Simple Generator - Calculating N Loops

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SUMMARY

The forum discussion focuses on calculating the number of loops required for a simple generator to achieve a peak output voltage of 26.0 V. The generator's square armature, with windings measuring 6.3 cm per side, rotates in a magnetic field of 0.440 T at a frequency of 60.0 revolutions per second. Using the formula ε / (2Blv sin θ) = N, the calculation yields 250 loops, assuming the angle θ is 90 degrees. The derived velocity of the armature is 1.89 m/s, confirming the correctness of the calculations presented.

PREREQUISITES
  • Understanding of electromagnetic induction principles
  • Familiarity with the formula ε / (2Blv sin θ)
  • Basic knowledge of rotational motion and angular velocity
  • Ability to perform unit conversions (e.g., cm to m)
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  • Study the principles of electromagnetic induction in depth
  • Learn about the effects of varying the angle θ in generator calculations
  • Explore the relationship between angular velocity and linear velocity in rotating systems
  • Investigate the design considerations for optimizing generator efficiency
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Students in electrical engineering, physics enthusiasts, and professionals involved in generator design and optimization will benefit from this discussion.

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Simple Generator - Calculating "N" Loops

Homework Statement



A simple generator is used to generate a peak output voltage of 26.0 V . The square armature consists of windings that are 6.3 cm on a side and rotates in a field of 0.440T at a rate of 60.0 rev/s. How many loops of wire should be wound on the square armature?
Express your answer as an integer.

Homework Equations



ε / 2Blv sin θ = N

The Attempt at a Solution



l = 0.063 m
ε = 26.0 V
B = 0.440 T
v = 1.89 m/s (Is this what is incorrect?)

v = ωr
v = (60.0 rev/s)(0.0315 m)
v = 1.89 m/s

(26.0 V) / (2)(0.440 T)(0.063 m)(1.89 m/s)

(26.0 V) / (.104 T m^2 / s) = 250 loops

ETA : I assumed the angle is 90 degrees.
 
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PeachBanana said:

Homework Statement



A simple generator is used to generate a peak output voltage of 26.0 V . The square armature consists of windings that are 6.3 cm on a side and rotates in a field of 0.440T at a rate of 60.0 rev/s. How many loops of wire should be wound on the square armature?
Express your answer as an integer.

Homework Equations



ε / 2Blv sin θ = N

The Attempt at a Solution



l = 0.063 m
ε = 26.0 V
B = 0.440 T
v = 1.89 m/s (Is this what is incorrect?)

v = ωr
v = (60.0 rev/s)(0.0315 m)
v = 1.89 m/s

(26.0 V) / (2)(0.440 T)(0.063 m)(1.89 m/s)

(26.0 V) / (.104 T m^2 / s) = 250 loops

ETA : I assumed the angle is 90 degrees.

ω=2pi times revolutions/second

ehild
 

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