Problem + Hard 52-5    Source:   Adapted from Problem 60 - List of Problems Circuits II - Electrical Circuits of the School of Engineering - UFRGS - 2017 - Prof. Dr. Valner Brusamarello.

The 20 kVA transformer, shown in Figure 52-05.1, operates with 62.5% of its nominal capacity and with FP = 0.8 overdue. The wattmeter W reads 4,000 W with the current lagging behind the voltage and the voltmeter V reads 200 V at your reading. Determine the values ​​of R, X and their nature.

Solution of the Problem + Hard 52-5

Note that the voltage V_bd = 200 V was supplied and the current that circulates through the impedance Z and the passive network. How two are in series it is possible to calculate the equivalent impedance. There are several ways to calculate this equivalent impedance. As we know the real power consumed by it, we can calculate the value of the equivalent resistance. Let's call it R_rp:

It is also possible to find the modulus of the equivalent impedance, or:

And with these two data it is possible to calculate the inductive reactance of the equivalent impedance, as we know that Z_rp² = R_rp² + X_rp². Therefore:

As the problem statement states that the current lags behind the voltage, we conclude that X_rp is an inductor. And Z_rp is:

Thus, between points b - d we have two impedances in parallel: 4 + j8 and Z_rp. Therefore, we can calculate this parallel and we will call it Z_bd.

Carrying out the calculation, we obtain:

Looking at the circuit in Figure 52-05.1, we see that there is a resistor of 0.4 Ω and a capacitor of - j 8 Ω in series with the impedance Z_bd. Therefore, it is possible to add them together and obtain a new impedance that we will call Z_k.

We must remember that a parallel R-L circuit has an equivalent series R-L circuit. Thus, in the circuit of Figure 52-05.1, we will replace the parallel R-L circuit with a series R-L circuit. Let's call it Z_s = R_s + j X_s . See how the circuit turned out in Figure 52-05.2.

Note that with the transformer data we can calculate I_T in module and phase. Since FP = 0.8, we know that φ = 36.87° (φ is the phase angle). On the other hand, let's assume that the transformer voltage is the reference. Then V = 200 √2 ∠0°. The statement states that the transformer operates with 62.5% of its rated power. Therefore, the apparent power, S_T, that the transformer supplies to the circuit is:

And the current supplied by the transformer is:

Note that the phase angle of the current is negative, as its power factor is inductive.

We can easily verify from the circuit shown in Figure 52-05.1 that:

Let's calculate the value of I_C.

Knowing I_T and I_C we can calculate the phase angle of I, because we already know the module. That way:

Carrying out the calculation, we find:

This means that the current I is 45° behind the voltage supplied by the transformer. With this information we can calculate the impedance equivalent that the transformer "sees" to the right of the capacitor, that is, excluding the capacitor from the circuit. We will call this impedance Z_eq.

Carrying out the calculation, we find:

Knowing Z_eq and Z_k we can easily find the value of Z_S, because:

Carrying out the calculation, we find:

Well, now that we have calculated the series impedance we need to transform it to a parallel impedance. To do this, we will use the following transformation based on the circuit shown in Figure 52-05.1:

Nota

At this point we will pause to calculate a relationship between R' and X'. This is possible because we know the real and reactive powers involved in the circuit. As the transformer supplies 12,500 VA to the circuit with a FP = 0.8, then we have that the real and reactive power supplied to the circuit is:

Let's calculate the real power dissipated by R_K = 2.8 Ω

And, naturally, the power dissipated by R' will be the difference between the power supplied by the transformer and the power dissipated by R_K. Soon:

As for reactive power, we can calculate the reactive power consumed by X_K. Soon:

And the reactive power supplied by the capacitor C is:

Then the reactive power consumed by X' is:

Referring to the circuit shown in Figure52-05.2, we can write that:

Obtaining the quotient R' / X' and making the numerical substitutions of P_R' and Q_X' we find the relationship we need, or:

After this parenthesis we can continue solving the problem by finding the parallel between R' and X' which we will call Z_P.

Remembering the properties of complex numbers, we know that division can be simplified by multiplying and dividing the fraction by the complex conjugate of the denominator. Like this:

After some algebraic operations with complex numbers and separating the real part from the imaginary part, we obtain:

Substituting X' = 0.75 R' in the equation above we find the values ​​of R' and X', or:

However, let's not forget that R' = R + 0.1 and X' = X + 0.7. So, the values ​​of R and X are:

And as is evident, the nature of X is an inductor.