Problem 83-3    Source:   Problem elaborated by the author of the site.

In the circuit of the Figure 83-3.1, assuming a direct sequence, we have:

V_AN = 127 ∠0° and Z_AB = Z_BC = Z_CA = 15∠-60° . Calculate:

a) the line voltages and their respective phases.

b) the phase and line currents.

c) the active power dissipated by the circuit.

Solution of the Problem 83-3

Item a

As the phase voltage was supplied, V_AN, then the diagram on the Figure 83-3.2 shows that V_AB should be advanced by 30° in relation to V_AN. In consequence, V_BC should be delayed by 120° in relation to V_AB and V_CA should be advanced by 120° in relation to V_AB. And so, the line voltages are perfectly defined. Not forgetting that the magnitude must be multiplied by √3. Then:

Item b

To calculate the phase currents, simply apply the law of Ohm, or:

As we know the phase currents, you can easily find the line currents, just use the eq. 83-02 shown below.

Hence, we must multiply the magnitude of the phase current by √3 and subtract 30° from the angle. Recall that the line current A is associated with the phase current AB. For the other phases it is the same. Like this:

Another way to calculate the line current is to apply the law of Kirchhoff to each node of the delta connection.

Item c

In order to calculate the active power dissipated in the circuit it is enough to calculate the power dissipated by the resistor of each phase and multiply the result by three because the circuit is balanced. By transforming the value of the impedance given in the polar form to the Cartesian form, one discovers the value of the resistor. So Z_AB = 15∠-60° = 7.5 - j13. Therefore, the value of the resistor is 7.5 Ω. Then:

Then the total active power of the circuit is: