Problem 83-5    Source:   Problem 11.17 - page 273 - Book: Circuitos Elétricos - NILSSON, James W. & RIEDEL, Susan A. - 5ª edição - Editora LTC - 1999.

In the balanced circuit of the Figure 83-5.1, assuming a direct sequence, we have

V_1n = 12,600 ∠0°  , V_2n = 12,600 ∠-120°  and V_3n = 12,600 ∠120°. In addition, knowing that Z = 2,514 - j 900 Ω calculate:

a) The value rms and the phase angle of the current I_BC.

b) What percentage of the average power provided by the source is dissipated by the load.

Solution of the Problem 83-5

Item a

Since the circuit is balanced, the easiest way to solve the problem is to do a transformation of the delta circuit to star. To find the new values of load impedance simply divide their original values by three. Below, the new circuit with its respective values is shown in the Figure 83-5.2. Because it is a balanced circuit, we know that the potential of point n is the same as the point N. Therefore, we can calculate the current of phase A as if it were a single-phase circuit.

Thus, we must calculate the total impedance of phase A between points 1 and N. Adding all impedances:

Then the current I_A will be:

As the circuit is balanced the other two currents have the same module that I_A but lagged by 120°. So:

On the other hand, in a balanced circuit it is known that the phase current in a delta circuit advances 30° in relation to the phase current of a star circuit, in addition to having its magnitude reduced by a factor √3. Thus:

Performing the calculation:

Another way to solve, a little more laborious, would be to calculate the voltage V_BN = V_2n - (2 + j20) I_B = 12 666∠-121.26° and the voltage V_CN = V_3n - (2 + j20). I_C = 12 666∠118.74°.With this data it is possible to calculate V_BC = V_BN - V_CN = 21 938.16∠-91.26°. Returning to the original circuit, with the loads in the delta configuration, and applying the Ohm's law, we get the current I_BC. Amount already calculated above.

Item b

In order to calculate the average power in the load, we can calculate the power in one phase and multiply it by three. Note that for the average power only the resistance value must be used. Then:

For the calculation of losses, the transmission line resistance, 2 ohms must be used per phase. However, it is necessary to use the line current rather than the phase current. To calculate the total loss, multiply the result by three, or:

Then to find the percentage of the average power that the source supplies to the load: