Problem 54-7
Source: Problem 7 - RLC Problem List - Discipline
Electrical Circuits of the School of Engineering - UFRGS - 2017 - Prof. Dr. Valner Brusamarello.
In the circuit shown in Figure 54-07.1 we know that E = A cos (10 t + α) volts and
the currents indicated in the circuit are valid.
Find the values of R and L.
Figure 54-07.1
Solution of the Problem 54-7
From the circuit shown above we can easily conclude that I = I1 + I2. On the other hand, as I2 passes through an inductive circuit, this current will suffer a delay in relation to I1. So, let's consider the current I1 as a reference. Therefore, we can write I1 = 7∠ 0° . This way, it is possible to construct a diagram showing the currents involved in the circuit. see the
Figure 54-07.2 below.
Figure 54-07.2
Note that we have a triangle where the values of all sides are known. In this case, it is possible to apply
the law of cosines and determine the angle φ. So, let's calculate this angle using
eq. 51-03, repeated below for clarity.
eq. 51-03
Keep in mind that, in this equation, x is the side opposite the angle we want to calculate. In this case, x = 15. So, substituting for numerical values, we obtain:
cos φ = (72 + 202 - 152) / (2 x 7 x 20) = 0.8
Applying the inverse function of the cosine function, we obtain the value of φ.
φ = cos-1 0.8 = 36.87°
Now, see Figure 54-07.3 below where we show the angle θ between the phasor I2 and the vertical axis. Note that by determining the angle θ, we can easily calculate the values of IR and IL.
Figure 54-07.3
So, making the projection of I
and I2 on the vertical axis, we have:
I sin φ = I2 cos θ ⇒ 20 x sin 36.87° = 15 . cos θ
And so we find the value of θ, or:
cos θ = 0.8 ⇒ θ = cos-1 0.8 = 36.87°
And now applying a little trigonometry, we calculate IR and IL. Soon:
IR = I2 x sin θ = 15 x sin 36,87° = 9 A
IL = I2 x cos θ = 15 x cos 36,87° = 12 A
Taking into account the current I1 and R2, we can calculate the value of Vab.
Vab = I1 . R2 = 7 x (3/7) = 3 V
So, knowing Vab, IR and IL we can calculate the values of R and XL.
R = Vab / IR = 3 / 9 = 1 / 3 Ω
XL = Vab / IL = 3 / 12 = 1 / 4 Ω
Knowing XL we can calculate the value of L. To do this we need to know the value of ω. However, in the problem statement
the value of E was provided. The term that accompanies the variable t is the value of ω = 10 rad/s. Then:
XL = ω . L ⇒ L = (1 / 4) / 10 = 1 / 40 Ω
Observation
Knowing the values of R and XL it is possible to calculate the values of A and α. To do so, let's calculate the equivalent impedance value that the voltage source "sees" in the circuit.
The reader can confirm that Zeq = 10.12 + j 0.09 = 10.12∠ 0.51°. Knowing that I = 20∠-36.87° and carrying out the calculation, we find:
E = Zeq . I = 10.12∠ 0.51° x 20∠-36.87°
E = 202.4 ∠ -36.36° Ω
Therefore, A = 202.4 V and α = -36.36°, that is, in trigonometric form we have E = 202.4 cos (10 t - 36.36°).
