Problem + Hard 55-7    Source:   Problem 7 - RLC 2008 Problem List IV - Discipline Electrical Circuits of the School of Engineering - UFRGS - 2008 - Prof. doctor Valner Brusamarello.

In the circuit below, determine the magnitude and phase of the voltage source E, and the value of Z (it is known that Z = X∠60°, where X is a constant to be determined). Also determine the value of the constant A, knowing that the current I₃ is 15 A and the voltmeter V reads 75 V. It is also known that the angle α is 60° ahead of V_ab. Note that the voltage over Z is provided in the circuit.

Resistors Values:   R₁ = 10√3Ω   R₂ = R₃ = 2.5√3 Ω

Solution of the Problem + Hard 55-7

To solve this problem, the voltage V_ab is assumed as the reference voltage, so V_ab∠0°. The voltage on Z will be called V_Z. And the impedance through which I₁ circulates, as Z₁. Likewise, where I₂ circulates, as Z₂. So, writing these impedances, in cartesian and polar form, we have:

As the two impedances are in parallel, that is, they are on the same potential difference, it is concluded that:

This is because, in module terms, Z₂ has half the value of Z₁. How do you know the angles of Z₁ and Z₂, as well as the angle of V_ab, then one can determine the angles of I₁ and I₂.

Therefore, it follows that I₁ is 30° ahead of V_ab and I₂ is 30° behind V_ab. Note that from the statement of the problem, the difference of potential between points 1 and 2 is 75 volts. On the other hand, the voltage across the capacitor is given by:

We used the fact that -j 10 = 10∠-90°. Similarly, one can write the voltage over R₃, or:

But, it is known that |I₂| = 2 |I₁|. Making this substitution in the equation above, we obtain:

The graph below illustrates this situation in a didactic way.

Based on the graph above, we know the angle formed by V_C and V_R3 (30 °). In addition, the size of the side opposite the angle is also known. And, in the equations above, V_C and V_R3 are in function of I₁. Therefore, the value of I₁ is found by applying the law of cosines. Then:

Working the above equation algebraically:

Carrying out the calculation, the value of |I₁| is:

And since it has already been determined that |I₂| = 2  |I₁|, then:

However, from the graph above, the phases of I₁ and I₂ are known. Like this:

With these values, you can determine the value of V_ab. Then:

Knowing the value of V_ab and V_Z = 180 ∠60°, one can calculate the value and phase of the source of voltage E. So, from the circuit:

Substituting for numerical values ​​and carrying out the calculations, we obtain:

Important -   From the statement of the problem, it is known that the impedance Z has a phase of 60°. However, the potential difference over it also has a phase of α = 60°. For this to happen, it is evident that the phase of the total current, I_T, current flowing through Z, must have a phase equal to zero. Having knowledge of this particularity, you can determine the phase of the current I₃.

Knowing that I₁ = 15∠+30° and I₂ = 30∠-30° is easy to understand that the phase of I₃ must be equal to 30°, so you get I₁ + I ₃= 30∠+30°. Note that this value is the symmetric of I₂. Adding the three currents together, the result is a total current with phase zero.

Carrying out the calculation, I_T is equal to:

With the value of I_T, it is easy to calculate the value of Z, or:

Carrying out the calculation, Z is equal to:

And from the figure showing the circuit we know that Z = X ∠60°. So, comparing with the equation above we have:

Important -   Knowing the values ​​of V_ab and I₃, you can calculate the impedance value Z₃ , this branch impedance where the current I₃ circulates.

Carrying out the calculation, Z₃ is equal to:

And, based on the statement of the problem, this impedance should be equal to:

Comparing the last two equations, it is concluded that: