Problem + Hard 55-5    Source:   Adapted from Problem 30 - List of RLC Problems - Discipline Electrical Circuits of the School of Engineering - UFRGS - 2017 - Prof. doctor Valner Brusamarello.

Knowing that V_ab = 60√3∠θ and that the voltmeter V measures 60 volts between the points 1 and 2, determine:

a) V_cb knowing that it is 30° behind V _ab.

b) The value of V_ac and I_T.

b) The values ​​of R and X.

Solution of the Problem + Hard 55-5

Item a

As you know, a voltmeter does not circulate current. In this way, the current I₁ circulates through branch c-2-b, while I₂ cycles through branch c-1-b. Then, following the path indicated by the orange arrow, you can write the following equation:

In the same way, we can follow the path indicated by the green arrow and obtain the equation:

Thus, we have a system of two equations with two unknowns that can be solved by substitution, Cramer's rule or by the Octave. This way we get the following value for I₁:

And for I₂, we get:

Knowing I₁ we can easily determine the value of V_cb. For that, let's calculate the impedance of branch c-2-b, which we will call Z₁.

Then V_cb will be:

Item b

Since we know the angle value of V_cb, we can now determine the angle θ of the voltage V_ab, because according to the problem statement it should be 30° ahead of V_cb< /x>. Soon:

To determine V_ac just apply the Kirchhoff law for voltage, or:

Carrying out the calculation, the value of V_ac is found, or:

To find I_T just add the currents I₁ and I₂.

Item c

The impedance between points a and c can be represented by:

As we know the value of the voltage between these points and the current flowing between them, applying the Ohm law we will find Z_ac, or:

Ora, comparando as duas últimas equações, parte real com parte real e imaginária com imaginária encontramos:

Therefore, it is concluded that the reactive element between points a and c is an inducer of reactance equal to √3 Ω

Final considerations

Finally, let's calculate the powers involved in the circuit. First, let's calculate the apparent power delivered by the voltage source.

Note that in the above equation, I_T^* is the complex conjugate of I_T. On the other hand, the angle of the apparent power is zero, meaning that the reactive power of the circuit is NULL. Let's check. The active power dissipated by the resistors is:

Note that the active power is equal to the apparent power, that is, we have no reactive power in the circuit. So we can say that the power factor of the circuit is unity. Let's calculate the reactive power of the circuit inductors.

Now let's calculate the reactive power in the capacitors.

It is evident that the reactive powers in the inductors and capacitors are NULL.