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Lesson : Induction, Transformers, and Generators
Question # Answer
- A 550-loop circular armature coil with a diameter of 8.0 cm rotates at 120 rev/s in a uniform magnetic field of strength 0.55 T. (a) What is the rms voltage output of the generator? (b) What would you do to the rotation frequency in order to double the rms voltage output?
Given:
Solution: (a)
Where , thus,
Where , thus,
(b) Based on this equation:
,we can see that is directly proportional to the rotation frequency. Therefore, in order to double the value of rms voltage output, the rotation frequency’s value must double.
Using the solution in (a), we will double the rotation frequency:
By doubling the rotation frequency, we can see based on the solution that the rms voltage output doubled.
Thus, the rotation frequency ( ) must double its value in order to double the rms voltage output.
- A simple generator has a square armature 6.0 cm on a side. The armature has 85 turns of 0.59-mm-diameter copper wire and rotates in 0.65-T magnetic field. The generator is used to power a lightbulb rated at 12.0 V and 25.0 W. At what rate should the generator rotate to provide 12.0 V to the bulb? Consider the resistance of the wire on the armature.
Given:
Solution:
First, determine the internal resistance of the wire on the armature.
The L in the equation is the length of the copper wire. To determine the length of the wire, we must multiply the total side length of the armature and the number of turns of the copper wire. ( )( )( )
And for the area,
change 117 V into 13,500 V, and there are 148 turns in the primary coil. How many turns are in the secondary coil?
higher voltage, it means that the transformer is a step-up transformer.
In step-up transformer, the number of turns in secondary coil is greater than the primary coil. And the voltage is directly proportional to the number of turns, and so, we can conclude that the highest voltage in the problem is the voltage of secondary coil and the lowest voltage is the voltage of primary coil.
Given:
Solution:
