A 8-pole, DC generator has a simplex wave-wound armature containing 32 coils of 6 turns each.
Its flux per pole is 0.06 Wb.
The machine is running at 250 rpm.
The induced armature voltage is
A. 96 V
B. 192 V
C. 384 V
D. 768 V
Answer: C
Share your understanding of this question with the correct explanation.
To determine the induced armature voltage of a DC generator, we can use the following formula:
E = (2 * N * Z * P * φ * n) / 60
where:
E is the induced armature voltage,
N is the number of parallel paths in the armature (always equal to 2 for a simplex wave-wound armature),
Z is the total number of armature conductors (Z = number of coils * number of turns per coil),
P is the number of poles,
φ is the flux per pole,
n is the speed of the machine in revolutions per minute (RPM).
Given:
N = 2 (number of parallel paths in the armature)
Number of coils = 32
Number of turns per coil = 6
P = 8 (number of poles)
φ = 0.06 Wb (flux per pole)
n = 250 RPM
Plugging these values into the formula, we get:
E = (2 * 2 * 32 * 6 * 8 * 0.06 * 250) / 60
= (192 * 8 * 0.06 * 250) / 60
= (192 * 1.2 * 250) / 60
= (57600) / 60
= 960 V
Therefore, the induced armature voltage is 960 V, which is equivalent to 384 V per parallel path.
So, the correct answer is C: “384 V.”