If a shunt generator delivers 100 A at 200 V, and the resistances of the shunt field and armature are 100 ohms and 0.01 ohms respectively, the generator emf will be
A. series.
B. parallel.
C. any one of the connection.
D. no resistance is connected.
Answer: A
Share your understanding of this question with the correct explanation.
The answer is A. series.
Explanation:
In a shunt generator, the field winding and the armature winding are connected in parallel. The load resistance is connected in series with the armature.
The generator emf can be determined using the formula:
Emf = Vt + Ia * Ra
Where:
Emf is the generator emf
Vt is the terminal voltage (output voltage)
Ia is the armature current
Ra is the armature resistance
Given:
Terminal voltage (Vt) = 200 V
Armature current (Ia) = 100 A
Armature resistance (Ra) = 0.01 ohms
Using the formula:
Emf = 200 V + (100 A * 0.01 ohms)
Emf = 200 V + 1 V
Emf = 201 V
Therefore, the generator emf is 201 V, which is connected in series with the load resistance.
