Problem + Hard 84-1    Source:  Problem prepared by the author of the site.

In the Figure 84-1.1 we have the circuit where the phase sequence is inverse or acb. It is known that V_an = 240∠0° V, V_bn = 240∠120° V and V _cn = 240∠-120° V. In addition we have Z_A = 8 + j27 Ω, Z_B = 18 + j17 Ω, Z _C = 18 - j43 Ω and Z_L = 2 + j3 Ω. Determine:

a) the line currents.

b) the voltages on the load of each phase.

c) the voltage V_Nn.

Solutiom of the Problem + Hard 84-1

Item a

To solve this problem, we will use the circuit shown in the Figure 84-1.2. This one circuit is a simplification of the original where we add the load impedance values ​​with the line impedance. Doing the algebraic sum of the impedances we have the values ​​of Z₁ = 10 + j30 Ω,  Z₂ = 20 + j20 Ω  and  Z₃ = 20 - j40 Ω

Using these values ​​we will establish the mesh equations to determine the value of I₁ and I₂. Starting with I₁ we have:

To I₂ we have:

We have a system of two equations with two unknowns. We know that subtracting two phase voltages gives a line voltage. In this way, performing the calculation we find the voltages:

Thus, with these data and solving the system, either by determinant, or by software like Octave, values ​​are found:

Now, comparing the two circuits above, it is concluded that:

Remember that I_C has the opposite direction to the sum of currents I₁ and I₂, so:

Carrying out the calculation, we find:

Item b

To determine the voltages on the load in each phase, one must calculate the voltages V_AN , V_BN and V_CN. Using the Ohm law we have:

Carrying out the calculation, we find:

Repeating the calculations for the phase voltages   B and C,   where V_BN = I_B Z_B   and   V_CN = I_C Z_C , we find:

Item c

To find the V_Nn voltage, you can loop through any of the three phases. Let's choose phase A. Then:

Substituting for numerical values ​​and carrying out the calculation, we find: