A demonstration of inductive reactance

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Discussion Overview

The discussion revolves around the demonstration of inductive reactance in an AC circuit, specifically focusing on the construction of an inductor and its behavior when connected to both DC and AC sources. Participants explore the necessary parameters for winding a coil, including turns and wire type, to achieve a specific inductance without tripping a circuit breaker.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests winding a coil to demonstrate inductive reactance and asks for guidance on the number of turns and wire type needed to avoid tripping a circuit breaker.
  • Another participant provides calculations indicating that approximately 300mH of inductance is required for a 250V AC mains at 50Hz to allow for a current of about 5 Amps, estimating around 1500 turns for an air coil of 6 inches in diameter.
  • This second participant notes that using an iron or ferrite core could significantly reduce the number of turns needed, possibly to 10-20 turns, depending on the core material.
  • A different participant warns against connecting the inductor directly to the AC mains without a load, suggesting that a light bulb should be used in series to prevent excessive current and provide a safer demonstration.
  • This participant also discusses the relationship between voltage, current, and impedance in the circuit, highlighting how the impedance changes when using AC versus DC.
  • Another participant suggests using the primary of a regular power transformer as an alternative approach.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to demonstrate inductive reactance, particularly regarding the use of loads in the circuit and the specifics of coil construction. No consensus is reached on the optimal method or parameters.

Contextual Notes

Participants mention various factors affecting inductance, including the number of turns, coil diameter, core permeability, and the frequency of the AC source. These factors introduce complexity and uncertainty in determining the exact specifications needed for the demonstration.

Who May Find This Useful

This discussion may be of interest to students and educators in electrical engineering or physics, particularly those exploring concepts related to inductance and AC circuits.

suhasm
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I'm conducting a seminar at college on application of AC to passive components.
To Demonstrate what inductive reactance is , i want to wind a coil , connect it to a DC source , find the resistance. Then i want to directly connect it the AC mains and find th effective resistance again and show that there is an extra reactive component that is preventing the AC mains from being shorted.

Can you tell around how many turns and of what wire i will have to wind to get the necessary inductance so that the circuit breaker does not trip?
 
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It would help if you gave some figures.
If you have 250V mains at 50Hz to get a current of about 5 Amps requires something around 300mH of inductance.

If we assume an air coil of say 6 inch diameter, something around 1500 turns will give you the necessary inductance.

That's quite a lot. If you use an iron or ferrite core it would be easy but the calculation then very strongly depends on the material. That would reduce the number of turns by a factor of a couple of hundred. - 10-20turns.

The wire is pretty irrelevant - use copper, as thick as you can get it, to reduce resistance and heating losses.
 
suhasm said:
I'm conducting a seminar at college on application of AC to passive components.
To Demonstrate what inductive reactance is , i want to wind a coil , connect it to a DC source , find the resistance. Then i want to directly connect it the AC mains and find th effective resistance again and show that there is an extra reactive component that is preventing the AC mains from being shorted.

Can you tell around how many turns and of what wire i will have to wind to get the necessary inductance so that the circuit breaker does not trip?
Not a good idea. You will have to put a load in line with the inductor - e.g. a light bulb. Using 110 VDC the light bulb in series with the inductor will give more light than 110 VAC in the same configuration. The current will be I = V/Z where:

[tex]Z = \sqrt{R^2 + (2\pi f L)^2}[/tex]

f is the frequency of the AC (0 for DC). R is the resistance in the circuit. L is determined by a number of factors: number of turns, diameter of coil, permeability of the core, length of coil

An inductor of .5 H in series with a 100 watt light bulb double the impedance when using 60Hz AC (ie. the DC current will be double the AC current).


AM
 
Use the primary of a regular power transformer?
 

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