Problem + Hard 58-1
Source: Problem 61 - List of RLC Problems - Discipline
Electrical Circuits of the School of Engineering - UFRGS - 2017 - Prof. doctor Valner Brusamarello.
When the circuit is in resonance IT = 15 A. Furthermore, the voltage drop
on the 3 Ω resistor is 30 volts and on the 1.5 Ω resistor is 19.84 volts,
as shown in the circuit below.
Determine X, XC and R' so that the
circuit is in resonance.
Attention -
In the list of exercises at UFRGS, the answer to this problem has a typo. The correct answer is: X = 12,
XC = 4.96 and R' = 2.57.
Solution of the Problem + Hard 58-1
Initially let's transform the parallel RL circuit (upper part of the circuit) into a series RL circuit,
calculating the parallel of R and L. Like this:
ZRL = 2.5 x j 1.25 / (2.5 + j1.25) = 0.5 + j Ω
Then we can rearrange the topology of the circuit as shown in the figure below. Note that the resistors that were in series,
had their values added together. The current flowing through the right branch was named I2∠φ. Notice that
we opted for a positive angle, as we are assuming XC > 1, that is, the circuit has predominance
capacitive. And the current flowing through the other branch was named I1∠-θ. Here, an angle was chosen
negative, since we assume that X > 5.07, that is, inductive predominance.
It is logical to assume that for there to be resonance one of the branches must have capacitive predominance while the other inductive.
On the other hand, as we want the circuit to be in resonance, we know that the current must be in phase with the voltage applied to the circuit. Choose for the total current, IT, phase zero, as well as for the voltage V.
As the phase of the currents was not specified in the problem statement, we will choose the one that is most convenient for the solution.
After these considerations, the Kirchhoff node law for this circuit can be written:
I1∠-θ + I2∠φ = IT∠0°
From the data provided in the statement, it is known that:
With this information we can find the values of θ and φ, because we know that the three phasors above
represent the three sides of a triangle. See in the figure below that we can determine the value of the angle φ
applying the law of cosines (as explained in chapter 51 click here!).
I12 = IT2 + I22 - 2 IT I2 cos φ
Substituting numerical data, we find:
cos φ = -300 / - 396.81 = 0.756
Applying the inverse function of cos we find the value of φ.
φ = cos-1 0.756 = 40.89°
To find the value of θ we will use the relationship below (taken from the graph above) that relates the function
sine with both angles.
-10 sin θ + 13.227 sin φ = 0
Substituting for numerical values, we have:
-10 sin θ + 13.227 sin 40.89° = 0
Applying the inverse function of sine we find the value of θ.
θ = sin-1 0.866 = 60°
Therefore, using the two equations obtained from the graph above, it is possible to verify if the angles found are correct.
10 cos 60° + 13.227 cos 40.89° = 15
-10 sin 60° + 13.227 sin 40.89° = 0
Note that if the current I1 is lagging by 60°, then the impedance through which it flows
I1 must be of type Z1 ∠ 60°. Then, it can be written that:
tan 60° = √3 = (X - 5.07) / 4
Performing the calculate, we get:
X = 12 Ω
With the value of X we can calculate the impedance through which I1 circulates
and from this data we find the value of V. Thus, Z1 = 4 + j6,93 = 8 ∠60°.
That way:
V = Z1 I1 = 8 ∠60° 10 ∠-60° = 80 ∠0°
With the value of V we can easily calculate the impedance through which I2 circulates, that is, Z2, since we know that I2 = 13.227 ∠40.89°. Soon:
Z2 = V / I2 = 6.048 ∠-40.89° = 4.57 - j 3.96 Ω
Equating this value of Z2 with the branch where I2 circulates, we have:
Z2 = 4.57 - j 3.96 = R' + 2 + j(1 - XC)
Equating the real part, we find:
4.57 = R' + 2 ⇒ R' = 2.57 Ω
Now equating the imaginary part, we have:
-j3.96 = j(1 - XC) ⇒ XC = 4.96 Ω
Final Considerations
We can make a power balance to verify the veracity of the values found.
The voltage source supplies an active power of:
PS = - V IT = - 80∠0° x 15∠0° = - 1,200 watts
Those who dissipate this power are the existing resistors in the circuit. Then the power dissipated in the resistor
of 4 ohms goes:
P4 = 4 |I1|2 = 4 x 102 = 400 watts
In the other branch the power dissipated by the 4.57 Ω resistor (sum of 2 + R') is:
P4.57 = 4.57 |I2|2 = 4.57 x 13.2272 = 800 watts
Therefore, the sum of the powers in the resistors is 400 + 800 = 1,200 watts.
Exactly the amount of power supplied by the voltage source. And to conclude, we must have the
total reactive power equal to zero. Let's check.
QT = 6.93 |I1|2 - 3.96 |I2|2 = 693 - 693 = 0
Therefore, we conclude that at resonance the impedance of the entire circuit must be equal to a resistor
whose value can be calculated by applying Ohm's law, or:
