Q. Two shunt wound generators running in parallel supply a total load current of 4000 A. Each generator has an armature resistance of 0.02? and a field of 40?. The fiels are excited so that the emf induced in one generator is 480 V and in another 490 V, find the bus - bar voltage and kilowatt output of each machine.
Sol. V = bus bar voltage
I_{1} = current output of generator 1
I_{2} = current output of generator 2
I_{1} + I_{2} = 4000
I_{sh} = V/R_{sh}
Bus - bar voltage
In each machine
Bus bar voltage + voltage
Drop in armature = EMF generated
for generated 1, E_{1} = V + I_{a1} R_{a}
480 = V + (I_{1} + V/40) × 0.02
(I_{a1} = I_{1} + I_{sh})
for generated 2 490 = V + (I_{2} + V/40) × 0.02
Subtrating both with each other we get,
0.021_{2} - 0.021_{1} = 10
or I_{2} - I_{1} = 500
or (4000 - I_{1}) - I_{1} = 500
I_{1} = 1750A and I_{2} = 4000 - 1750 = 2250A
Now 480 = V + (I_{1} + V/40) × 0.02 or 490 = V + (I_{2} + V/40) × 0.02
On solving V = 444.8 volts
Kilowatt output
kW output of gen. 1, VI_{1}/1000 = 444.8 × 1750/1000 = 778.4kW
kW output of gen. 2, VI_{1}/1000 = 444.8 × 2250/1000 = 1001kW