Question

A wye-delta transformer has a turns ratio of 2:1. It has a secondary phase to phase voltage of 480 V. The secondary is connected to a wye-connected three-phase resistive load of 50 ohms in each phase. How much current flow is there in each load resistor?

Answer 1

Answer:

4.8 A

Explanation:

We have given phase voltage that is 480 volt

We know that for transformer $\frac{v_1}{v_2}=\frac{N_1}{N_2}$ where $v_1\ and \ v_2$ are phase voltage and $N_1\ and\ N_2$ are primary and secondary turns respectively

We have given 480 v as phase voltage

So $\frac{480}{v_2}=\frac{2}{1}$

$v_2=240 v$

We have given that $z_{phase}=50\ ohm$

so current through each load resistor that phase current $i=\frac{v_{phase}}{z_{phase}}=\frac{240}{50}=4.8\ A$