Determining terminal voltage, field current,etc.

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The discussion revolves around calculating the terminal voltage and field current of a shunt generator with an induced emf of 596 V and armature resistance of 0.100 Ω. It emphasizes that the terminal voltage cannot exceed the generated emf, requiring sign changes in the armature current equation. The armature current is equal to the sum of the field current and the external circuit current, as dictated by Kirchhoff's Current Law (KCL). A circuit diagram is recommended for clarity, and the importance of understanding the relationship between the armature, field, and external circuit currents is highlighted. Overall, the conversation focuses on correcting the calculations and understanding the principles governing shunt generator operation.
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Homework Statement


The emf induced in the armature of a shunt generator is 596 V. The armature resistance is 0.100 Ω. (a) Compute the terminal voltage when the armature current is 460 A. (b) The field resistance is 110 Ω. Determine the field current, and the current and power delivered to the external circuit.

Homework Equations


Armature Current= (Line voltage - Back emf)/Armature resistance
R=V/I
P=(V)(I)

The Attempt at a Solution


I felt good about my work until I checked the key. Where did I go wrong?
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A DC generator's terminal voltage can not be greater than its generated emf (induced).
Draw the circuit diagram for shunt generator. Your armature current equation needs a couple of sign changes.
 
cnh1995 said:
A DC generator's terminal voltage can not be greater than its generated emf (induced).
Draw the circuit diagram for shunt generator. Your armature current equation needs a couple of sign changes.
I drew a circuit diagram and re-did the problem. I got the right answers, but I still have a few questions. Are the sign changes only
necessary in order to ensure that the terminal voltage was less than the generated emf? Why is the armature's current equal to the sum of the field's and external circuit's current? I wrote down that was true because I found solutions to a similar problem that stated this fact. Thank you for your time.
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Vladi said:
Are the sign changes only
necessary in order to ensure that the terminal voltage was less than the generated emf?
DC generator supplies power to the external circuit, hence, armature current flows out of the armature winding through its positive terminal.
Vladi said:
Why is the armature's current equal to the sum of the field's and external circuit's current?
Because the field is current is supplied by the generator terminal voltage. Hence, as per KCL, the armature current splits into field current and external circuit current.
 
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cnh1995 said:
DC generator supplies power to the external circuit, hence, armature current flows out of the armature winding through its positive terminal.

Because the field is current is supplied by the generator terminal voltage. Hence, as per KCL, the armature current splits into field current and external circuit current.
Thank you for all your help.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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