Problem 84-1    Source:   Problem elaborated by the author of the site.

Assuming a direct sequence in an unbalanced source, we have:

V_AN = 100 ∠30°,  V_BN = 150 ∠-60°   and   V_CN = 120 ∠180°. Calculate:

a) The modulus of the line voltages and their respective angles.

b) Make diagram of voltages.

Solution of the Problem 84-1

Item a

According to the values of  V_AN = 100 ∠30°  and  V_BN = 150 ∠-60° we clearly notice that the two voltages are out of phase by 90° (see figure below). In this case we can use the theorem of Pythagoras to find the modulus of V_AB. So:

To find the angle that V_AB does with the horizontal axis, we can use the sine law applied to the angle BÂN. Like this:

Substituting the numerical values and performing the calculation:

Adding the angle of 30° that V_AN does the horizontal (review property of the outer angle in triangles), we find the angle that V_AB makes with the horizontal axis, that is, 56.3° + 30° = 86.3°. Then:

Now let's calculate the magnitude and angle of V_BC. As V_BN and V_CN make an angle to each other other than 90°, we must use the cosine law. Then:

Substituting the numerical values and performing the calculation, we find:

Using the sine law it is possible to calculate the angle BĈN that the line voltage V_BC does with the horizontal. Note that this angle must be negative because it points down.

Substituting the numerical values and performing the calculation:

Therefore this angle, with negative sign, is the angle between V_BC and the horizontal axis. Soon:

And finally to calculate V_CA, we can use the same methodology, remembering that the angle between V_AN and V_CN is of 150° (see figure above). Like this:

Substituting the numerical values and performing the calculation:

Using the sine law it is possible to calculate the angle AĈN that the line voltage V_CA makes the axis horizontal. Notice that this angle must be negative because V_CA points down.

Substituting the numerical values and performing the calculation:

How do we calculate the angle between V_CA and the horizontal axis just subtract 180° of AĈN;. Then:

Summarizing the values found:

Item b

See in the Figure 84-1.1 the diagram of the voltages.