Problem 55-16    Source:   Problem 21 - List of RLC Problems - Discipline Electrical Circuits of the School of Engineering - UFRGS - 2011 - Prof. Dr. Valner Brusamarello.

In the circuit shown in the Figure 55-16.1, it is known that |I₁| = |I₂|. Besides that, V_ac is delayed of 45° in relation to the current source I.  Determine the values of R and X.

Solution of the Problem 55-16

Say that V_ac is delayed of 45° in relation to I is the same as saying I is advanced of 45° in relation to V_ac. In this way, it is assumed that V_ac as the reference voltage, that is, V_ac∠0°. Note, in the circuit above, which were called the currents that circulate through the capacitor and of the resistor, of I_C and I_R, respectively. Therefore, we know that I_C is advanced by 90° in relation to V_ac, while V_R does not suffer lags. On the other hand, from the circuit, I₁ = I_C + I_R. It is also possible to calculate the equivalent impedance between points a-c, since the values of the components that are in parallel are known. Like this:

Based on the value of Z_ac, we can determine the phase difference between I₁ and V_ac, that is :

Therefore I₁ is 18.43° advanced in relation to V_ac, because Z_ac is a capacitive circuit. By the statement of the problem, it is known that |I₁| = |I₂|. Then, we can write that:

From the path a-c-b we can write:

Equaling these equations and eliminating |I₁|, we obtain:

Using the properties of the complex numbers and effecting the module of the second member of the equality above, we find:

Thus, we find a relation between R and X_L, which will be used later.

In the graph in the Figure 55-16.2, we show the voltage V_ac as reference. Note that I_R has the same module and phase of V_ac, because R= 1Ω. But, I_C is advanced by 90° in relation to V_ac. Making the phasor sum of I_R and I_C, we obtain I₁ advanced of 18.43° in relation to V_ac, already calculated previously. By the statement of the problem, it is known that I is advanced by 45° in relation to V_ac. This way it is easy to calculate the angle between I and I₁, that is, 45° - 18.43° = 26.57°, as shown in the graph. On the other hand, it is known that |I₁| = |I₂|. And fron the circuit, I = I₁ + I₂. Then, we concluded that I₂ have to make an angle of 26.57° with I. Therefore, we can calculate the phase difference between I₁ and I₂ which is equal to 2 (26.57°) = 53.14°.

We can use the latter conclusion to find the value of X_L. It is known that the impedance Z_acb = Z_ac + j X_L = 0.9 + j(X_L - 0.3) and in addition, this impedance is responsible for the phase difference of θ_I = 53.14° between I₁ and I₂. So:

As tan (53.14°) = 4/3, then:

Performing the calculation:

Using the relation eq. 55-16.1 found previously, we find the value of R, or: