Question

A $3-\varphi ,Y-\Delta$ transformer has secondary to primary turns ratio per phase of 5. If the primary side voltage of transformer is 400 V, then value of secondary voltage will be

• A. $400\sqrt{3}V$
• B. 400 V
• C. 2000 V
• D. $\frac{2000}{\sqrt{3}}V$

Answer 1

The correct option is D $\frac{2000}{\sqrt{3}}V$

$\text{We know,}\frac{N_{\Delta \left(P\right)}}{N_{Y\left(P\right)}}=5$

$\text{Also,}$ $\frac{N_{\Delta \left(P\right)}}{N_{Y\left(P\right)}}=\frac{V_{\Delta \left(P\right)}}{V_{Y\left(P\right)}}$

$V_{Y\left(P\right)}=\frac{400}{\sqrt{3}}$V
$V_{\Delta \left(P\right)}=\frac{N_{\Delta \left(P\right)}}{N_{Y\left(P\right)}}{V}_{Y\left(P\right)}=5\times \frac{400}{\sqrt{3}}=\frac{2000}{\sqrt{3}}V$

$\text{For delta side,}$ $V_{\Delta \left(L\right)}=V_{\Delta \left(P\right)}=\frac{2000}{\sqrt{3}}V$