cant understand the idea of maximal power..

[nabliat6]

**broken link removed**
we have R_s>0
R_L>0
X_L is real
Z_L=R_L+jX_L
we are playing with Z_L so in order to have the optimal power on Z_L is the (Z_S*) (the complex opposite of Z_s where the imaginary part is multiplied by minus)

but i cant understand if Z_L should be connected in parallel or otherwise
and the optimal power is 1/(8R)|v_s|^2

when this law is not working?

 

[Hayato]

You have some types of power.
Reactive power (VAr).
Apparent power (VA).
Real power (W).

Only the real power is able to generate work.

Let's imagine:
Zs = 1 + 1j, Zl = 1 + 1j. Vs = (1,0) Volts.

Zt = 2 + 2j omhs
It = 0.5 - 0.5j A

Total power = 0.5 - 0.5j VA = 707.11 /_ -45° mVA

But from those 707.11 mVA that you drained from the supply, you are only be able to use 0.5W to generate work.

Observe that you drained 707.11 mVA of apparent power, but, again, you just use 0.5 W from it.


Now lets imagine:
Zs = 1 + 1j, Zl = 1 - 1j. Vs = (1,0) Volts.

Zt = 2 omhs
It = 0.5 A

Total power = 0.5 W = 500 /_ 0° mW

You drained 500 mW and will use those 500mW to generate work.

 

[nabliat6]

does ZL have to be connected in a circuit way?

 

[Hayato]

Yes, you have to have ZL, otherwise there will be no current flow, therefore, no power.

 

[nabliat6]

sorry i ment
does Zl have to be connected in parallel or in column
?

 

[Hero999]

ZL is always in series with ZS.

The idea is ZS represents the internal impedance of the voltage source i.e. the wiring resistance, ZL represents the load i.e. a light bulb, motor etc.

You can't change the value of ZS because it's built-in to the battery.

If ZL lower than ZS all the power will be dissipated in the battery.

Maximum power transfer from the voltage source to ZL occurs when ZL = ZS. If ZL is greater than ZS the current will be lower so less power will be dissipated.

In general maximum power transfer is undesirable because it's generally only 50% efficient as half the power is dissipated in ZS but this isn't always true, for example the voltage is AC and ZS is reactive. As a general rule of thumb, ZS should be as low as possible relative to ZL to make the system as efficient as possible.

 

[nabliat6]

how can ZL be bigger then ZS
they both are complex numbers?

 

[Hero999]

How does that make any difference?

Forget the complex numbers for now, assume that both are resistive. - it's basic Ohm's law either way.

 

[Hayato]

how can ZL be bigger then ZS
they both are complex numbers?

Complementing what Hero said.

For example, think of a 9V battery.

If you short the positive and negative terminals, then your ZL < ZS.

If you connect a 10k resistence between the terminals, for sure, ZL > ZS.

sorry i ment
does Zl have to be connected in parallel or in column
?

Deppends, just look at Norton and Thevenin theorems.

For voltage sources, ZS is ALWAYS in series with VS, so ZL will always be in SERIES with ZS.

For current sources, ZS is ALWAYS in parallel with IS, so ZL will always be in PARALLEL with ZS.

 

[nabliat6]

still mathematically you cant say that one complex number is bigger then the other?

 

[Hero999]

It see what you mean but don't allow that to confuse you.

Ignore the reactive part which can be easily be removed by connecting an inductor or capacitor in series.

 

[Hayato]

still mathematically you cant say that one complex number is bigger then the other?

You can compare 2 complex numbers by the magnitude or the phase or the both.

For example, 1 + 5j and 5 + 1j, both has the same magnitude = sqrt(26).

3 + 10j has a magnitude of sqrt(109) and 5+9j has a magnitude of sqrt(106).

So, yes, you can say if one is bigger than another.

Remember that a complex number is written in the form of a + b.j :
Magnitude = sqrt (a² + b²)
Phase = arctan (b/a).

 

[nabliat6]

so comparing the real part?

 

[Hayato]

so comparing the real part?

I did not get it. What exactly are you trying to do?

 

[Hero999]

My point is that maximum power transfer only applies to the real part of the impedance because the reactive part doesn't dissipate any power and can easily be canceled.

Suppose you have a 50Hz power supply with an series inductance of 10mH, say it's a transformer or generator and this is the leakage inductance and a series resistance of 1.5Ω.

The reactive internal impedance is 2π50×0.01 = 3.14Ω

If a capacitor with a reactive impedance of 3.14Ω (844µF) is connected in series with the supply, the inductance will be canceled so the output impedance will be a purely resistive 1.5Ω.

Because it's a series resonant circuit, care should be taken with the voltage ratings of the components, for example if 100A flows, the voltage across the capacitor will be 314V.

 

[MrAl]

Hi,


Here's a way to get the max power in a resistor connected in series with another resistor connected to a voltage source.
We call the upper resistor R1 and the lower resistor R2. We want to see what value R2 should be to get max power
in R2.

Start with:
v=Vs*R2/(R1+R2)
i=Vs/(R1+R2)
p=v*i

Then the derivative with respect to R2:
dp/dR2=-(Vs^2*(R2-R1))/(R2+R1)^3

The set dp/dR2 equal to zero:
-(Vs^2*(R2-R1))/(R2+R1)^3=0

Solving, we get:
R2=R1

So the max power occurs in R2 when R2 is equal to R1.

Just to try this in a circuit, we'll say that Vs=20v and R1=10 ohms, and now change R2 to several values and see what happens:

R2=10, P=10, V=10, I=1
R2=9, P=9.9723, V=9.4737, I=1.0526
R2=11, P=9.9773, V=10.4762, I=0.95238

From the above we can see that the power always goes lower when R2 is either a lower or higher value than R1. The power is greatest
when R2 equals R1 exactly.

The attached drawing shows the power as R2 changes from 0 to 50 ohms with R1=10 ohms and Vs=10 volts.

 

[Hayato]

Agreed.

I was not taking in account that you were talking about the corrected power factor impedance. (I was thinking generally).