Thevenin's Theorem: Solving Homework Statement on Load Current

[magician]

Ok.

I haven't checked your maths, but using that, you will now have a Voltage, Vth connected to j2.4, connected to Z_load, with current, i through the load.

Calculate Z_load (pretty simple)

Hence current in load is:

Vth = ...

 

[The_daddy_2012]

The current in the load will be

Vth/(Zth+ZL) which from this I'm going to go for.

664 + j249/ j2.4 + 40.95 + j28.68

Therefore

IL= 7.36 - j3.95
In polar 8.35angle28.2

Therefore

IL= root2 x 8.35 sin (100 pi t + 28.2)

Is that the right way?

 

[magician]

Try:

Vth = i * 2.4j + i * Z_load

 

[magician]

And PF is lagging...

50< - cos^-1 . 0.7

 

[The_daddy_2012]

Ok, I'm going to start again and work my way through making sure I get each stage right. Then I'll post my calculations and results.

Thank you magician, you've been a massive help! :)

 

[The_daddy_2012]

Just a quick one, the -v1+ j4 x I + j6 x I + v2 that's from nodal analysis isant it?

 

[The_daddy_2012]

Ok, what I have from the stage where the voltages are converted to polar/rectangular

V1= j415
V2= 415

Using

-v1 + (j4 x i) + (j6 x i) + v2

i = 41.5 + j41.5

Therefore:

Vth = j6 x i + v2

Now I have:

Vth = 166 + j249

Zth = j2.4

ZL = 35 + j 35.7

Current flowing through load is

iL = Vth / (Zth + ZL)

iL = 5.71 + j0.89 or 5.77angle8.86 or root2 x 5.77 sin (100 pi t + 8.86)

I thought I had it right, but then did the superposition theorem in the next question and the answers are completely different.

 

[gneill]

Okay, so your Thevenin approach looks good, and your answer looks fine.

For the superposition version, can you first describe in broad strokes what your approach is?

 

[dave pallamino]

Hi Guys,

I've pretty much got my head around everything up to finding the current for the Thevenin model... -v1 + (j4 x i) + (j6 x i) + v2 = 0 ... -415 + (j4 * I) + (j6 * I) + 415 = 0

but I'm really struggling with how to approach the complex math to achieve i = 41.5 + j41.5

I'm not looking for a complete breakdown but please could someone show me how to approach this?

Thank you

 

[KatieMariie]

Hello,

I too am having a terrible time with this question, I've had a look through this post but the only thing i can gather is that the folders I've been sent really are glorified paper weights!

If someone could talk me though and teach me the theory behind this that would be amazing, as i really am at a loss and am too stubborn to move on until this question is completed.

Thank you!

 

[KatieMariie]

I've found the equivalent Rt as j2.4 but I'm afraid I'm unsure as what to do next?

 

[KatieMariie]

Okay so I've got up to..

V1 = j415
V2 = 415

But when substituted into the circuit equation for 'i' ( -V1+(j4*i)+(j6*i)+V2 = 0) I'm getting i = -41.5 + j41.5, where am i going wrong?

 

[KatieMariie]

I've completed parts a) and b) thanks to some heavy swatting and a whole load of coffee.

now to part c) anyone willing to lend me a hand with this one?

 

[Gremlin]

Okay so I've got up to..

V1 = j415
V2 = 415

But when substituted into the circuit equation for 'i' ( -V1+(j4*i)+(j6*i)+V2 = 0) I'm getting i = -41.5 + j41.5, where am i going wrong?

I too am getting I = -41.5 + j41.5.

-V1 + (j4 * I) + (j6 * I) + V2 = 0
-j415 + (j4 * I) + (j6 * I) + 415 = 0
j10 * I = -415 + j415
I = (-415 + j415) / j10
I = -41.5 + j41.5

Why am i going wrong there?

 

[The Electrician]

I too am getting I = -41.5 + j41.5.

-V1 + (j4 * I) + (j6 * I) + V2 = 0
-j415 + (j4 * I) + (j6 * I) + 415 = 0
j10 * I = -415 + j415
I = (-415 + j415) / j10
I = -41.5 + j41.5

Why am i going wrong there?

I = (-415 + j415) / j10 can be rewritten as:

I = -415 / j10 + j415/j10 and further as:

I = 1/j * -415/10 + j/j * 415/10

If 1/j = -j and j/j = 1, what do you get if you carry out the arithmetic?

It really helps to have a calculator that can do complex arithmetic.

 

[Gremlin]

Thanks.

 

[Gremlin]

Thevenin equivalent voltage is

Vth = 6j * i + v2
Vth = 299.26∠56.31

What formula have you used to calculate this?

 

[NascentOxygen]

What formula have you used to calculate this?

Are you asking how to go from this rectangular form: Vth = 6j * i + v2
to this polar form: Vth = 299.26∠56.31 ?

Or are you asking about the substituted values for i and v2?

 

[Gremlin]

No i can convert rectangular to polar, I'm looking to find out how you go about getting Vth when you've multiple sources of voltage.

 

[Gremlin]

It's ok, I worked it out. This site was useful for anyone struggling:

http://www.electronics-tutorials.ws/dccircuits/dcp_7.html

 

[Gremlin]

Homework Statement

FIGURE 1 shows a 50 Ω load being fed from two voltage sources via
their associated reactances. Determine the current i flowing in the load by:

(a) applying Thévenin’stheorem
(b) applying the superposition theorem
(c) by transforming the two voltage sources and their associated
reactances into current sources (and thus form a pair of Norton
generators)
View attachment 74253
[/B]

Would i be correct in saying that i = 5.71 + j0.892 A?

That's from Vth / (Rth + Rl) = 166 + j249 / (0 + j2.4 + 35 + j35.71).

 

[The Electrician]

That's correct; now all you have left is part (b) and (c).

 

[MrBondx]

Here is a link for anyone still struggling, it will cover the grey areas. I found it helpful, I hope you will too. Check out the 2nd/last example he gives -http://youtu.be/25axDabtoFk

 

[KatieMariie]

How did you guys get on with part c? I've got so far as a refined circuit with a current source in parallel with Rn (j2.4) and the 35+j35.707 Load, but can't seem to get past this as my numbers don't add up! I'm not getting the same as i did in a or b?

 

[The Electrician]

How did you guys get on with part c? I've got so far as a refined circuit with a current source in parallel with Rn (j2.4) and the 35+j35.707 Load, but can't seem to get past this as my numbers don't add up! I'm not getting the same as i did in a or b?

What is the value of the current source? How did you determine it? It''s easier to see where you've gone wrong when you show your work.

 

[KatieMariie]

Hi, apologies for the lack of detail in the first post.

So i have 103.75-j69.16 for the current source, as stated above, and also have the circuit written out as the current source with Rn on one parallel branch, and the Load on the other.

To convert the current to voltage i used (103.73-j69.16)*(j2.4) = 165.98 + j248.95, so this becomes the voltage and the circuit is then rewritten in series. Using ohms law, the load resistance and this voltage, the current value comes out at 5.87 + j1.11, which is close, but not exactly what I'm looking for?

 

[KatieMariie]

Hi, apologies for the lack of detail in the first post.

So i have 103.75-j69.16 for the current source, as stated above, and also have the circuit written out as the current source with Rn on one parallel branch, and the Load on the other.

To convert the current to voltage i used (103.73-j69.16)*(j2.4) = 165.98 + j248.95, so this becomes the voltage and the circuit is then rewritten in series. Using ohms law, the load resistance and this voltage, the current value comes out at 5.87 + j1.11, which is close, but not exactly what I'm looking for?

Sorry, not 'as above', that applies to another thread!

 

[The Electrician]

With the angle and the magnitude you have the load impedance in polar form. Thus
$$Z_L = 50 \Omega ~~\angle 45.57°$$
You can convert this to rectangular form if you need it that way.

By the way, before you dive into the calculations for the Thevenin equivalent, take a close look at the definitions of the voltage supplies. While they both have the same frequencies they won't have the same phase angle (why is that do you think?).

Hi, apologies for the lack of detail in the first post.

So i have 103.75-j69.16 for the current source, as stated above, and also have the circuit written out as the current source with Rn on one parallel branch, and the Load on the other.

To convert the current to voltage i used (103.73-j69.16)*(j2.4) = 165.98 + j248.95, so this becomes the voltage and the circuit is then rewritten in series. Using ohms law, the load resistance and this voltage, the current value comes out at 5.87 + j1.11, which is close, but not exactly what I'm looking for?

You need to include the j2.4 ohms in series with the load impedance, all driven by the voltage of 166 + j249 volts.

 

[KatieMariie]

Thankyou, I've realized my mistake, I've had the correct method all along, just one of my values was out by 0.1. Typical.

Thanks again.

 

[Gremlin]

Homework Statement

FIGURE 1 shows a 50 Ω load being fed from two voltage sources via
their associated reactances. Determine the current i flowing in the load by:[/B]

(c) by transforming the two voltage sources and their associated
reactances into current sources (and thus form a pair of Norton
generators)
View attachment 74253

Straight question: would it be seen as incorrect here if i just converted the thevenin's equivalent voltage that i calculated in thevenin's equivalent circuit in part a) using the I_N = V_TH/R_TH formula?