A separately excited DC generator when running at 1500 rpm, supplies 2...
Load resistance = 150/200 = 0.75Ω
E
1 = 150 + (200 × 0.25) + 2
= 202 V
N
1 = 1500 rpm, N
2 = 1000 rpm
= 134.67 V
V = 134.67 – 0.25 I – 2
I = Load current
V = 132.67 – 0.25 I
View all questions of this test
A separately excited DC generator when running at 1500 rpm, supplies 2...
Assuming that the armature resistance remains constant at 0.25 ohms, we can use the formula:
Load current = (terminal voltage - back emf) / armature resistance
At 1500 rpm, the back emf can be calculated using the formula:
Back emf = K * N * Ф
where K is a constant (dependent on the machine's design), N is the number of turns in the armature winding, and Ф is the magnetic flux in the air gap. Since these values are not given in the problem, we cannot calculate the back emf directly.
However, we can use the fact that at 1500 rpm, the generator supplies 200 A at 150 V to find the product of K * N * Ф:
K * N * Ф = (terminal voltage - armature voltage) * generator current
At 1500 rpm, the armature voltage is equal to the product of armature resistance and load current:
Armature voltage = armature resistance * load current
Substituting these values, we get:
K * N * Ф = (150 - 0.25 * load current) * 200
Simplifying, we get:
K * N * Ф = 30000 - 50 * load current
Now, at 1000 rpm, the back emf can be calculated using the formula:
Back emf = K * N * Ф'
where Ф' is the new magnetic flux in the air gap. Since the generator is separately excited, we can assume that the field current remains constant and therefore the magnetic flux is proportional to the speed. Thus:
Ф' / Ф = 1000 / 1500
Ф' = Ф * 2/3
Substituting this into the formula for back emf, we get:
Back emf = K * N * Ф * 2/3
Using the formula for load current, we get:
Load current = (terminal voltage - back emf) / armature resistance
Substituting the given values and simplifying, we get:
Load current = (150 - K * N * Ф * 2/3) / 0.25
We still cannot solve for load current directly, because we do not know the value of K * N * Ф. However, we can use the earlier equation relating K * N * Ф to load current, and solve for K * N * Ф:
K * N * Ф = 30000 - 50 * load current
Substituting this into the formula for load current, we get:
Load current = (150 - (30000 - 50 * load current) * 2/3) / 0.25
Simplifying and solving for load current, we get:
Load current = 105.88 A
Therefore, the load current decreases from 200 A to 105.88 A when the speed drops from 1500 rpm to 1000 rpm, assuming that the armature resistance remains constant.
A separately excited DC generator when running at 1500 rpm, supplies 2...
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).