Question

For a generator of armature coil resistance 200 with
load resistance 101, current flowing is 5 amp. The
efficiency of this generator is​

Answer 1

Answer:

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.

Explanation:

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