Thevenin's Theorem: Solving Homework Statement on Load Current

[gneill]

am I correct then at saying j4 and j6 are the impedances respectively?

Yes. Those are the impedances of the two inductors.

this equation is giving me a hard time haha they only show examples of thevenins theorem using resistors which I totally get but then they throw the odd ball like this and I am completely lost like the other chaps on here...

oh and you multiply j4&j6 to get j24?

The 'j' represents the square root of negative one. That is, the impedances are imaginary values (and are complex in general). So the arithmetic you do with them must follow the rules of complex arithmetic.

 

[magician]

Hey Ebies,

As gneill has already said, "The 'j' represents the square root of negative one."

To help you get started, below is the calculations I used;

1/Z_th = 1/j4 + 1/j6
1/Z_th = (4+6)/j24
1/Z_th = 10/j24

Therefore;
Z_th = j24/10
Z_th = j2.4

Now following the procedure I did above, [post 30] you'll be there in no time, aslong as you take your time with the complex arithmetic.

 

[Ebies]

thanks gneil and magician, your help is much appreciated... I want to try and crack this one out before Wednesday morning as I am heading offshore and won't be able to do any of this whilst I am away and trying to come back to a half completed equation after a few weeks at sea is pants as you forget half of what you have done before you left...

 

[magician]

thanks gneil and magician, your help is much appreciated... I want to try and crack this one out before Wednesday morning as I am heading offshore and won't be able to do any of this whilst I am away and trying to come back to a half completed equation after a few weeks at sea is pants as you forget half of what you have done before you left...

Completly understand Ebies.

Put up what you've got.

Ive had mine marked and it is correct

 

[Electest]

Hi Guys

I'm really trying to get my head round this and would appreciate some pointers. I understand obtaining the Thevenins equivalent impedance (Zt) but am struggling obtaining The Thevenins equivalent Voltage (Vt).

Looking at the circuit drawing the j4Ω & j6Ω inductances act as a potential divider between the 2 voltage sources, do they not?

The voltage across the j6Ω is therefore given by $$\left (v1-v2 \right ) \frac{6}{4+6}$$

where:

v1 = 415∠90 v or 0+j415 v
v2 = 415∠0 v or 415+j0 v

Then I could minus the voltage across the j6Ω from the v2 Source to get my answer?

Thanks in advance

 

[Ebies]

Hi guys, just got back from my off shore trip, been a short one to say the least. I attempted this during my time away and to be completely honest I am still at a loss,

I understand calculating the Zth using product over sum rule,

I have used the p.f. of 0.7 lagging to calculate theta=45.572996 deg
I understand that v1 = sqrt2 * 415sin(100pi.t +90) and v2 = sqrt2 * 415sin(100pi.t) respectively

what I do not get however is:

- How do you get from v1 = sqrt2 * 415sin(100pi.t +90) and v2 = sqrt2 * 415sin(100pi.t) to either Polar or Rectangular form? I an change from Polar to Rect and vice versa but not completely sure how to get there from what's given by V1 and V2 so just a nudge in the right direction here will help...

thank you kindly

 

[gneill]

If you have a time signal of the form $A sin(\omega t + \phi)$ or $A cos(\omega t + \phi)$ then you can use the phasor form $A ∠ \phi$. So then you have the polar form of the signal in the frequency domain. You should already know how to convert a polar form to rectangular form.

Note that it is critical that when you convert a phasor result back to the time domain that you employ the same trig function that you started with otherwise you'll introduce a spurious phase difference (the phase difference between a sine and cosine function).

 

[Ebies]

thanks for that gneill

 

[Ebies]

out of interest can I make the following statement and would it ring true for imaginary numbers? this would give me my Thevein Voltage when the circuit has been redrawn as originally posted but without the Load of 50Ω

the voltage drop across j6Ω would be 1.5x more than the voltage drop across j4Ω and as such write it as:

Vj6Ω= (V2-V1)x(j6/j10)...
I know this will work with normal numbers for both voltage supplies and resistances but not sure if it will work when you use imaginary numbers...

 

[gneill]

I think it should work if the numbers are purely real or purely imaginary (as in this case). Else there will be "interaction" between the real and imaginary components, and the phase of the result will change.

 

[Ebies]

not completely sure it works this way when complex arithmic is involved... I Vth=352.14∠-45 and clearly it is not the same as Vth in Magician's earlier post... could however be down to me miscalculating... Can anyone please confirm my working out on this...

(V2-V1)*(j6/j10)
=[(415+j0)+(-0-j415)]*(0.6)
=[(415-j0)+(0-j415)]*(0.6)
=415-j415*(0.6)
=249-j249

which in turn gives me: Vth=352.14∠-45

Sorry I keep on posting on this post but I am trying to calculate this equation and need to ensure I understand it...

 

[gneill]

You've calculated the potential across the j6 Ω inductor with an assumed polarity of - on its left and + on its right end (since you've assumed current flow to be counterclockwise by your choice of "V2 - V1"), but you've not yet found the Thevenin voltage. For that you need to take into account the fact that V2 is between that inductor and the reference node (ground).

 

[Ebies]

so does that mean I need to calculate the p.d over the j4Ω and then sum the two resultant p.d's together? I am a bit lost here now... can you point me in the right direction here please gneill?

 

[gneill]

You are looking for the open terminal voltage where the load sits. That is, the Thevenin voltage. To do that you should take a "KVL walk" from one terminal to the other, summing the potential changes. You've found the potential change across the j6 Ohm impedance, so you could choose the path that includes it (in blue below):

 

[Ebies]

so in essence I stick to the blue path to calc j4Ω as well yes? or do I follow the red path for the second part?

 

[gneill]

so n essence I stick to the blue path to calc j4Ω as well yes?

You've already taken into account the effect of the j4Ω inductor when you found the potential drop across the j6Ω one. Your calculation effectively found the total current due to the sources in series with the inductors and multiplied that by the j6Ω impedance (a voltage divider scenario).

Either path will yield the same result, thanks to KVL holding around the loop.

I think my preferred approach would have been to just use nodal analysis at the junction of the inductors, so one equation with one unknown (Vth) to solve.

 

[Ebies]

This is as far as I have managed o get and now I am stuck with the equation. not sure how to deal with and equation that has a mix of unknowns and imaginary numbers in it... also attached is my diagram that I am using so the equation makes sense... I spose if I can learn where I am going wrong now I won't make the same mistake again... by the way X is the node between the two inductors or in other words thevenins voltage

my equation thus far:

I1 + I2 +I3 =0

Therefore: (V1 - X) / j4 + (V2 - X) / j6 + (0 - X) / 50∠45.6 (which is the load and lag angle)

(0 + j415 - X) / j4 + (415 + j0) / j6 + (0 - X) / 67.7 + j42.3 (Load and angle converted to Rectangular form)

this is where I am stuck now, hence not trying nodal analysis as I was not 100% sure about it...

 

[Ebies]

I understand division with complex numbers but not when there's an unknown involved... not sure ha to do with the unknown to get it to the other side of the equation...

 

[gneill]

∠This is as far as I have managed o get and now I am stuck with the equation. not sure how to deal with and equation that has a mix of unknowns and imaginary numbers in it... also attached is my diagram that I am using so the equation makes sense... I spose if I can learn where I am going wrong now I won't make the same mistake again... by the way X is the node between the two inductors or in other words thevenins voltage

my equation thus far:

I1 + I2 +I3 =0

Therefore: (V1 - X) / j4 + (V2 - X) / j6 + (0 - X) / 50∠45.6 (which is the load and lag angle)

Okay, the Thevenin voltage is the unloaded output voltage of the network. That means you want to remove the load impedance, so the last term should be removed. Set the remaining expression to equal zero. Solve for X.

(0 + j415 - X) / j4 + (415 + j0) / j6 + (0 - X) / 67.7 + j42.3 (Load and angle converted to Rectangular form)

this is where I am stuck now, hence not trying nodal analysis as I was not 100% sure about it...

With the load term out of the picture you should find an expression for X that involves just some complex arithmetic.

 

[Ebies]

thanks mate, your right the arithmetic is complex haha

 

[Ebies]

last question... j0 * j = 0 or j^2 ??

 

[gneill]

Anything multiplied by zero is zero.

 

[Ebies]

Anything multiplied by zero is zero.

I didn't know if I should treat it as zero sorry. bone question...

 

[Ebies]

I finally worked out Vth gneill... thanks for the help and guidance along the way

 

[Ebies]

gneill, can I ask this? why do we use the rms value for the calculations? should 415 not be multiplied with sqrt 2? when working with thev theorem you want the max values right?

 

[gneill]

gneill, can I ask this? why do we use the rms value for the calculations? should 415 not be multiplied with sqrt 2? when working with thev theorem you want the max values right?

You can use either peak or rms values if you're just looking for voltages or currents. You can write out the time domain functions for the results easily enough with or without the root 2 . If you're looking to calculate power though, it's easier to start with the rms values for the voltages and currents.

 

[Ebies]

thank you very much buddy.. you are a star by far... bloody genius I tell you!

 

[Ebies]

ok so I used my thev equivalent voltage and the total impedance (which was j4 & j6 combined and then added with the load of 50Ω @ 0.7p.f.) to calculate the current I with Vth/Zth and came up with a completely different answer to what it should be thus making me think I made a mistake somewhere... I get i=5.8∠8.9°

 

[gneill]

The magnitude is about right. The angle's off though. What are your (rectangular format) values for V_th, Z_th, and Z_L?

 

[Ebies]

Vth=166+j249 , Zth=35+j38.10714214 and ZL=35+j35.70714214

I have checked my calculations for calculating the current and its correct so one of my actual values must be up the duffer...

 

[gneill]

Vth=166+j249 , Zth=35+j38.10714214 and ZL=35+j35.70714214

Zth can't have any real component as it consists of the reduction of two imaginary values (inductance impedances). Should be something closer to 3j Ohms.

Your ZL looks fine, some resistance and a positive imaginary value implying inductance, which is what you'd expect for a load with a lagging power factor.

Your Vth looks a bit odd to me, if you started with v1 being 415V and v2 as 415V ∠ -90°. Looks like the real and imaginary component magnitudes have been swapped.

 

[Ebies]

I had V1=415∠90 and V2=415∠0... thanks for the hand there again gneill... with regards to Zth - I see what you mean by Zth, sorry I labelled em different in the equation... my Zth=j2.4

That explains my phase shift as I put myself out a few degrees by not reading the question properly... thanks again gneill

 

[Ebies]

ok so going off what gneill suggested last night I get a new revised figure for Vth=247.02-j164.69... what I do not understand however is why using a cos trig function or sine trig function yield different results...? ie the real and imaginary parts being swapped...

 

[gneill]

ok so going off what gneill suggested last night I get a new revised figure for Vth=247.02-j164.69... what I do not understand however is why using a cos trig function or sine trig function yield different results...? ie the real and imaginary parts being swapped...

The difference is in the phase of the complex number. The magnitude should be the same either way. A 90° phase difference is equivalent to rotating your phasor through 90°, which effectively swaps the real and imaginary components.

 

[Ebies]

yes you are correct the magnitude is very much the same sin rig funct (299∠56) and cos trif funct (296∠-33). makes sense now thanks...