Problem + Hard 83-1
Source: Question 2 of the 2nd exam of the School of Engineering - UFRGS - 1976.
In the Figure 83-1.1 the circuit is symmetrical, balanced and the phase sequence is inverse or BCA. It is known that the line voltage
VCB = 500∠0° V and the current IA = 8.66∠65° A. Furthermore, it is known that
V1 = 450 V and V2 < |VCB|. Determine:
a) phase currents.
b) the values of r, X1 and X 2.
c) the phase of V1.
Figure 83-1.1
Solution of the Problem + Hard 83-1
Item a
Because the sequence is inverse and the circuit is balanced, we can write the
other two line currents, as they are 120° out of phase. Then:
IB = 8.66∠185° = 8.66∠-175° A
IC = 8.66∠-55° A
It is known that in the inverse sequence the phase current lags 30° in relation to the line current and
its magnitude is divided by √3. Soon:
Iab = 5∠35° A
Ibc = 5∠-205° = 5∠155° A
Ica = 5∠-85° A
Item b
As the problem provided the value of VCB = 500∠0° V, this implies that
VBC = 500∠180° V, VAB = 500∠60° V and VAC = 500∠-60° V.
Now, as we know the value of the voltages and currents that circulate through the branches, then we can calculate
the impedances of each branch. As the circuit is balanced, the values will be the same for the three phases.
Doing for the BC phase:
ZBC =VBC /Ibc = 500∠180° / 5∠155° = 100∠25°
Converting to rectangular form:
ZBC = 90.63 + j42.26 Ω
Comparing this result with the elements of the BC branch, we determine the values of R1 and X. Then:
R1 = 90.63 Ω and X = 42.26 Ω
Now, comparing this result with the elements of branch AB, it is concluded that R1 = 3 r and X is the same value as calculated above. Then:
R1 = 3 r = 90.63 Ω ⇒ r = 30.21 Ω
To determine the values of X1 and X2, in branch CA , and taking into account
that the circuit is balanced, it can be concluded that:
X = X1 + X2 = 42.26 Ω
From this result, it is possible to see that X1 and X2 cannot, simultaneously , to assume negative values. Let's use the given data where V1 = 450 V. How do you know the value of
Iab and the value of r, you can calculate the voltage between points A and u, which will be called VAu. Like this:
VAu = r Iab = 151.05∠35° V
On the other hand, the voltage between the points A and x can be calculated, because:
VAx = j X1 Ica = 5∠-85° X1∠90° = 5 X1∠5°
Note that the value of j has been replaced by the angle of 90°. VAx is calculated
just to determine which angle exists between VAu and VAx.
Simply subtract the angle values of the two voltages, or θ = 35° - 5° = 30°.
Recalling the equation that must be used to add phasors when they form an angle
other than 90° , 180° or 270°:
V12 = VAu2 + VAx2 + 2 VAu VAx cos θ
Working this equation algebraically and replacing it with known numerical values, we obtain the following equation of the second
degree:
VAx2 + 261.63 VAx - 179.684 = 0
And so, one finds two values for VAx, or:
VAx = 312.8 V and VAx = -574.43 V
However, according to the problem statement, the condition V2 < |VCB| = 500 should be satisfied.
By circuit, V2 = VAx. Therefore, the second value found must be discarded, or
VAx = - 574.43 V. So the searched value of VAx is:
VAx = 312.8 ∠5° V
Now we can calculate the value of X1, or:
|X1| = |VAx| / |Ica| = 312.8 / 5 = 62.56 Ω
As the module was used, the reactance X1 can assume a positive or negative value, that is, it can be a
inductor or a capacitor. As a consequence, X2
it can also assume two values: one positive (inductor) and the other negative (capacitor). You must calculate both values.
X1 = +62.56 Ω ⇒ X2 = 42.26 - 62.56 = -20.3 Ω
In this case, the reactance X2 must be represented by a capacitor, while X1 is
represented by an inductor.
X1 = -62.56 Ω ⇒ X2 = 42.26 + 62.56 = 104.82 Ω
If X1 is represented by a capacitor, then X2
must be represented by a inductor.
Item c
To calculate the phase of V1 just perform the phasor sum of VAu and
VAx. Then:
V1∠φ =VAu+ VAx= 151.05∠35°+ 312.8∠5° = 450∠14.66° V
