The percentage reduction in speed of a generator working with constant excitation on 500 V bus-bars to decrease its load from 500 to 250 kW. The armature resistance is 0.015 Ω. Effect of armature reaction can be neglected

Load current I₁ for the load 500 kW \(= \frac{{500\; \times \;{{10}^3}}}{{500}} = 1000\;A\)

Load current I₂ for the load 250 kW \(= \frac{{250\; \times \;{{10}^3}}}{{500}} = 500\;A\)

E_g1 = V + I_a1 R_a = 500 + (1000) (0.015) = 515 V

E_g2 = V + I_a2 R_a = 500 + (500) (0.015) = 507.5 V

E_g ∝ N

\(\Rightarrow \frac{{{E_{g1}}}}{{{E_{g2}}}} = \frac{{{N_1}}}{{{N_2}}}\)

\(\Rightarrow \frac{{{N_2}}}{{{N_1}}} = \frac{{{E_{g2}}}}{{{E_{g1}}}} = \frac{{507.5}}{{515}} = 0.09854\)

Percentage reduction in speed \(= \frac{{{N_1} - {N_2}}}{{{N_1}}} \times 100\)

\(= \left( {1 - \frac{{{N_2}}}{{{N_1}}}} \right) \times 100\)

= 1.45%