Problem 53-2    Source:   Problem 4 - RLC Problem List - Discipline Engineering School Electrical Circuits - UFRGS - 2011 - Prof. Dr. Valner Brusamarello.

In the circuit of Figure 53-02.1, V = - 300 sin (100 t - 20°) and I = 20 sin(100t - 146.9°). Determine R and X.

Attention:    In the Ufrgs list there are some typing errors. In current I, rather than ωt, is 100t. In the drawing the current I is not represented. See circuit above with correction. In the answer appears X = C = 1250 uF.   X is the reactance in ohms. C is the capacitance in microfarads. Therefore, the correct answer is: R = 4 Ω,   X = 8 Ω   and    C = 1250 µF.

Solution of the Problem 53-2

Initially, the Thévenin equivalent is calculated for the source and the two known resistors of the circuit. Thus, the remodeled circuit is shown in the Figure 53-02.2, where Thévenin voltage is V_th = - 200 sin(100t - 20°) and Thevenin's resistance is the parallel between the resistors of 3 ohms and 6 ohms, resulting in one of 2 ohms. The current I continues the same, circulating for R and X.

To solve this problem, we must first understand the relation between V and I. Note that the function - sin ωt can be represented by the function + sin (ωt + 180°) or by + sin (ωt - 180°). In this particular case, the second alternative was chosen. Thus, by writing V as:

Note that both V and I are written as a positive sine function. Then, the phase difference between the two values can be determined. Do not forget that the equivalent impedance for the source is given by the quotient between the voltage and the current. In this way, we can write:

From this result we conclude that the circuit has capacitive predominance, since the equivalent impedance angle is negative. Recall that the tangent of the angle is given by the quotient between the imaginary part of the impedance and its real part. Based on this information:

By performing the calculation we find the following relationship between X and R:

From the found equivalent impedance value, it is easy to conclude that |Z| = 10 ohms. On the other hand, we know that the modulus of an impedance is given by the square root of the sum of the squares of the real and imaginary part. Like this:

Substituting X for the value found in the equation above and developing, we find a equation of the second degree as a function of only R. See below the equation:

Solving this equation determines two values: R = - 2 ohms, this value is discarded, because it is impossible physically to have resistances (passive element) of negative value; and R = 4 ohms. Therefore, the value of R is:

As we already know the relation between X and R, easily calculate the value of X, or:

By the negative value of X it is perfectly defined that this reactance is due to a capacitor. To find the capacitance of this capacitor, use the relation below: