Increasing number of turns in a Solenoid?

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Discussion Overview

The discussion revolves around the effects of increasing the number of turns in a solenoid on its magnetic field strength, power requirements, and current draw from a battery. Participants explore concepts related to resistance, Ohm's law, and the characteristics of batteries in a DC circuit context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that increasing the number of turns in a solenoid strengthens the magnetic field, but question whether this requires more power input from the power source.
  • There is uncertainty about how the resistance of the solenoid changes with the number of turns, with some assuming that resistance increases as the number of turns increases, which may lead to a decrease in current.
  • Participants discuss the relationship between voltage, current, and power in the context of a battery, with some suggesting that the power input can be calculated as V*I.
  • One participant expresses confusion about the implications of a battery's amp-hour rating, questioning whether it limits the current draw from the battery.
  • Another participant highlights that the internal resistance of the battery and the load's resistance affect the actual current drawn, suggesting that the maximum current draw is not simply determined by the battery's rating.
  • Some participants express a desire to measure the solenoid's resistance and current draw using a meter to clarify their understanding.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the effects of increasing the number of turns on resistance and current draw. There are competing views regarding the relationship between battery ratings and current draw, and the discussion remains unresolved.

Contextual Notes

Participants acknowledge limitations in their understanding of RL circuits and the behavior of batteries under load, indicating that further study is needed to clarify these concepts.

Who May Find This Useful

This discussion may be useful for individuals interested in electrical engineering, circuit design, and the principles of electromagnetism, particularly those seeking to understand the relationship between solenoid design and power requirements.

Wiz700
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Hello!

I understand that by increasing the number turns the magnetic field is strengthened. But would that "extra" turns demand more power input from the power source?
 
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Another thing!

I'm using a double AA battery rated at 2A/H, so will my Solenoid use the battery's 1.5V and draw 2 Amps?
Thus power input will be 3W? Or it depends on the Solenoid's design?
 
Wiz700 said:
I understand that by increasing the number turns the magnetic field is strengthened. But would that "extra" turns demand more power input from the power source?
The strength of the magnetic field is proportional to N*I, where N is the number of turns of the solenoid and I is the current through it.

What happens to the resistance of the solenoid as you increase the number of turns on it? How will it affect the current?

If V is the voltage of your battery and I is the current through it, what is the power delivered to your circuit?

I don't know how familiar you are with RL circuits but let's assume we're talking DC only and disregard the initial transient state of your circuit.

Wiz700 said:
I'm using a double AA battery rated at 2A/H, so will my Solenoid use the battery's 1.5V and draw 2 Amps?
Thus power input will be 3W? Or it depends on the Solenoid's design?
It seems a bit odd that your battery should determine both the voltage across your circuit and the current through it. Shouldn't the load have something to say about it?
 
milesyoung said:
The strength of the magnetic field is proportional to N*I, where N is the number of turns of the solenoid and I is the current through it.

What happens to the resistance of the solenoid as you increase the number of turns on it? How will it affect the current?

If V is the voltage of your battery and I is the current through it, what is the power delivered to your circuit?

I don't know how familiar you are with RL circuits but let's assume we're talking DC only and disregard the initial transient state of your circuit.

Well I'm not sure what happens to R as N increases... I assume R increases. So I only can assume that I decreases... Based on Ohm's law.
I'm not sure how much power is delivered. I'm not familiar with RL circuits, yes its DC only.
I just has basic knowledge of circuits.


milesyoung said:
It seems a bit odd that your battery should determine both the voltage across your circuit and the current through it. Shouldn't the load have something to say about it?

You're right. I just assumed the maximum output that the battery will deliver. I don't know how much the load will draw, and I can't measure it right now. I can only guess!
 
Wiz700 said:
Well I'm not sure what happens to R as N increases... I assume R increases. So I only can assume that I decreases... Based on Ohm's law.
Those are good assumptions :smile:

A solenoid is, after all, just a piece of conducting wire wound into an odd shape. Everything else being equal, you'd expect a longer wire to have a larger resistance:
http://en.wikipedia.org/wiki/Resistivity

Wiz700 said:
I'm not sure how much power is delivered.
The power delivered to your circuit is V*I. If you continue your reasoning, what happens to the power as you increase the number of turns on the solenoid?

Wiz700 said:
You're right. I just assumed the maximum output that the battery will deliver. I don't know how much the load will draw, and I can't measure it right now. I can only guess!
If you hook up your battery, with know voltage, V, to a resistive load, what determines how much current, I, it will draw? You mentioned Ohm's law?
 
I got my meter so I'll make a Solenoid and measure the resistance.
Then use Ohm's law to figure out how much current that coil is going to draw!

Thanks.
 
I can assume that the Solenoid MAX current draw is 2 Amps @ 1.5V
The power = 3W
However, more turns = higher R = less current drawn out.
 
Wiz700 said:
I can assume that the Solenoid MAX current draw is 2 Amps @ 1.5V
No you can't. The 2 amp-hour rating does not mean your battery somehow limits the current you can draw from it to 2 amps. Your battery has an internal resistance (not by design) that is, in series with the resistance of your connecting wires and that of the solenoid itself, what limits the current you can draw from it.

Further, your battery is rated at 2 amp-hours, but this does not mean you can pull 2 amps from it at 1.5 V for one hour. You might have a hope of pulling out 1.5*2 Wh worth of energy if you discharge it at very low power, i.e. very low current. For an alkaline cell at 2 amps, its terminal voltage will drop like a rock and you'll probably only be able to pull out a fraction of those 1.5*2 Wh.

If available, the discharge curves of your battery might give you a much better idea of what to expect.
 
Woah... I have a lot of things to study!
 
  • #10
Wiz700 said:
Woah... I have a lot of things to study!
The difference between Amps and Amp-Hours would be a helpful place to start.
 

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