# Evaluating indication of the wattmeter in three-phase system

1. Feb 10, 2015

### pizzico85

• Member warned about posting without the HW template
Hi all, I am new to this forum and I hope you can help me.

I have the following three-phase system where I have to evaluate the indication of the wattmeter W. Available data is: V = 380 V, R = 12 Ohm, R' = 120 Ohm.

Can you make me understand the process for obtaining W?

2. Feb 10, 2015

### Staff: Mentor

You have 5 energy sinks here. Three identical resistors, one in each phase---that's fortunate, it maintains balance! There's a fourth resistor, and a 0.8pf load of Q=600VAr .

It looks like the wattmeter measures the power in R', in addition to the power in one phase of a balanced load.

3. Feb 10, 2015

### pizzico85

My thoughts are that I don't know how to calculate V' or I'.

I am completely in trouble.

I would like to understand how to proceed step by step to solve the problem.

Last edited: Feb 10, 2015
4. Feb 10, 2015

### Staff: Mentor

Since you know V, you should be able to calculate the line current going into that box on the right.

5. Feb 10, 2015

### pizzico85

OK, the line current going into the box should be the following:

Since
Q = sqrt(3) * V * I * sin(phi)
cos(phi)
= 0,8 => phi = 36,87°

Then
I = Q / ( sqrt(3) * V * sin(phi) ) = 1,52

I = I1 = I2 = I3 because the load is balanced without considering R'.

Is it correct? If so, how should I proceed?

6. Feb 10, 2015

### Staff: Mentor

Once line current is known, you can determine the voltage across R. You'll need to start drawing a phasor diagram.

7. Feb 10, 2015

### pizzico85

OK, maybe voltage across R should be the following:

Since
R = R1 = R2 = R3
I = I1 = I2 = I3

Then
Vr = V1 = V2 = V3 = R * I = 12 Ohm * 1,52 A = 18,24 V

Now I should be able to use Kirchhoff for tensions in the mesh where the voltmeter V is present, this way (V12 refers to the voltage of the three-phase system):

V12 - V1 - V + V2 = 0

Since V1 and V2 are equal:
V12 = V => V12 = 380 V
and since the three-phase system is symmetrical:
V12 = V23 = V31 = 380 V

Is it correct? If so, how should I proceed?

8. Feb 10, 2015

### Staff: Mentor

Time to start putting this on a phasor diagram, I think.

9. Feb 13, 2015

### pizzico85

No too much help.

10. Feb 13, 2015

### Staff: Mentor

Are you able to draw a phasor diagram? Decide which phasor you will use as a reference, then draw others in relation to it. This is the type of problem where a phasor diagram is essential, IMO.

Can you explain in words how you will go about finishing this problem?

If you have not learnt about phasor diagrams, I expect you will ask further questions.