AC circuits, Inductance and phase shift.

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SUMMARY

The discussion focuses on calculating the self-inductance of an electric motor drawing 1A (rms) from a 230V (rms) mains supply, with a phase shift of 30% of the period at 50Hz. The user correctly applies Kirchhoff's loop rule and integrates to derive the relationship between voltage, current, and inductance. The final calculation yields a self-inductance of 0.732 H, confirming the accuracy of the approach taken. The phase shift is effectively utilized in the calculations to determine the inductance value.

PREREQUISITES
  • Understanding of AC circuit theory
  • Familiarity with Kirchhoff's loop rule
  • Knowledge of phase shift in alternating current
  • Basic calculus for integration
NEXT STEPS
  • Study the implications of phase shift on power factor in AC circuits
  • Learn about the role of inductors in AC circuit design
  • Explore advanced topics in electromagnetic theory related to inductance
  • Investigate the effects of varying frequency on inductive reactance
USEFUL FOR

Electrical engineers, students studying AC circuit theory, and professionals involved in motor design and analysis will benefit from this discussion.

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Homework Statement



An electric motor draws a 1A (rms) from the 230V (rms) mains. and is found to shift the phase of the 50Hz alternating current by 30% of the period. What is the self inductance of the coils in the motor?

Homework Equations




The Attempt at a Solution



i've started by finding the phase shift:

\delta = \frac{3}{10}\cdot \frac{1}{f} = \frac{3}{10}\cdot\frac{1}{50} = 0.006

i think the circuit has only an inductor and power source in it, so using kirchhoffs loop rule:

V_{max}cos(\omega t) - L\frac{dI}{dt} = 0

separate variables and integrate:

V_{max} \int cos(\omega t) dt = \int L dI

\frac{V_{max}}{\omega}\cdot sin(\omega t) = LI

sin(wt) = cos(wt - d)

L = \frac{V_{max}}{\omega I} cos(\omega t - \delta)

I know everything here apart from t, is this the right way to do this?
 
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I just thought of another way to do this, not sure it's right because it doesn't use the phase shift at all:

I_{max} = \frac{V_{max}}{X_L} = \frac{V_{max}}{\omega L}

L = \frac{V_{max}}{\omega I_{max}}

where

V_{max} = \sqrt{2}V_{rms}

I_{max} = \sqrt{2}I_{rms}

so:

L = \frac{V_{rms}}{\omega I_{rms}}

L = 0.732 H
 
Last edited:

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