Problem + Hard 55-8    Source:   Adapted from Problem 4 - RLC 2008 Problem List IV - Discipline Electrical Circuits of the School of Engineering - UFRGS - 2008 - Prof. doctor Valner Brusamarello.

In the circuit below, the voltmeter V measure 70 volts. In addition, the voltage E is 30° behind V_ab. Determine the magnitude and phase of the voltage source E, as well as the value and nature of X. Also determine the value of V_ab , I and the voltage V_ca. Make one phasor diagram of the main voltages in the circuit.

Solution of the Problem + Hard 55-8

The two impedances that are in parallel can be determined and named as follows:

The voltage phase V_ab was not mentioned in the problem statement. Therefore, it is appropriate to take it as a reference, making V_ab = V_ab ∠0°. Thus, using this reference, one can calculate the currents I_L and I_C. Then:

From the circuit it is verified that I = I_L + I_C. Note that I_L and I_C are 90° out of phase with each other. So it is possible to use the Pythagoras theorem to calculate I. Carrying out the calculation:

Observing the circuit very carefully and taking into account the information given in the problem statement, it is possible to build the phase diagram of the voltages and currents of interest for the solution of the problem. This diagram is shown in figure below.

Note that the points of interest are marked on the circuit and on the graph above. From the circuit it is possible to calculate the impedance between points a-b when solving the parallel existing between these points. Calling this impedance Z_ab and carrying out the calculation, we obtain:

And the existing impedance between the points c-a and calling it Z_ca:

Thus, the voltage between these points and calling it V_ca is:

Performing the calculate:

Now it is necessary to calculate the voltage V_am. Note that this voltage is across the 5 ohms resistor where current flows I_L. Soon:

Performing the calculate:

Note that the two calculated voltages are in phase (-60°), as shown in the graph above. The sum of these voltages is exactly the same as the voltmeter reading V (70 volts). Thus, one can write:

Substituting the numerical values ​​calculated above and considering only the voltage modules V_ca and V_am, we have:

Now you can easily calculate the value of V_ab, because:

With this data it is possible to calculate the value of V_ca as requested in the problem statement, or:

With the value of V_ab it is possible to calculate the value of I, or:

To proceed with the solution of the problem, it was decided to reduce the circuit showing the 4 ohms resistor with the reactance X. Note that the point n is the junction of the two components. The figure on the side shows this configuration agreeing with the graph shown above. Note that all components are in series and current I flows through them.

It is possible to calculate the voltage across the 4 ohms resistor, which is the voltage V_nc = 4 I. making up to numerical substitution and calculating, we find:

Looking at the graph above, note that from the point n to the voltage E, what you have is the voltage on the element X, which will be named V_x. This voltage must be 90° behind the I (same phase as V_nc) so that the polygon closes at point E. Thus, it is evident that the element X is a capacitor.

Right now there are two unknowns to determine: the value of E and the value of V_x. Can be determined graphically or analytically.

Graphically, we know the value of the voltage V_nc and the angle of -15° in relation to V_ab . At the point n, draw a line perpendicular to V_nc until you find the phasor E, which in turn has a -30° angle with respect to V_ab. Check that V_x has the same value as V_nc. As the two components (resistor and capacitor) are in series and carry the same current, it is concluded that:

Analytical Calculation

We will present two different ways to solve the problem analytically.

Alternative I

To find the value of E we know that V_nc = 56.56 ∠-15°. From the graph above we see that V_nc makes an angle of 45° with the voltage phasor V_Ec . Then, we conclude that |V_x| = |V_nc|. In this way, the phasor V_x closes the right triangle c-n-E. Soon, we can easily calculate the value of V_Ec, because:

Substituting for numerical values, we have:

Notice from the graph that V_Ec is in phase with V_ac. So we can add the two phasors and obtain the value of V_Ea = 80 + 20 = 100 V. Now we know two sides of the triangle b-a-E, that is, V_ab and V_Ea. And we know the angle between them which is 120°. With that and using the law of cosines we can calculate the value of E.

ubstituting for numerical values, we have:

Performing the calculate, we have:

Note that we added the angle of E, as the problem statement states that E is lagging by 30° in with respect to V_ab and we define at the beginning of the problem solution V_ab = V_ab∠ 0°.

Alternativae II

For an analytical calculation, you must work with the line equation. Remembering that the reduced equation of the line is given by y = a x + b, and that a = tan φ, where φ is the angle that the line makes with the reference line (in this case, V_ab). For the voltage V_nc, it is known that the angle is -15°. Consequently a = tan (-15°) = - 0.268. And to find the equation of the line, you must know a point on the line. Using the point n, x_n and y_n are calculated as follows:

Ready. Knowing a point on the line and its angular coefficient, it is possible to determine the equation using the relationship below:

Substituting for numeric values ​​and performing the calculation:

As it is known that V_x is perpendicular to V_nc, the next step is to find the equation of the line perpendicular to the V_nc passing through the point (x_n , y_n ) . So that two lines are perpendicular, their angular coefficients must obey the relation a_p = - (1 / a). So, it is determined that the value of the angular coefficient of the perpendicular line is a_p = 3.73. Returning to the relationship ⓵, find the equation of the line perpendicular to V_nc, or:

To determine the values ​​of E and V_x, you must know the equation of the line E. like this line passes through the origin, just know its angular coefficient , or a_E = tan (-30°) = - 0.577. So the equation is:

Now that you know the equation of the two lines of interest, you must determine the point where they intersect. Then, subtracting ⓷ from ⓶ and remembering that in point of intersection y_E = y_p and x_E = x_p , you get

From this relation one obtains the point of intersection of E and V_x. Like this:

For the calculation of y_E, the equation ⓷ was used. With these data, one can calculate the length of the phasor E. As it passes through the origin, then:

Thus, the phasor E is:

Finally, an analytical calculation of X can be carried out, remembering that V_x = 56.56 ∠-105°. Soon

Carrying out the calculation, we find:

Note that, analytically, it follows that X is a capacitor.