Working with complex numbers

[snarkysparky]

i ran in LTSpice also.  I get 10A there

[orolo]

Here is my simulation. 586.8986V is 415*sqrt(2). The inductances are for the given reactances at 50 hertz. The complex load as calculated by the OP.

[IanB]

ZL = 35.5 + 35.7J   # Complex load.

If the complex load is 50 \$ \Omega \$ with 0.7 p.f. lagging, shouldn't this mean that \$ \cos \phi = 0.7 \text{ (lagging)} \$ ?

So \$ Z_L = 50(\cos \phi - j \sin \phi) = 50(0.7 - j0.714) \$

[orolo]

ZL = 35.5 + 35.7J   # Complex load.

If the complex load is 50 \$ \Omega \$ with 0.7 p.f. lagging, shouldn't this mean that \$ \cos \phi = 0.7 \text{ (lagging)} \$ ?

So \$ Z_L = 50(\cos \phi - j \sin \phi) = 50(0.7 - j0.714) \$

In fact, yes. I've been using the OP calculated values for the circuit, to match his original simulation, and since the discussion was around the Thevenin equivalents. The load should be capacitive, and in that case the RMS current, if I'm not mistaken, is 6.15 amps RMS. It just requires changing the code from 35.5 + 35.7J to 35.5 - 35.7J. Once the algebra of the problem has been solved, the rest is quite easy.

[Benta]

By the way, what's the significance of the "0.7 p.f. lag" on the diagram? Is that an important piece of information that needs to be worked into the solution?

In other words, is that not actually a resistor, but rather a "black box" with a complex impedance?

That's an extremely good question, I wondered about that myself. But not knowing the symbol usage in Simon's textbook, impossible to answer.
It could be a simple resistor, but more likely a "black box" symbolizing, eg, a motor with PF. No way of knowing without more information.

[Simon]

Yes the load is complex, but I was of the understanding that a lagging PF is inductive, ie the current follows or lags the voltage applied.

[Simon]

By the way, what's the significance of the "0.7 p.f. lag" on the diagram? Is that an important piece of information that needs to be worked into the solution?

In other words, is that not actually a resistor, but rather a "black box" with a complex impedance?

That's an extremely good question, I wondered about that myself. But not knowing the symbol usage in Simon's textbook, impossible to answer.
It could be a simple resistor, but more likely a "black box" symbolizing, eg, a motor with PF. No way of knowing without more information.

The 50 ohm impedence and lagging power factor of 0.7 iss that is needed to convert it to a resistor and inductor or capacitor.

[Simon]

Using Thevenin I get 5.99A so not far off the superposition method giving 6.02, reasuring.

[mstevens]

Correct.

In other words, is that not actually a resistor, but rather a "black box" with a complex impedance?

[Simon]

Using Thevenin I get 5.99A so not far off the superposition method giving 6.02, reasuring.

Actually a bit less because I forgot to add the i2.4 to the impedence at the end but hey ho I think I am done with it.

[sibeen]

I'll throw my two cents worth in.

With the diagram showing that the complex impedance is 50 ohms 0.7 lagging this requires that the impedance have a positive j figure. So when we have a simple ohms law I = V/Z, then when the Z has a +J the current ends up with a -j answer, and is therefore lagging. So the correct impedance should be Z = 35 + 35.7J.

V1 = 415. V2 = -j 415 (setting the voltages)

A simple Nodal analysis then gives us the equation:

(N - V1)/4j + N/(35 + 35.7J) + (N-V2)/6j = 0

Solving for the single node gives us N = 235.28 - 168.14j

The current through the complex load is therefore:

Iz = (235.28 - 168.14j)/(35 + 35.7J) = 0.893 - 5.72J

|Iz| = 5.78


Now to do a sanity check and make sure that the load current is actually lagging by the amount we need it to be.

The angle of the node voltage N is arg(N) = -35.55 deg. The angle of the current is arg(Iz) = -81.12 deg.

arg(N) - arg(Iz) = 45.57 deg.

cos (45.57) = 0.7 and the current is lagging the voltage, so all looks good.

BTW, I'm an old fart who uses mathcad, so that's why my maths is laid out as it is.

[The Electrician]

This seems to be a popular problem.  It has been beat to death on another forum: https://www.physicsforums.com/threads/thevenins-theorem.775385/

[sibeen]

Just to be pedantic,and to try to avoid some confusion, in my solution I set V1 = 415. V2 = -j 415, thereby setting V1 as the reference phasor.

This then gives the result as I showed:

Iz = (235.28 - 168.14j)/(35 + 35.7J) = 0.893 - 5.72J

If the reference phasor is given as V2 then we have V1 =j* 415. V2 = 415

This then gives the result of:

N1= 168.14 + 235.28j

Iz = (168.14 + 235.28j)/(35 + 35.7J) = 5.72 + 0.89j.

Your reference phasor makes a difference to what answer you calculate.

[Simon]

This seems to be a popular problem.  It has been beat to death on another forum: https://www.physicsforums.com/threads/thevenins-theorem.775385/

thats because it is a course assignment question for the only university that can be bothered to do a distance learning HNC so there is only one assignment people are trying to do at home so always the same question is asked about. Which is why no one has asked for an outrigfht solution just confirmstion on the direction they are going in.

I take "i" to mean that it is 90 degrees ahead of the real term voltage. To a degree the questions seem made up in such a way that it's hard to get them wrong or not be able to assign some competence to the student, but then this is yhe british education system.

[Simon]

and the guy in the physics forum really is clueless and has not read the course material very well. Granted it is a bit of a mess (long winded mathematical proofs with little preactical explanation on what to do with it) but power factor and what it means were clearly explained, although when you have dundreds of pages of drivel on PDF or in HTML locating the crucial few sentences that explain what to go do with it is quite hard. I bought a book by john bird to use as a reference.

[The Electrician]

To complete the computation:

Veq = 166 + 249J
Req = 2.4 Ohm

The resistor sees a voltage v_load = Veq * Zload / (Zload + Req) = 152.6 + 245.9J    (Zload = 35.5 + 35.7J)

Some of your numerical results are a little off because you have Zload = 35.5 + 35.7J, when in fact it's Zload = 35.0 + 35.7J

[Simon]

yes it looks like they didn't know which way to go. 0.7 is just shy of 0.707 which gets you 45 degrees phasor angle.

[sibeen]

yes it looks like they didn't know which way to go. 0.7 is just shy of 0.707 which gets you 45 degrees phasor angle.

*Cough*

Or perhaps 0.7, or 0.8, or even 0.9 are what are common power factor style questions, or even industry standards. Nearly every generator in the world is rated at 0.8 lagging PF. The industry didn't do this because it was close to 37 degrees.