting indeed.

13. Jun 18, 2015

### Staff: Mentor

The thing that is so simplifying, yet easy to overlook is that. P=IV always on an instantaneous basis. That applies to resistiors, inductors and capacitors alike. No real or imaginary components arise.

Only when we want to take averages across a whole number of cycles (i.e. AC analysis), do we encounter complex values with real and imaginary components.

So, one could spin the semantics either way. VARs are very "real" in my world, but choosing AC analysis over instantaneous analysis is a "trick" that makes my job easier.

I love what Professor Suskind says, "Physicists are not concerned with what is true, they are interested in what is useful."