Question

A 50 kVA, 22 kV/220 V, $\Delta -Y$ distribution transformer has a resistance of 1 percent and reactance of 8 percentage.The magnitude of transformer's phase impedance referred to hv side will be ____

• A. $(290.4+j2323.2)\Omega$
• B. $(290.4+j1162.40)\Omega$
• C. $(280.25+j2242)\Omega$
• D. $(145.2+j1161.6)\Omega$

Answer 1

The correct option is A $(290.4+j2323.2)\Omega$

$\text{Given,}$
$\text{Transformer rating 50 kVA, 22 kV/220 V}$

$\text{percent resistance = 1\%}$

$\text{percent reactance =8\%}$

$\text{Taking hv side voltage as base voltage is 22 kV.}$

$\text{Base apparent power,}{S}_B=50kVA$

$\text{As primary is}\Delta \text{connected.}$

$\text{Phase voltage,}\left(V_P\right)\text{= line voltage}\left(V_L\right)$

$\therefore$ $\text{Base impedance,}$ $Z_{base}=\frac{3(V_{\varphi ,base}{)}^2}{S_{base}}=\frac{3\times (22000{)}^2}{50\times {10}^3}=29040\Omega$

$\text{The per unit impedance of transformer,}$
$Z_{eqpu}=0.01+j0.08pu$

$\therefore$ $\text{High voltage side impedance,}$
$Z_{phhv}=Z_{eqpu}\times Z_{base}$
$=(0.01+j0.08)\times 29040$
$=290.40+j2323.2\Omega$