Phase relationships in Thévenin circuit

[Weaver]

Homework Statement

2. Homework Equations

3. The Attempt at a Solution

[/B]
For part (c), should the phasor diagram just have everything on a straight line, save for the Rt and Vrt?

Honestly not sure how to attempt part (d) as we never really covered phase relationships in lectures, just looking specifically at phase diagrams of a capacitor or inductor.

 

[cnh1995]

It would be convenient if you chose the load voltage as the reference phasor in your phasor diagram. You can solve all the questions upto e) once you draw this phasor diagram.
Start with the horizontal load voltage phasor V_L∠0°.

 

[Weaver]

It would be convenient if you chose the load voltage as the reference phasor in your phasor diagram. You can solve all the questions up to e) once you draw this phasor diagram.
Start with the horizontal load voltage phase V_L∠0°.

Ok. So will all the Voltages, save for the Thévenin resistance voltage, be on the horizontal axis? Or is V_L shifted as result of the complex impedance of R_T before it.

 

[Weaver]

Ok. So will all the Voltages, save for the Thévenin resistance voltage, be on the horizontal axis? Or is VL shifted as result of the complex impedance of RT before it.

Actually, after thinking about it some more, would the voltage of V_L be V_TH - V_R given the voltage from V_TH isn't 'dropped' across anything else other then those two resistances?

 

[cnh1995]

would the voltage of VL be VTH - VR given the voltage from VTH isn't 'dropped' across anything else other then those two resistances?

You are forgetting the j0.25 ohm reactance. Yes, V_L and V_R will add algebraically since they are in phase but that won't give you Vth. You need to draw the reactance drop phasor too.

 

[Weaver]

You are forgetting the j0.25 ohm reactance. Yes, V_L and V_R will add algebraically since they are in phase but that won't give you Vth. You need to draw the reactance drop phasor too.

Ok, I am still not sure how the phasor diagram should look. Prior to this, the only phasor diagrams we have been taught is the general one for a capacitor and inductor.

Would it look like this?

 

[cnh1995]

Ok, I am still not sure how the phasor diagram should look. Prior to this, the only phasor diagrams we have been taught is the general one for a capacitor and inductor.

Would it look like this?

View attachment 110304

No. Vth and V_L are not in phase. V_R and V_L are in phase. Recall KVL. Sum of the voltage drops across Thevenin resistance, Thevenin reactance and load is equal to the Thevenin voltage. You know the magnitudes of three of them and their phase relation.

Edit: Just to confirm, are we using the same terminology? I am taking V_R as the voltage across the thevenin resistance, V_L as the voltage across the storage heater load and Vth is the mains voltage.

 

[Weaver]

No. Vth and V_L are not in phase. V_R and V_L are in phase. Recall KVL. Sum of the voltage drops across Thevenin resistance, Thevenin reactance and load is equal to the Thevenin voltage.

Ok, that's what I meant when I said

would the voltage of VL be VTH - VR

I was taking V_R to be the voltage drop for both the reactance and resistance.

Would the phasor diagram look like this then?

 

[cnh1995]

I was taking VR to be the voltage drop for both the reactance and resistance.

Ok. Let's call it the Thevenin impedance drop V_z.

Edit: Just to confirm, are we using the same terminology? I am taking VR as the voltage across the thevenin resistance, VL as the voltage across the storage heater load and Vth is the mains voltage.

In your diagram, you have shown V_z leading V_th by 45°, which is not true. Phase difference between V_z and V_th is not 45°. You need to show the Thevenin reactance drop V_x too in the diagram (drop across the line reactance).
Start with the horizontal load voltage phasor V_L. How will Thevenin resistance drop V_R and Thevenin reactance drop V_x be added to V_L? You know the magnitudes of these three and their phase relationship. You also know that their phasor addition will give you Vth. Could you proceed with this?

 

[Weaver]

Ok. Let's call it the Thevenin impedance drop V_z.

In your diagram, you have shown V_z leading V_th by 45°, which is not true. Phase difference between V_z and V_th is not 45°. You need to show the Thevenin reactance drop V_x too in the diagram (drop across the line reactance).
Start with the horizontal load voltage phasor V_L. How will Thevenin resistance drop V_R and Thevenin reactance drop V_x be added to V_L? You know the magnitudes of these three and their phase relationship. You also know that their phasor addotion will give you Vth. Could you proceed with this?

Ok. There's been confusion from terminology, so I'll use yours (V_th, V_r, V_x,V_L)

Right. So V_th = V_r + V_x + V_L

The current in the circuit is 80A

This means V_r = 80(0.25) = 25

And V_x 80(j0.25) = j25

V_L = V_th - V_r - V_x
V_L = (240 +j0) - 25 - j25
V_L = 215 - j25

Is this right so far?

 

[cnh1995]

This means Vr = 80(0.25) = 25

And Vx 80(j0.25) = j25

20 and j20...

VL = Vth - Vr - Vx
VL = (240 +j0) - 25 - j25
VL = 215 - j25

We are taking V_L as reference here.
So, V_th= (V_L+j0)+20+j20.
i.e.
V_th=(V_L+20)+j20.

Can you draw the phasor diagram from this green equation?

 

[Weaver]

20 and j20...

Sorry, do know what I was thinking there

We are taking VL as reference here.
So, Vth= VL+j0+20+j20.
i.e.
Vth=(VL+20)+j20.

I'm not sure. Do I take the V_L and the 20(V_r) both to be on the horizontal axis and just put j20(V_x on the Vertical axis?

Where do I draw V_th?

 

[cnh1995]

I'm not sure. Do I take the VL and the 20(Vr) both to be on the horizontal axis and just put j20(Vx on the Vertical axis?

Exactly. The reactance drop j20 will be vertical, on the tip of the horizontal phasor (V_L+20). It is in quadrature with the resistive drop.

Where do I draw Vth?

Vth is the "phasor addition" of voltages on horizontal and vertical axes. Don't they form a right angled triangle? Vth is its hypotenuse. You know the magnitude of the hypotenuse and the vertical side. You can compute the length of the horizontal side and get the load voltage.

 

[Weaver]

Exactly. The reactance drop j20 will be vertical, on the tip of the horizontal phasor (V_L+20). It is in quadrature with the resistive drop.

Vth is the "phasor addition" of voltages on horizontal and vertical axes. Don't they form a right angled triangle? Vth is its hypotenuse. You know the magnitude of the hypotenuse and the vertical side. You can compute the length of the horizontal side and get the load voltage.

So, is this it?

 

[cnh1995]

So, is this it?
View attachment 110309

Your calculations are all correct, but in your phasor diagram, you should swap Vr and V_L since you have assumed V_L as the reference phasor. It's a standard practice to draw the reference phasor first and then draw the remaining ones.

 

[Weaver]

Your calculations are all correct, but in your phasor diagram, you should swap Vr and V_L since you have assumed V_L as the reference phasor. It's a standard practice to draw the reference phasor first and then draw the remaining ones.

Ok. I'll do that in future. Can I draw the circuit currents in the diagram also? Is the 80A just along the V_L?

 

[cnh1995]

Ok. I'll do that in future. Can I draw the circuit currents in the diagram also? Is the 80A just along the V_L?

Yes.

 

[Weaver]

Yes.

Ok, so for part (d) Calculate the phase relationship between the Thevenin equivalent circuit voltage and the current

Do I work out the angle the current is lagging by?

 

[cnh1995]

Ok, so for part (d) Calculate the phase relationship between the Thevenin equivalent circuit voltage and the current

Do I work out the angle the current is lagging by?

Yes. You can do it using elementary trigonometry.

 

[Weaver]

Yes. You can do it using elementary trigonometry.

Grand. So it is:

ϕ = sin^-1(20/240) ≈ 4.78°.

I take it for part(e) Calculate the total resistance of the storage heaters, it is:
V/I

V_L / 80∠4.78°?

 

[cnh1995]

VL / 80∠4.78°?

Load voltage V_L and current are in phase. The phase angle of both is 0°.

 

[Weaver]

Load voltage V_L and current are in phase. You need not use the phase angle.

Ok, so it is (20√143 - 20)/80 ≈ 2.74 Ω ?

Do I have to use 2.74∠-4.78° to get it relative to the V_th ?

 

[cnh1995]

Ok, so it is (20√143 - 20)/80 ≈ 2.74 Ω

Yes.

Do I have to use 2.74∠-4.78° to get it relative to the Vth ?

No.

 

[Weaver]

Yes.

No.

Right. And then for (e) What happens to the power dissipated in the storage heaters if the supply frequency is raised above 50 Hz? Is it just:

Given the V_x is dependant on ω (which is 2πf), an increase in f will lead to an increase in V_x and so the current in the system will decrease and with that, the power output in the storage heater also decreases?

 

[cnh1995]

Given the Vx is dependant on ω (which is 2πf), an increase in f will lead to an increase in Vx and so the current in the system will decrease and with that, the power output in the storage heater also decreases?

Right.

 

[Weaver]

Right.

Great. Thank you very much!

 

[cnh1995]

Great. Thank you very much!

You're welcome!