Gate 2016 EE – Set 2 – Question 43

Given rated values, $$V_{1}=415V, I_{L}=120A, V_{2}=110V$$

$$I_{2}=x(I_{st})_{rated}$$

$$x=\frac{V_{reduced}}{V_{rated}}=\frac{110}{415}$$

$$I_{2}=\frac{110}{415}\times 120=31.807A$$