A demonstration of inductive reactance

[suhasm]

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?

 

[AJ Bentley]

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.

 

[Andrew Mason]

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

 

[schip666!]

Use the primary of a regular power transformer?

 

[sardar]

Inductive reactance is a fundamental concept in the study of AC circuits and is a crucial factor in understanding the behavior of passive components. Your demonstration is a great way to showcase the effects of inductive reactance in a practical manner.

To determine the necessary number of turns and wire gauge for your coil, you will need to calculate the inductance required using the formula L = (μ₀N²A)/l, where μ₀ is the permeability of free space, N is the number of turns, A is the cross-sectional area of the coil, and l is the length of the coil. The resistance of the coil can also be calculated using the formula R = ρ(l/A), where ρ is the resistivity of the wire.

The specific number of turns and wire gauge will depend on the specifications of your AC mains and the desired inductive reactance. It is important to consider the voltage and frequency of the AC mains, as well as the maximum current that the circuit breaker can handle. You may also want to consider using a variable inductor to adjust the inductance and observe the effects on the overall resistance.

I would also suggest discussing the concept of inductive reactance in more detail, such as how it is affected by frequency and how it can be cancelled out by a capacitor in a series circuit. Overall, your demonstration is a great way to illustrate the practical application of inductive reactance and its role in AC circuits.