Problem 52-10
Source: Problem 3 of the Electrical Circuits Test II - UFRGS - 1975.
In the circuit of Figure 52-10.1 we have:
1 - Load Z1 ⇒ 40 √2 kVA with PF = 0.707 in advance.
2 - Load Z2 = R2 + j X2.
The reading of the instruments is:
V1 = 1,000 √2 V, V2 = 125 V and W = 100
kW.
It is known that the global circuit is predominantly inductive. Determine:
a) the values of I1 and I2
b) the values of R2 and X2
.
Figure 52-10.1
Solution of the Problem 52-10
Item a
Analyzing the load Z1 we can find the active and reactive power of the load. Soon:
P1 = 40 √2 x 0.707 = 40 kW
Q1 = - 40 √2 x 0.707 = - 40 kVAR
Note the negative sign of Q1, because according to the statement the power factor is advanced. Then the load
Z1 can be written as Z1 = R1 - j X1. In addition, the real powers and
reactive are equal, meaning that R1 = X1.
On the other hand, knowing the value of V2 we can find the modulus value of IT .
Let's write the impedance that is in parallel with V2 in polar form. Then
1 - j 0.75 = 1.25 ∠- 36.86°. So the modulus value of IT it's:
| IT | = V2 / 1.25 = 125 / 1.25 = 100 A
Knowing the module of IT it is possible to find the apparent power that the circuit consumes,
because we know the reading value of V1. Soon:
ST = V1 IT = 1,000 √2 x 100 = 141,421 VA
Having the value of the apparent power of the circuit and knowing that the active power was given and is equal to 100,000 W, we can calculate the power factor of the circuit, that is:
FPcirc = W / ST = 100,000 / 141,421 = 0.707
And with this value it is possible to calculate the phase shift angle between the applied voltage and the current flowing through the circuit. Then:
θ = cos- 1 (0.707) = 45°
Let's take the measured voltage V1 as a reference. Then, we have:
V1 = 1,000 √2 ∠0°
The problem statement states that there is a global inductive predominance. Therefore, the current is lagging by 45° in
to the voltage applied to the circuit. With that we can write:
IT = 100 ∠ - 45°
After determining the main variables involved in the solution, we can redraw the circuit as seen in the Figure 52-10.2.
Figure 52-10.2
Note that between the points 1 - 2 appears the impedance 2 + j 4, which is the sum of the components that appear in the Figure 52-10.1.
Now we want to determine the values of Z1 and Z2. For this, we need to know the value of
V2b. So, first let's determine the value of V12. Soon:
V12 = IT (2 + j 4) = 100 ∠- 45° x 4.47 ∠ 63.43°
In this equation we use the fact that 2 + j 4 = 4.47 ∠ 63.43°. Doing the math, we find:
V12 = 447 ∠ 18.43° V
In this way, the value of V2b is given by the difference between V1 and V12, or:
V2b = V1 - V12 = 1,000 √2 ∠ 0° - 447 ∠ 18.43°
Doing the calculate, we find:
V2b = 1,000 ∠ - 8.12°
At this point in the solution to the problem, we should pay close attention to load 1. As we concluded at the beginning of
solution, we have R1 = X1. Then the voltage drop across R1 and X1 will be equal to VR1 = VX1 = 707 V. As we know the power in both elements and they are equal in values
absolute, we get:
R1 = X1 = (VR1)2 / P1 = 7072 /
40,000
Doing the calculate, we find:
R1 = X1 = 12.5 Ω
Now we can write Z1 = 12.5 - j12.5 = 17.68 ∠- 45°. So, applying the Ohm's law we can
determine the value of I1, that is:
I1 = V2b / Z1 = 1,000 ∠ - 8.12° / 17.68 ∠ - 45°
Doing the calculate, we find:
I1 = 56.56 ∠ 36.88° A
And to calculate the current I2 just calculate the difference between IT and I 1, because we know by the law of Kirchhoff for currents that IT = I1 + I2. Then:
I2 = IT - I1 = 100 ∠- 45° - 56.56 ∠ 36.88°
Doing the calculate, we find:
I2 = 107.71 ∠- 76.32° A
Item b
To find the values of R2 and X2, we must consider that, based on the circuit shown
in Figure 52.10.2, the power dissipated by R = 2 Ω is equal to 2 x 1002 = 20 kW. And the power dissipated by R1 of load 1, already calculated, is 40 kW. Therefore, obligatorily
R2 must dissipate a power of 40 kW, given that the power meter measured a total power of
100 kW. Knowing the power that the resistor dissipates and the current flowing through it, we can easily determine its value, or:
R2 = P2 / I22 = 40,000 / 107.712
Doing the calculate, we find:
R2 = 3.45 Ω
To calculate the value of X2, we consider that the inductor j 4 has a reactive power of
4 x 1002 = 40 kVAR. Thus, the reactive power of this inductor is canceled by the reactive power of the capacitor in the load 1 which equals Q1 = - 40 kVAR, a value already calculated at the beginning of the problem solution. Therefore, as the problem states that there is an inductive predominance in the circuit, so we conclude that X2 is an inductor and has a reactive power equal to 100 kVAR. This value is the result of the apparent power of the circuit, already calculated, multiplied by the angle of delay of the total current in relation to the voltage applied to the circuit. And we know that this angle is equal to 45°. Therefore, similarly to the calculation of R2, we have:
