GATE EE 2014 Set 3 — Question 63

The bridge draws quasi-square line currents from the delta secondary. Each line current ($i_a$, $i_b$, $i_c$) is $+I_0$ for $120^\circ$, zero for $60^\circ$, $-I_0$ for $120^\circ$ and zero for $60^\circ$, and the three are displaced by $120^\circ$. The firing angle $\alpha = 30^\circ$ only shifts these waveforms in time; it does not change their shape.For the delta winding, $i_a = i_{ab} - i_{ca}$ and $i_b = i_{bc} - i_{ab}$. With no circulating current ($i_{ab} + i_{bc} + i_{ca} = 0$):$$i_{ab} = \frac{i_a - i_b}{3}$$Over successive $60^\circ$ intervals, $i_a - i_b$ takes the values $I_0, 2I_0, I_0, -I_0, -2I_0, -I_0$. So the delta phase current is a six-step wave with levels $\pm\frac{I_0}{3}$ and $\pm\frac{2I_0}{3}$.The star primary phase has 1 turn for every $K$ turns of the secondary phase. Ampere-turn balance gives $i_R = K\, i_{ab}$. With $I_0 = 1$ p.u., the input phase current is a six-step wave with levels $\frac{K}{3}$ and $\frac{2K}{3}$ in each half-cycle.Option (A) has the right shape, but its levels are divided by $K$ instead of multiplied by it. Options (C) and (D) are quasi-square waves ($120^\circ$ blocks), which a delta-connected secondary does not reflect to the primary. Only option (B) matches.Answer: (B)