Question

A balanced $\mathbf{∆}$-load can be converted to an equivalent balanced-Y load by dividing the $\mathbf{∆}$-load impedance by

(a) role="math" localid="1652699155976" $\sqrt{\mathbf{3}}$

(b) 3

(c) $\frac{\mathbf{1}}{\mathbf{3}}$

Short answer

Answer

Therefore, the correct option is (b) 3.

Step-by-step solution

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

Consider the expression for the line current of the delta load.

$$\begin{gathered} {\mathit{I}}_{\mathbf{A}}\mathbf{=}\sqrt{\mathbf{3}}{\mathit{I}}_{\mathbf{A}\mathbf{B}}\mathbf{\angle }\mathbf{-}\mathbf{30}\mathbf{°} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{\sqrt{\mathbf{3}}{\mathbf{E}}_{\mathbf{A}\mathbf{B}}\mathbf{\angle }\mathbf{-}\mathbf{30}\mathbf{°}}{{\mathbf{Z}}_{\mathbf{∆}}} \end{gathered}$$

Consider the expression for the line current of the delta load.

$$\begin{gathered} I_A=\frac{E_{AN}}{Z_Y} \\ =\frac{E_{AB}\angle -30°}{\sqrt{3}{Z}_Y} \end{gathered}$$

Compare the two equations for the impedance of Y load.

$Z_Y=\frac{Z_{∆}}{3}$ …….. (1)

Thus, the relation between the Y load and $∆$ load impedance is obtained.

Define if statement is true or false.

From equation (1) it is clear that to convert delta load to equivalent start load, divide the delta load by 3.

Thus, the correct option is (3).