Answered

IDNLearn.com connects you with a community of experts ready to answer your questions. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.

A simple generator is used to generate a peak output voltage of 23.0 V . The square armature consists of windings that are 5.1 cm on a side and rotates in a field of 0.500 T at a rate of 55.0 rev/s .
How many loops of wire should be wound on the square armature?

Sagot :

Answer: 51

Explanation:

Given

Output is 23 V

The square armature side is [tex]a=5.1\ cm[/tex]

Magnetic field [tex]B=0.5\T[/tex]

Rate of revolution [tex]n=55\ rev/s[/tex]

Angular speed

[tex]\omega =2\pi n\\\omega=2\pi \times 55=110\pi\ rad/s[/tex]

Peak voltage is given by

[tex]E_{peak}=NB\omega A\quad [\text{N=Number of windings; A=area of cross-section}]\\\\N=\dfrac{E_{peak}}{B\omega A}\\\\N=\dfrac{23}{0.5\times 110\pi\times (0.051)^2}\approx 51[/tex]

So, there are approximate 51 loops

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.