Home » Uncategorized » Gate 2016 EE - Set 2 - Question 43

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

Translate »