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A 24 kW, 250 V, 1600 rpm separately-excited d.c. generator has armature circuit resistance of 0.1Ω. The machine is first run at rated speed and the field winding current is adjusted to give an open circuit voltage of 260 V. Now, when the generator is loaded to deliver its rated current, the speed of the driving motor is found to be 1500 r.p.m. Assuming flux to remain constant, the terminal voltage of the generator under these conditions is
  • a)
    250 volt
  • b)
    226 volt
  • c)
    238 volt
  • d)
    234 volt
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A 24 kW, 250 V, 1600 rpm separately-excited d.c. generator has armatur...
We know that,

Now, rated armature current is 

So, the terminal voltage is
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A 24 kW, 250 V, 1600 rpm separately-excited d.c. generator has armatur...
Ohms. The field winding resistance is negligible. The generator is connected to a load that requires a constant voltage of 250 V. Calculate:

a) The full-load current and output power of the generator.
b) The open-circuit voltage of the generator.
c) The percentage voltage regulation of the generator when delivering full-load current.

a) Full-load current (I) can be calculated using the formula:

I = P / V

where P is the output power and V is the voltage.

Output power (P) = 24 kW

Voltage (V) = 250 V

Therefore, Full-load current (I) = 24,000 / 250 = 96 A

b) Open-circuit voltage (E0) can be calculated using the formula:

E0 = V + IaRa

where V is the voltage, Ia is the armature current and Ra is the armature circuit resistance.

At open-circuit, Ia = 0

Therefore, E0 = V + 0 = 250 V

c) Percentage voltage regulation can be calculated using the formula:

% Regulation = (E0 - V) / V x 100%

At full-load, the armature current (Ia) can be calculated as:

Ia = P / V = 24,000 / 250 = 96 A

The armature voltage drop (IaRa) can be calculated as:

IaRa = Ia x Ra = 96 x 0.1 = 9.6 V

Therefore, the terminal voltage (E) at full-load is:

E = V - IaRa = 250 - 9.6 = 240.4 V

% Regulation = (E0 - E) / E x 100%

% Regulation = (250 - 240.4) / 240.4 x 100% = 3.98% (approx.)
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A 24 kW, 250 V, 1600 rpm separately-excited d.c. generator has armature circuit resistance of 0.1Ω. The machine is first run at rated speed and the field winding current is adjusted to give an open circuit voltage of 260 V. Now, when the generator is loaded to deliver its rated current, the speed of the driving motor is found to be 1500 r.p.m. Assuming flux to remain constant, the terminal voltage of the generator under these conditions isa)250 voltb)226 voltc)238 voltd)234 voltCorrect answer is option 'D'. Can you explain this answer?
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A 24 kW, 250 V, 1600 rpm separately-excited d.c. generator has armature circuit resistance of 0.1Ω. The machine is first run at rated speed and the field winding current is adjusted to give an open circuit voltage of 260 V. Now, when the generator is loaded to deliver its rated current, the speed of the driving motor is found to be 1500 r.p.m. Assuming flux to remain constant, the terminal voltage of the generator under these conditions isa)250 voltb)226 voltc)238 voltd)234 voltCorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 24 kW, 250 V, 1600 rpm separately-excited d.c. generator has armature circuit resistance of 0.1Ω. The machine is first run at rated speed and the field winding current is adjusted to give an open circuit voltage of 260 V. Now, when the generator is loaded to deliver its rated current, the speed of the driving motor is found to be 1500 r.p.m. Assuming flux to remain constant, the terminal voltage of the generator under these conditions isa)250 voltb)226 voltc)238 voltd)234 voltCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 24 kW, 250 V, 1600 rpm separately-excited d.c. generator has armature circuit resistance of 0.1Ω. The machine is first run at rated speed and the field winding current is adjusted to give an open circuit voltage of 260 V. Now, when the generator is loaded to deliver its rated current, the speed of the driving motor is found to be 1500 r.p.m. Assuming flux to remain constant, the terminal voltage of the generator under these conditions isa)250 voltb)226 voltc)238 voltd)234 voltCorrect answer is option 'D'. Can you explain this answer?.
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