Problem + Hard 83-2    Source:  Question 3 of the 2nd exam of the School of Engineering - UFRGS - 1975.

In Figure 83-2.1 the circuit is symmetrical, balanced and the phase sequence is direct or ABC. It is known that V₁ = 346.1 V,   V₂ = 200 V   and   A = 20 A.   Determine:

a)  the values ​​of R, X and the line voltage.

b)  the power factor of the star circuit.

Solution of the Problem + Hard 83-2

Item a

For the solution of this problem we will base ourselves on a diagram of voltages and currents as the one shown in Figure 83-2.2. We point out that the solution will be a mixture of algebra and geometry.

Now let's analyze this diagram by parts. Note that the voltage measured by the voltmeter V₂ is represented by the phasor V_kN = 200 V. From the figure, it follows that V_kN = V_CN sin 30°. Hence, the value of V_CN is:

As the circuit is symmetrical then V_CN = V_BN = V_AN = 400 volts and the line voltage It is:

To find the values ​​of R and X, calculate the voltages V_xN and V_Ax. Notice the circle in red where the diameter is the phase voltage V_AN = 400 V. This circle delimits the voltages V_xN and V_Ax, as indicated in the figure, where at the point x we have an angle of 90°. Note that to find the value of V_xN it is necessary to calculate the value of V_yN. Note that the point y is equal to one third of the value of V_CA. This means that it is at the same height as point N. Therefore, the segment Oy is parallel to voltage V_CB. This allows us to say that V_yN = V_Ay cos 60°, given that the angle between V_Ay and V_yN   is   60°.   And  V_Ay = 2/3 V_AC = 461.88 volts. Like this:

Then, the values ​​of the two phasors that allow the calculation of V_xN are known. For this it is necessary to employ the law of cosines. However, it is missing to know the angle between V_Ay and V_yN. The following trick should be used: let's draw a segment connecting the points O and y. In this way, the triangle NOy is created, which is a right triangle. Using the Pythagorean theorem it is possible to calculate the value V_Oy, remembering that V_ON = 200 volts, as it is half the value of V _AN.

And since the triangle NOy is a rectangle, then the angle φ is:

Now it remains to calculate the θ angle. Note that when tracing the segment Oy, triangle Oxy is created, which is not a right triangle. In this case, you must use the law of cosines to calculate the θ angle. So, you must use the eq. 51-03 reproduced below.

Remember that x is the side opposite the angle you want to calculate. So x = V_Ox = 200 volts, as it is equal to the radius of the circle.

Carrying out the calculation:

Adding the two angles found, the result is the angle between V_xy and V_yN, that is, φ + θ = 75.58°. Again using the law of cosines it is possible to calculate the value of V_xN. As we want to calculate the third side of a triangle, we use eq. 51-01 reproduced below.

Here x = V_xN. So:

Efetuando-se o cálculo, obtém-se:

Using the Pythagorean theorem, the value of V_Ax is obtained, as:

Performing the calculation, we obtain:

As the phase current I_AN is known, which was measured by the ammeter A = 20 A, then the values ​​of R< /x> and X will be:

Item b

To find the power factor, calculate the angle φ between the current I_AN and the voltage V_AN. However, as you know the values ​​of R and X this is very easy, because:

So the power factor is: