A 6-pole lap-connected DC generator has 480 conductors and armature circuit resistance is 0.06 ohm. If the conductors are reconnected to form wave winding, other things remaining unchanged, the value of armature circuit resistance will be

Given that:

DC generator wit Pole (P) = 6

Conductor = 480 (Z)

For lap winding armature resistance Ra = 0.06 Ω

When the generator is connected as lap winding then

\({R_a} = {R_{Lap}} = \frac{{Z/P}}{{m.p}} \cdot \frac{{\rho \ell }}{A} = \frac{Z}{{{P^2}}}\frac{{\rho \ell }}{A}\)        (Parallel path = m.p and m = 1)

Now, when the generator is connected as wave winding.

\({R_{wave}} = \frac{Z}{{\left( {{A_p}} \right)}} \cdot \frac{{\rho \ell }}{A}\)

In wave winding, the no. of parallel paths = A = 2

No. of conductors in each parallel path (AP) = \(\frac{Z}{2}\)

\({\left( {{R_a}} \right)_{wave}} = \frac{Z}{4}\frac{{\rho \ell }}{A}\)

\(\frac{{{{\left( {{R_a}} \right)}_{lap}}}}{{{{\left( {{R_a}} \right)}_{wave}}}} = \frac{{\left( {\frac{Z}{{{P^2}}}\frac{{\rho \ell }}{A}} \right)}}{{\left( {\frac{Z}{4} \cdot \frac{{\rho \ell }}{A}} \right)}} = \frac{4}{{{P^2}}} = \frac{4}{{{6^2}}} = \frac{1}{9}\)

⇒ (Ra)wave = 9 × (Ra)lap

= 9 × 0.06 = 0.54 Ω 

Alternate Method

 A²_lap × R_lap = A²_wave × R_wave

In lap winding A = P = 6

R_lap = 0.06 Ω

In wave winding A = 2

6² × 0.06 = 2² × Rwave

Rwave = 9 × 0.06 = 0.54 Ω