Problema 85-3    Source:   Adapted from the Exercise 22.8 - page 677 - BOYLESTAD, Robert L. - Book: Introdução à Análise de Circuitos - 10ª Edição - Ed. Prentice Hall - 2008.

In the circuit shown in the figure below, V_AB = 208∠0°, V_BC = 208∠-120°, V_CA = 208∠120°, Z_AB = 10 Ω, Z_BC = 15 + j20 Ω and Z_CA =12 - j12 Ω. This unbalanced load, connected in the delta configuration, has two wattmeters connected as shown in the figure below. Find:

a)  the modulus and the angle of the phase currents.

b)  the modulus and the angle of the line currents.

c)  the reading of wattmeters.

Solution of the Problem 85-3

Item a

In order to calculate the phase current, I_AB, just apply the law of Ohm. Like this:

For the calculation of I_BC, the impedance value of the corresponding phase must be passed to the polar form, or Z_BC = 15 + j20 = 25∠53.13°. Then:

For the calculation of I_CA, Z_CA = 12 - j12 = 16.97∠-45° . Then:

Item b

In order to calculate the line currents, the law of Kirchhoff is applied to each node of the circuit. Like this:

To calculate I_B:

And for I_C:

Performing the calculation:

Item c

For the calculation of the wattmeter readings, great care must be taken in the angle of lag between the voltage and the current that the wattmeter reads. In case of W₁, it is reading the current I_A and the voltage V_AB. As V_AB has angle zero and current I_A has an angle of 5.55°, then the discrepancy between them is 5.55°. Therefore, the reading of W₁ will be:

In case of W₂, it is reading the current I_C and the voltage V_CB. Note that the supplied voltage was V_BC which is out of phase by 180° in relation to V_CB. Then, the angle of V_CB will be -120° + 180° = 60°,that is, the phase difference between current and voltage is of 130.65 - 60° = 70.65°. Therefore, the reading of W₂ will be:

Note that the signal of the phase angle between the voltage and the current has not been taken into account, since it is known that cos (- φ) = cos φ.