Question

A 30kw, 300 volt DC shunt generator has armature and field resistance of 0.05 ohm and hundred ohm respectively calculate the total power developed by the armature when it delivers full load output​

Answer 1

Answer:

The generator is supplying the Load of 20 kW, plus the Field Winding of 200 ohms. The total current drops an internal voltage across the armature winding of resistance 0.05 ohms. The EMF = Terminal Voltage + Internal drop.

Given

Terminal Voltage = 200 V; Field Resistance = 200 ohms; Load = 20 kW.

Load Current I = 20 kW/200 V = 100 Amps; Field Current If = 200 V/200 ohm = 1 Amp; Therefore, Total Current It = 101 Amps.

Internal drop = Armature Resistance * Total Current = 0.05 * 101 = 5.05 Volts

EMF = Terminal Voltage + Internal Drop = 200 + 5.05 = 205.05 Volts.

Answer 2

Terminal Voltage = 200 V; Field Resistance = 200 ohms; Load = 20 kW.

Load Current I = 20 kW/200 V = 100 Amps; Field Current If = 200 V/200 ohm = 1 Amp; Therefore, Total Current It = 101 Amps.

Internal drop = Armature Resistance * Total Current = 0.05 * 101 = 5.05 Volts

EMF = Terminal Voltage + Internal Drop = 200 + 5.05 = 205.05 Volts.