Problem 55-15    Source:   Problem 26 - List of RLC Problems - Discipline Electrical Circuits of the School of Engineering - UFRGS - 2011 - Prof. Dr. Valner Brusamarello.

In the circuit shown in the Figure 55-15.1, it is known that V₁ = 60 V. Determine the reading of the voltmeter V₂, as well as the value of V_ab.

Solution of the Problem 55-15   

Calling the impedance of the branch where it circulates I₁, of Z₁, and the impedance of the branch where it circulates I₂, of Z₂, we can write:

Assuming the voltage V_ab as the reference voltage, that is, V_ab∠0° and based on this information, we can write:

Likewise for |I₂|:

We can find a relation between |I₁| and |I₂| if we divide |I₁| by |I₂|, that is:

With this information, we can find the values of |I₁| and |I₂|. Note, in the graph below, that we know the value of the angle between |V_R1| and |V_C|, whose value is 15°. Then we can find the values of these voltages as a function of a current, for example |I₂|. As V_R1 = R₁  |I₁| = 4.33  |I₁| y |I₁| = 2 √2 |I₂|, then we can write:

Likewise for |V_C|:

By the statement of the problem, it is known that |V₁| = 60 volts. Now, looking at the graph in the Figure 55-15.2, it is noted that |V_C|, |V_R1| and |V₁| form a triangle. What's more, one of the angles and its opposite side are known. Therefore, we can use the law of cosines and determine the value of |I₂|, because the equations relating to |V_C| and |V_R1| with |I₂| already been deducted. Then:

After an algebraic arrangement and the calculation is performed, we obtain:

Easily find the value of |I₁| using the relation eq. 55-15.1 found above, or:

It should be noted that we can determine the phase of I₁ and I₂, remembering that it was used V_ab as reference. Soon:

These values are in accordance with the construction of the graph above. We can now calculate the values of V_C and V_R3.

Making the phasor sum of V_C and V_R3, we get the value of V_ab. So:

This is in line with the proposal for a V_ab be a reference voltage. To find the value of V₂, we should evaluate the graph shown in the figure above. Note that the 5 point is exactly the midpoint of V_ab. And the point 4 is the midpoint of the voltage between the points 5-a. Therefore, it is concluded that the voltage V_4-5 is the fourth part of the voltage V_ab. That is:

On the other hand, we can find the value of |V_L|, that is:

As |V_R1| = |V_R2| and both volages have the same angle with respect to the V_ab, it is noticed that the point 5 is the midpoint of V_L. Then we can write the voltage between the points 2 - 5 as:

It is concluded that |V_2-5| = |V_4-5| and more, these voltages make an angle of 60° between them, as indicated in the graph. Considering this and with some basic knowledge of geometry, it is obvious that we have an equilateral triangle formed by the voltages |V_2-5|, |V_4-5| and |V₂|. Therefore, there is no need for calculations to conclude that

There is a small discrepancy of values in V₂ between the list and the one found here. Perhaps in the list the answer was found graphically, while the result here was found analytically. Therefore, this small discrepancy can be disregarded.