What is the voltage regulation of a transformer while supplying a current of 10 A at 480 V, at a power factor of 0.8 lag , if the equivalent resistance and reactance of the transformer referred to the secondary of a single-phase transformer is 0.6Ω and 0.8Ω respectively?

Correct option is C

​$$\text{Voltage regulation of a transformer (approximate) is given by:} \\[8pt]\%\!VR = \frac{I\left(R_{\text{eq}}\cos\phi + X_{\text{eq}}\sin\phi\right)}{V} \times 100 \\[12pt]\textbf{Given:} \\[6pt]\bullet \ \text{Load current } I = 10 \text{ A} \\[4pt]\bullet \ \text{Secondary voltage } V = 480 \text{ V} \\[4pt]\bullet \ \text{Power factor } = 0.8 \text{ lag} \\[6pt]\cos\phi = 0.8, \quad \sin\phi = 0.6 \\[8pt]\bullet \ \text{Equivalent resistance } R_{\text{eq}} = 0.6 \, \Omega \\[4pt]\bullet \ \text{Equivalent reactance } X_{\text{eq}} = 0.8 \, \Omega \\[14pt]\textbf{Calculation:} \\[8pt]\%\!VR = \frac{10(0.6 \times 0.8 + 0.8 \times 0.6)}{480} \times 100 \\[10pt]= \frac{10(0.48 + 0.48)}{480} \times 100 \\[10pt]= \frac{10 \times 0.96}{480} \times 100 \\[10pt]= \frac{9.6}{480} \times 100 = 2\%$$​​