GATE EE 2023 Set 1 — Question 22

Field produced by the stator currents. In figure (i) the conductors at the top ($A$) and at the bottom ($B$) carry equal currents into the page, while those at the right ($A'$) and at the left ($B'$) carry the same current out of the page. Going round the periphery at $0^\circ$, $90^\circ$, $180^\circ$ and $270^\circ$ the current directions are out, in, out, in.Taking the space vector of the fundamental (two-pole) m.m.f. as the sum of the conductor contributions $i_k\,e^{j(\phi_k + 90^\circ)}$:$$\bar{F} \propto I e^{j180^\circ} + I e^{j360^\circ} - I e^{j90^\circ} - I e^{j270^\circ} = (-I + I) + (-jI + jI) = 0 .$$So the DC excitation produces no two-pole field at all; the pattern in-out-in-out repeating every $90^\circ$ is a four-pole field.Flux linking the rotor coils. From figure (ii), coil $a$–$a'$ is formed from conductor $a$ (top) and conductor $a'$ (bottom), i.e. the two coil sides are $180^\circ$ apart; coil $b$–$b'$ ($b$ at the right, $b'$ at the left) likewise spans $180^\circ$ mechanical.In a four-pole field a span of $180^\circ$ mechanical covers exactly two pole pitches. The flux entering the rotor surface over the first pole pitch is therefore equal and opposite to the flux leaving it over the second, so the net flux linked by each coil is zero:$$\lambda_{a\text{-}a'}(\theta) = \lambda_{b\text{-}b'}(\theta) = 0 \quad \text{for every rotor position } \theta = \omega t .$$Induced e.m.f. Since the linkage is identically zero (not merely zero at one instant), its rate of change is zero as the rotor turns at $\omega$ rad/s:$$e = -\frac{d\lambda}{dt} = 0 \quad \text{in both coils.}$$Answer: (D) $E_{a\text{-}a'} = E_{b\text{-}b'} = 0$