hello

I got another one here im not sure about

I have a circuit




values

v1 = 20 + j0
R1 = 15 + j0
L1 = 0 + j10
C1 = 0 - j5

first I added R1 and L1, R1 + L1 = (15+j0) + (0+j10) = 15+j10 = Req

then I found C1 in parallel with this Req, like this

(0-j5) (15-j10)
(15+j10)(15-J10)

so,
15 x 15 = 225
15 x -j10 = -j150
j10 x 15 = +j150
j10 x -j10 = -j^2 100, if -j^2 is equal to -1 it would leave -1 x -100 leaving 100

so that leaves 225 -j150 +j150 +100

-j150 +j150 cancel leaving 225 + 100 = 325 ohms

as far as I know that's it. it says "a)" which leads me to believe that there is a "b)" but its not on there
so, any help or confirmation would be great. I don't know how to confirm this using multisim

thanks

simon

 

hello

I got another one here im not sure about

I have a circuit


View attachment 74855

values

v1 = 20 + j0
R1 = 15 + j0
L1 = 0 + j10
C1 = 0 - j5

first I added R1 and L1, R1 + L1 = (15+j0) + (0+j10) = 15+j10 = Req

then I found C1 in parallel with this Req, like this

(0-j5) (15-j10)
(15+j10)(15-J10)

What is this supposed to be? Are you saying that the equivalent impedance is (0-j5)(15(-j10) divided by (15+j10)(15-j10)? I can't think of what else it might be.

so,
15 x 15 = 225
15 x -j10 = -j150
j10 x 15 = +j150
j10 x -j10 = -j^2 100, if -j^2 is equal to -1 it would leave -1 x -100 leaving 100

so that leaves 225 -j150 +j150 +100

-j150 +j150 cancel leaving 225 + 100 = 325 ohms

You need to track your units instead of just taking on the units that you wish they would be at the end. If you had, you would know that you've made an error?

(15 Ω + j10 Ω)(15 Ω + j10 Ω) = 225 Ω² + j150 Ω² - j150 Ω² +100 Ω² = 325 Ω²

The units do not work out, so you know this is NOT an impedance.

Similarly, if what you started with IS supposed to be a division of those terms, then that turns out to be unitless and, again, you know that it is wrong.

Always, always, ALWAYS track your units THROUGHOUT your work!

The formula for parallel impedances is a direct counterpart to the formula for parallel resistances:

Zeq = (Z1·Z2)/(Z1+Z2)

Your Z1 = -j5 Ω and your Z2 = 15 Ω + j10 Ω.

But even then you are not finished. You were not just asked for the Thevenin impedance, but for the Thevenin equivalent circuit. So you need to find the Thevenin voltage that is in series with the Thevenin impedance.

 

What is this supposed to be? Are you saying that the equivalent impedance is (0-j5)(15(-j10) divided by (15+j10)(15-j10)? I can't think of what else it might be.



You need to track your units instead of just taking on the units that you wish they would be at the end. If you had, you would know that you've made an error?

(15 Ω + j10 Ω)(15 Ω + j10 Ω) = 225 Ω² + j150 Ω² - j150 Ω² +100 Ω² = 325 Ω²

The units do not work out, so you know this is NOT an impedance.

Similarly, if what you started with IS supposed to be a division of those terms, then that turns out to be unitless and, again, you know that it is wrong.

Always, always, ALWAYS track your units THROUGHOUT your work!

The formula for parallel impedances is a direct counterpart to the formula for parallel resistances:

Zeq = (Z1·Z2)/(Z1+Z2)

Your Z1 = -j5 Ω and your Z2 = 15 Ω + j10 Ω.

But even then you are not finished. You were not just asked for the Thevenin impedance, but for the Thevenin equivalent circuit. So you need to find the Thevenin voltage that is in series with the Thevenin impedance.

WOOPS AFLOGGINGDAISY!!!!

I FORGOT SOMETHING!!!!

CHEERS LOVE!!!! BACK IN A MINUTE!!!!


I forgot the 0-j5 and 15 -j10

0 x 15 =0
0 x -j10 = 0
-j5 x 15 = -j75
-j5 x - j10 = j^250?

I then did this j^250 -j75 = -j25?

 

WOOPS AFLOGGINGDAISY!!!!

I FORGOT SOMETHING!!!!

CHEERS LOVE!!!! BACK IN A MINUTE!!!!


I forgot the 0-j5 and 15 -j10

0 x 15 =0
0 x -j10 = 0
-j5 x 15 = -j75
-j5 x - j10 = j^250?

I then did this j^250 -j75 = -j25?

This makes no sense.

-j5 x - j10 = j^250?

How on earth does multiplying -j5 by -j10 result in j raised to the 250th power?

And how do you come by this: j^250 -j75 = -j25?

And you still have a fundamentally wrong answer because you are doing something that will not yield the correct units and you still are failing to catch it because you still won't track units.

 

This makes no sense.

-j5 x - j10 = j^250?

How on earth does multiplying -j5 by -j10 result in j raised to the 250th power?

And how do you come by this: j^250 -j75 = -j25?

And you still have a fundamentally wrong answer because you are doing something that will not yield the correct units and you still are failing to catch it because you still won't track units.

hello

I don't know how to track units, never heard of that.
I thought that -j x -j = j^2 (squared) the two minus's cancel so 5 x 10 = 50
so j (squared) = -1 so -1 x 50 = -50,
so the -j5 x 15 is -j75, I think

so -j75-50 = -j125
this over 325 = -0.384615 this is the equilvalent circuit

I think I have to use polar notation as well so I put 325 and -0.384615 into a calculator and got
r = 325 and theta = -0.0678

 

So what you MEANT to write was (j^2)*50, not j^(250), which is what j^250 is. You need to be much more careful with your expressions.

Units are a factor and, in almost all instances, be treated just like any other factor. For instance, if you have

x = a·b + c·b

You can write this as

x = (a+c)b

So if you have

x = 5 V + 4 V

you simply have a=5, b=V, and c=4, which allows you to go

x = (5+4) V = 9 V

But if you have

x = 5 V + 4 W

Then you can't add these any more than you could add

x = a·b + c·d

Just treat units like they are a variable and follow all the normal rules and you will be fine.

When you have a physical quantity, the units ARE an integral part of the value. If someone asks me how tall I am, I can't just say 72. That's not a height. I am 72 inches.

Similarly, 72 != 6 != 182.88 != 2. But, 72 in = 6 ft = 182.88 cm = 2 yd.

When you work a problem, you WILL make mistakes. Most of those mistakes are algebraic and will usually mess up the units, allowing you to catch the mistake right when you make it and fix it. Even if you don't catch it right away, it will mess up the units on the final answer allowing it to catch the mistake at that point and also make it much easier to track down where the mistake was made. If you just throw units aside and tack on what you think the answer should be at the end, you throw away what is very probably the single most effective error detection tool available to the engineer. That is gross negligence and incompetence as far as I am concerned and, on occasion, courts have agreed.

Now, let's go back to your original post where you had:

(0-j5) (15-j10)
(15+j10)(15-J10)

I asked you what this was and you never answered. Did you mean

Zeq = [(0 - j5) (15 - j10)] / [(15 + j10)(15 - j10)]

If so, then this is GUARANTEED to be wrong, because it is dimensionless. The units, where they there, would completely cancel out.

Zeq = [ (0 - j5) Ω · (15 - j10) Ω ] / [ (15 + j10) Ω · (15 - j10) Ω]

But you need units of impedance. So that means that something is wrong. Not point going any further until that is fixed. I gave you the proper equation for computing the effective impedance of two parallel impedances:

Zeq = (Z1·Z2) / (z1 + z2)

The numerator has units of Ω² while the denominator has units of Ω, and Ω²/Ω = Ω, which is what you need.

 

im sorry but my tutor has shown me this method.
Zeq = [ (0 - j5) Ω · (15 - j10) Ω ] / [ (15 + j10) Ω · (15 - j10) Ω] this is what I meant yes.

sorry that I didn't make it clear.

no the denominator isn't squared in any of the examples my tutor gave me.

is the equation i wrote above correct for finding this out?

 

im sorry but my tutor has shown me this method.
Zeq = [ (0 - j5) Ω · (15 - j10) Ω ] / [ (15 + j10) Ω · (15 - j10) Ω] this is what I meant yes.

sorry that I didn't make it clear.

no the denominator isn't squared in any of the examples my tutor gave me.

is the equation i wrote above correct for finding this out?

No, it is not. The answer is dimensionless and you HAVE to have units of impedance. You don't. So it is wrong.

On top of that, you are multiplying the numerator and denominator both by (15-j10). Why? It just cancels out leaving you with Z1/Z2.

If someone told you that the formula for calculating the area of a football field was area=length/width, you would KNOW it is wrong because that yields a dimensionless result you have HAVE to have units of area-squared.

I gave you the correct formula (twice already):

Zeq = (Z1·Z2) / (z1 + z2)

You have

Z1 = (15 + j10) Ω (the resistor in series with the capacitor)
Z2 = (0 - j5) Ω (the capacitor)

Plug them in:

Zeq = [ (15 + j10) Ω · (0 - j5) Ω ] / [ (15 + j10) Ω + (0 - j5) Ω]

And do the math.

Zeq = [ (15·0) Ω² - j(5·15) Ω² + j(0·10) Ω² - j^2(5·10) Ω² ] / [ (15 + 0) Ω + j(10 - 5) Ω]

Zeq = ( 50 Ω² - j75 Ω² ) / (15 Ω + j5 Ω )

We can factor 5 Ω out of each term in both the numerator and denominator and cancel them, leaving:

Zeq = ( 10 Ω - j15 Ω ) / (3 + j )

If we want to get rid of the complex number in the denominator, then we can multiply both the numerator and the denominator by the complex conjugate of the denominator:

Zeq = [( 10 Ω - j15 Ω )(3 - j)] / [(3 + j )(3 - j)]

Zeq = ( 15 Ω - j55 Ω ) / 10

Zeq = 1.5 Ω - j5.5 Ω = (1.5 - j5.5) Ω

Now we can ask if this answer makes sense.

If we have two impedances in parallel, we expect the equivalent impedance to be dominated by the smaller impedance (just like in the purely resistive case). Here the capacitor impedance is significantly smaller than either the resistor or the inductor, so we expect the result to be dominated by the capacitor and to actually be fairly close to the value of the capacitor's impedance. Which it is. So we have fair confidence that the answer is correct. By the way, when I first did this I got an answer of (10.5-j2.5)Ω, which didn't pass this test, so I went back and looked at my work and found not one, but two mistakes I had made, both of them simple transcription errors that I didn't catch. These checks work!

 

I was using division of complex conjugate numbers

 

Regardless of what you were trying to do, your formula was wrong as indicated by the fact that the units didn't work out.

 

hello WBahn

thanks for getting back to me about this!

im not a math wizard, if honest im not very good at this at all. I had a look at some of my notes and think that I got the wrong idea entirely. we have used this method (division of conjugate) in college but I think it was for something else, components in series perhaps. sorry for any confusion, the fact that you return to help regularly is great I appreciate it, and I imagine its irritating when the person on the other end gets it wrong. which I do, a lot!
thanks for your help and support

all the best

simon

 

Hello there simon,

If you want to get the right answer you have to know the math. If you dont know the math, then you may have to review before you try to calculate anything new.

To help check your result you can use a simulator. The simplest way to do this is to model the Thevenin voltage using two sine voltage sources and one complex impedance, along with some value for the load resistance RL. You must first calculate the Thevenin voltage and series impedance.

The Thevenin voltage source will resolve into two components, one real and one imaginary, each with their own amplitude:
VT=Va+Vb*j

and the series impedance Zs will resolve into two components:
Zs=A+B*j

To model the impedance make A a resistor and B according to the sign of B:
If B is negative, make it a capacitor with value C=1/B,
if B is positive, make it an inductor with value L=B.

To model the Thevenin voltage, make Va a sine wave source with amplitude equal to the real part of VT, and make Vb another sine source with amplitude equal to the imaginary part of VT, and phase shift Vb by pi/2. Make both sources angular frequency w equal to 1. Use a summer to sum the two voltages, and the output becomes the Thevenin voltage VT. If either or both Va and Vb are negative, you'll have to invert them too which can be done several ways.

Also draw in the original circuit and look at the output voltage across the load RL. The time scale will be several seconds because we are working at w=1 which means the frequency f is only 1/(2*pi).

Now when you use VT with Zs in series with the load RL, the load voltage will have the same amplitude and phase shift as the original circuit if you calculated VT and Zs correctly.

You can also do the same thing using current sources if you want to convert the source into a current source. One current source for the real part, one for the imaginary part with a phase shift of pi/2.

It is also a good idea to simply analyze the circuit a second time with the new values VT and Zs and the load Rs and see if it comes out the same as the original circuit when you calculate the output voltage.

Example Thevenin source for VT=-v/10-3*v*j/10:
Here we have:
Va=-v/10
Vb=-3*v/10
To start, we can make Va=1/10 and Vb=3/10 and phase shift Vb by pi/2. Flip both sources on the schematic, and add the two using a summer. The output will be the Thevenin voltage source VT.

Example for Zs=3-5*j:
Rs=3 ohms
C=1/5 Farads
The two are connected in series.

Example for Zs=4+5*j:
Rs=4 ohms
L=5 Henries
The two are connected in series.

Also note that the exponential part of the response is not always going to be the same for the Thevenin circuit, so you need to allow time for the circuit to reach steady state if you do the simulation in the time domain. Once steady state has been reached the two outputs should look identical both in amplitude and in phase.

 

Not a problem -- I appreciate that you keep trying!

The use of the complex conjugate has a few applications in circuit analysis. As used here, the intent is to take a fraction that has a complex number in the denominator and get rid of it. If you multiply both the numerator and the denominator by the complex conjugate of the denominator, you haven't changed the value of anything since you've merely multiplied the whole fraction by one. But now the denominator consists of a complex number multiplied by its conjugate, which is a real number. The other big use is in complex power computations for AC circuits where it creeps into the formulas because we are interested in the differences in the phase angles between the voltage and current waveforms and using the conjugate of the current phasor is a convenient way to perform the subtraction.