Correct Answer - Option 3 :

\(\frac{5}{6}th\) of pole pitch

For eliminating n^{th} harmonic coil span required is

\( = 180 \times \left( {\frac{{n - 1}}{n}} \right)\)

To eliminate 5^{th} harmonic coil span required is

\(= 180 \times \left( {\frac{4}{5}} \right) = 144^\circ\)

To eliminate 7^{th} harmonic coil span required is

\(= 180 \times \left( {\frac{6}{7}} \right) = 154.285^\circ\)

To eliminate both 5^{th} and 7^{th} harmonic, we should take average of both i.e.

\(= \frac{{144^\circ\; + \;154.28^\circ }}{2} = 149.14^\circ \approx 150^\circ\)

150° is \(\frac{5}{6}\) parts of 180° i.e. pole pitch

Hence, the coil span in three phase AC machines must be

\(\frac{5}{6}th\) of pole pitch.