Problem + Hard 85-1    Source:  problem prepared by the author of the site.

In the circuit of the Figure 85-1.1 we have V_ab = 440∠30° V, V_bc = 440∠150° V , V_ca = 440∠-90° V, ie an inverse sequence,  and   Z = 45 + j 26. Calculate:

a) the line currents.

b) the source line voltages.

c) the reading on the two wattmeters and the total power.

d) the reactive power.

e) the apparent power.

Solution of the Problem + Hard 85-1

Item a

Passing the value of Z to polar form we have Z = 45 + j 26 = 52∠30°. Soon, we can calculate the value of the phase currents.

As the circuit is balanced, the two other phase currents have the same magnitude as I_ab, but shifted by 120°. Soon:

We know that in an inverse sequence the line currents are 30° ahead of the phase currents and their magnitude must be multiplied by √3. Soon:

Item b

To calculate the source voltages we need to calculate the voltage drop across the line impedance. It is impedance is formed by the series circuit Z_line = 2 + j = √5∠26.56°. Then V_Aa is given by:

Carrying out the calculation we find:

In the same way we can find the value of V_Bb.

Carrying out the calculation we find:

To calculate the line voltage V_AB we use the Kirchhoff voltage law, or:

Doing the numerical substitution and performing the calculation we find:

Naturally, the other two line voltages will be 120° out of phase with V_AB and, remembering that we are facing an inverse sequence:

Item c

To find the values ​​measured by the wattmeters we must use the load angle which is φ = 30°. For W₁, we will use the eq. 85-03. Like this:

Carrying out the calculation we find:

For the value of W₂, we will use the eq. 85-04, or:

Making the numerical substitution and carrying out the calculation we find:

So the total real power supplied by the voltage source is:

Item d

We can calculate the reactive power using the eq. 85-06. Then:

Item e

To calculate the apparent power we will use the eq. 85-08. Then: