Q. Series generator having a combined armature and field resistance of 0.4? is running at 1000 r.p.m. and delivering 5.5kw at terminal voltage of 110 V. If the speed is raised to 1500 rpm and load is adjusted to 10 kw. Find the new current and terminal voltage. Assume that machine is working on the straight line portion of the magnetization characteristics.

Sol.    Given        P₁ = 5.5kw

                            V₁ = 110V

        Thus load current P₁/V₁ = 5500/110 = 50A

        Now generated voltage will be  (when N₁ = 1000rpm)

                             E_g1 = V₁ + I₁ (r_a + r_sc) = 110 + 50 × 0.4

      Let at 1500 rpm, current supplied by generator at terminal voltage V₂ is I₂

                                       V₂ = P₂/I₂ = 10000/I₂

            Generated voltage E_g2 = V₂+ I₂(r_a + r_sc) =10000/I₂ + 0.4I₂

            As          E_g ?Φ

             or          E_g2/E_g1 = Φ₂/Φ₁ × N₂/N₁ = N₂/N₁  × I₂/I₁

             or          10000/I₂ + 0.41/130 = 1500/1000 × I₂/50

             or          10000 + 0.4 I₂² = 3.9 I₂²

             or           3.5 I₂² = 10000

             thus        I₂ = 53.45 A

             and         V₂ = 10000/53.45 = 187.14