The ratio of the capacity of the star—star bank to the delta—delta bank of a transformer is:

Correct option is C

​$$\text{Delta–Delta capacity} = \sqrt{3} \, V_L I_L \\\qquad = \sqrt{3} V_L \left( \sqrt{3} I_P \right) \quad \{ I_L = \sqrt{3} I_P \} \\\qquad = 3 V_L I_P \\[10pt]\text{Star–Star capacity} = \sqrt{3} V_L I_L \\\qquad = \sqrt{3} V_L I_P \quad \{ I_L = I_P \} \\[10pt]\frac{\text{Star–Star capacity}}{\text{Delta–Delta}} = \frac{\sqrt{3} V_L I_P}{3 V_L I_P} \\\qquad = \frac{\sqrt{3} \cdot V_L \cdot I_P}{\sqrt{3} \cdot \sqrt{3} V_L I_P} = \frac{1}{\sqrt{3}} \\[10pt]= 0.577 \text{ or } 57.77\%$$​​