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 P_{1} = 5.5kw
V_{1} = 110V
Thus load current P_{1}/V_{1} = 5500/110 = 50A
Now generated voltage will be (when N_{1} = 1000rpm)
E_{g1} = V_{1} + I_{1} (r_{a} + r_{sc}) = 110 + 50 × 0.4
Let at 1500 rpm, current supplied by generator at terminal voltage V_{2} is I_{2}
V_{2} = P_{2}/I_{2} = 10000/I_{2}
_{ }
Generated voltage E_{g2} = V_{2}+ I_{2}(r_{a} + r_{sc}) =10000/I_{2} + 0.4I_{2}
As E_{g} ?Φ
or E_{g2}/E_{g1} = Φ_{2}/Φ_{1} × N_{2}/N_{1} = N_{2}/N_{1} _{ }× I_{2}/I_{1}
or 10000/I_{2} + 0.41/130 = 1500/1000 × I_{2}/50
or 10000 + 0.4 I_{2}^{2} = 3.9 I_{2}^{2}
or 3.5 I_{2}^{2} = 10000
thus I_{2} = 53.45 A
and V_{2} = 10000/53.45 = 187.14