Question

For a balanced- $\mathbf{∆}$ load supplied by a balanced positive-sequence source, the line currents into the load are $\sqrt{\mathbf{3}}$times the $\mathbf{∆}$-load currents and lag by$30°$ .

(a) True

(b) False

Short answer

Answer

The statement is true.

Step-by-step solution

Define relation between line-to-line current and line-to-neutral current

The line-to-line voltages are defined as follows:

$$\begin{gathered} E_{ab}=\sqrt{3}\left(10\angle 30°\right)V \\ {E}_{bc}=\sqrt{3}\left(10\angle 90°\right)V \\ {E}_{ca}=\sqrt{3}\left(10\angle 150°\right)V \end{gathered}$$

Consider the load impedance is of $Z_y=2\angle 30°\Omega$. Then, the load currents are as follows:

$$\begin{gathered} I_{AB}=\sqrt{3}\left(\frac{10\angle 30°}{5\angle 30°}\right) \\ =3.46\angle 0°V, \end{gathered}$$

Solve further as,

role="math" localid="1652698303023" $$\begin{gathered} I_{BC}=\sqrt{3}\left(\frac{10\angle -90°}{5\angle 30°}\right) \\ =3.46\angle -120°A, \end{gathered}$$

and

$$\begin{gathered} I_{CA}=\sqrt{3}\left(\frac{10\angle -90°}{5\angle 30°}\right) \\ =3.46\angle +120°A, \end{gathered}$$

Using the KCL equation at bus a, we have

$$\begin{gathered} I_a=I_{AB}-I_{CA} \\ =3.464\angle 0°-3.464\angle 120°A \\ =\sqrt{3}\left(3.464\angle -30°\right) \end{gathered}$$

Using the KCL equation at bus b, we have

$$\begin{gathered} I_b=I_{BC}-I_{AB} \\ =3.464\angle 120°-3.464\angle 0°A \\ =\sqrt{3}\left(3.464\angle -150°\right) \end{gathered}$$

Using the KCL equation at bus c, we have

$$\begin{gathered} I_c=I_{CA}-I_{BC} \\ =3.464\angle 120°-3.464\angle 120°A \\ =\sqrt{3}\left(3.464\angle -90°\right) \end{gathered}$$

The above calculations show that for the balanced load system, the line current into the load is $\sqrt{3}$ times the load currents that lag by $30°$.

$$\begin{gathered} I_a=\sqrt{3}\left(I_{AB}\angle -30°\right) \\ {I}_b=\sqrt{3}\left(I_{BC}\angle -30°\right) \\ {I}_c=\sqrt{3}\left(I_{CA}\angle -30°\right) \end{gathered}$$

Determine if the statement is true or false

It is clear from the relation between the line-to-line current and the line-to-neutral current that the neutral current is $\sqrt{3}$ times the line-to-line current.

Thus, the statement is true.