Correct Answer - Option 3 : 164 V

In a 3-ϕ transformer,

Primary turns, N_{1} = 420 turns

V_{1} = 3300 V

Secondary turns, N_{2} = 30 turns

For star delta transformer,

\({V_L} = \sqrt 3 \;{V_p}\) (star)

V_{L} = V_{P} (Delta)

Primary phase voltage, \({V_{p1}} = \frac{{{V_L}}}{{\sqrt 3 }} = \frac{{3300}}{{\sqrt 3 }}\)

V_{p1} = 1905.255 V

Now, \(\frac{{{V_{p1}}}}{{{V_{p2}}}} = \frac{{{N_1}}}{{{N_2}}}\)

\( \Rightarrow {V_{p2}} = {V_{p1}}\frac{{{N_2}}}{{{N_1}}}\)

\(= 1905.255 \times \frac{{36}}{{420}}\)

⇒ V_{2} = 164 V