A separately excited dc generator when running at 1200 rpm, supplies 200 A at 125 V to a circuit constant resistance. What will be the current when the speed is dropped to 1000 rpm and the field current is reduced to 80%? The armature resistance is 0.04 ohm and total brush drop is 2 V.
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- A separately excited dc generator when running at 1200 rpm supplies 200 A at 125 V to a circuit of constant resistance. Armature circuit resistance is 0.10 ohm. Determine the terminal voltage when the speed is dropped to 1000 rpm. Assume that the field current is unaltered.Suppose we have a dc motor that produces a back emf of EA=240 V at a speed of 1500 rpm. a. If the field current remains constant, what is the back emf for a speed of 500 rpm? b. For a speed of 2000 rpm?A 110V shunt generator has a full load current of 100A, shunt field resistance of 55 ohms and constant losses of 500W. If full load efficiency is 88%, find the armature resistance. Assuming voltage to be constant at 110V, calculate the efficiency at half load. Find the load current.
- A separately excited generator,when running at 1000rpm supplied 200A at 125V. What will be the load current when the speed drops to 800rpm if current is unchanged? Given the armature resistance of 0.04 ohms and brush drop of 2V.A DC motor has stall torque of 2 N-m (Newton*meters) and a no-load speed of 2000 RPM. With a load of 0.5 N-m, (a) what is the motor speed?(b) what it the motor power? (c) What is the maximum power the motor can deliver?A 200V dc shunt generator can deliver 10 kW at half load with efficiency of83 %. If the armature resistance is 0.05 ohms and shunt field resistance of 50ohms. What will be the efficiency of the generator when running at full load?
- A shunt excited DC motor consumes 8Kw by feeding it from a 500V line and driving a load at 1000RPM. The counter electromotive force generated is 496V, the resistance of the excitation winding is 250Ω and its total losses account for 17% of the absorbed power. Determine: a) The resistance of the armature winding b) The useful torque c) Total copper losses..A separately excited d.c. generator, when running at 1200 r.p.m. supplies 200 A at 125 V to a circuit of constant resistance. What will be the current when the speed is dropped to 1000 r.p.m. and the field current is reduced to 80%? Armature resistance, 0.04 ohm and total drop at brushes, 2 V. Ignore saturation and armature reactionA 110-V shunt generator has a full load current of 100 A, shunt field resistance of 55 Ω and constant losses of 500 W. If F.L. efficiency is 88%, find armature resistance. Assuming voltage to be constant at 110 V, calculate the efficiency at half-F.L. and at 50% overload. Find the load current corresponding to maximum efficiency.
- A 110-V shunt generator has a full-load current of 100 A, shunt field resistance of 55 Ω and constant losses of 500 W. If F.L. efficiency is 88%, find armature resistance. Assuming voltage to be constant at 110 V, calculate the efficiency at half F.L. And at 50% overload. Find the load currentA dc shunt motor runs at 900 rpm from a 400V supply when taking an armature current of 25A. Calculate the speed at which it will run from a 230V supply when taking an armature current of 15A. The resistance of the armature circuit is 0.75ohm. Assume the flux per pole at 240V to have decreased to 75% of its value at 400VA shunt generator delivers full-load current of 200A at 240V, the shunt field resistance 60Ω and full load efficiency of 90%. Determine the armature resistance.