Problem + Hard 55-1    Source:   Problem 12 - List of RLC Problems - Discipline Electrical Circuits of the School of Engineering - UFRGS - 2011 - Prof. doctor Valner Brusamarello.

For the circuit in the figure below we have V = 5,   I = 3 + j4 and E = 40√2∠45°. Determine   Z,   X   and   V_ab.

Solution of the Problem + Hard 55-1

Since the current I is given in the problem statement, we can calculate the voltage between the points a - x, that is, V_ax.

Carrying out the calculation:

On the other hand, it is known that I = 3 + j 4 = 5 ∠53.13°. Now, knowing the voltage between the points a-x and the current that the voltage source supplies to the circuit, the equivalent impedance (Z_eq) of the circuit can be calculated. Soon:

Carrying out the calculation:

Alternatively, we could have found the value of Z_eq by dividing the source voltage (of the first circuit) by the electric current I. The value found would be 11.2 - j1.6. Subtracting the resistor value from 10 ohms, we would arrive at the same value of Z_eq = 1,2 - j1,6. We opted for the above method to show that we can reduce the circuit, so that it is more "readable" to the student, as can be seen in the figure below.

Analyzing the circuit between the points b-x it is said that the module of the potential difference between these points is 5 volts. But the current, in module, flowing through it is 5 A. Then, one can easily calculate the equivalent impedance modulus between these two points. Like this:

For an R-L circuit, the impedance module is given by:

Adapting for this case and replacing by known values, we obtain:

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

As the value of Z_bx was determined, the value of Z_ab is easily calculated using the relation below.

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

With this data in hand, the value of Z can be calculated. Initially, the parallel of Z is calculated with the impedance 2 - j 6. Compare this result with the one obtained above and determine the value of Z.

After some algebraic work, the value of Z is obtained, or:

And finally let's calculate the value of V_ab, because applying Ohm's law:

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

Complementing the problem, the value of V_bx will be:

Carrying out the calculation, the value of V_bx is found:

Final considerations

Finally, a graph is presented showing the voltages involved in the circuit and the electric current I. See on figure below how it turned out.

The voltage V_ax is taken as reference, as it has an angle of zero degrees. Note that this voltage is the result of the phasor sum of V_ab and V_bx. In the graph on the side, this looks perfectly evidenced.

On the other hand, we have that the phasor sum of E with - V_R (this voltage, over the resistor of 10 ohms) results in V_ax as well. Note that V_R (not shown in the graph) is in phase with I, because at resistors there is no voltage or current phase shift. Thus, - V_R is in the opposite sense to I.

Finally, it should be noted that the current I leads the voltage E by an angle equal to 53.13° -45° = 8.13°. This is justified because the circuit has a capacitive predominance, given that the equivalent impedance of the entire circuit is    11.2 - j1.6 = 11.31∠-8.13°, as stated at the beginning of the problem solving.