Previous in Forum: testing mothods and procedures   Next in Forum: Reactance Changes with Fault Currents
Close
Close
Close
6 comments
Rate Comments: Nested
Participant

Join Date: Jan 2011
Posts: 1

Basic Understanding of Electromagnetism

01/17/2011 10:17 AM

If I have a long insulated copper wire between a + and - terminal, there will be short as there is no resistance. However, if I wind the wire over a ferrous metal, there is no short. How does winding the same copper wire over a metal create a load? Is load also synonymous to resistance?

Register to Reply
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.

Good Answers:

These comments received enough positive votes to make them "good answers".
Guru
United Kingdom - Member - New Member

Join Date: Aug 2010
Location: Resting under the Major Oak
Posts: 4347
Good Answers: 181
#1

Re: Basic Understanding of Electromagnetism

01/17/2011 10:36 AM

You would be right with an AC supply where the reactance of the ferrous core restricts the current. With DC you have to rely on the resistance of the wire to limit the current.

__________________
The reports of my death are greatly exaggerated.
Register to Reply
4
Guru
India - Member - New Member

Join Date: Sep 2009
Location: Temporarily at Ashburn, VA
Posts: 2744
Good Answers: 164
#2

Re: Basic Understanding of Electromagnetism

01/17/2011 10:40 AM

If I have a long insulated copper wire between a + and - terminal, there will be short as there is no resistance.

Not quite.Any length of wire will have resistance R = ρ*L/A.

However, if I wind the wire over a ferrous metal, there is no short.

If wound around a ferrous (or any) core, it will have Inductance in addition, a more complex formula that we will ignore for now. A ferrous core will give a far higher inductance to the coil than a core of non-magnetic material (or air). So the coil may or may not be a "short"

The inductance will become a "Reactance" only if the supply is AC. Not if it is DC as you say. In such an event, the current will rise exponentially with time, according to the formula I = Im*(1-e^-Rt/L). And I will be equal to E/R. No help from the inductance.

__________________
Nothing worthwhile can ever be taught, it can only be learnt.
Register to Reply Good Answer (Score 4)
Anonymous Poster
#3
In reply to #2

Re: Basic Understanding of Electromagnetism

01/17/2011 11:14 AM

Thanks.

Register to Reply
Guru

Join Date: Oct 2009
Posts: 2061
Good Answers: 169
#4
In reply to #2

Re: Basic Understanding of Electromagnetism

01/18/2011 12:18 PM

Excellent KVS Sir! GA to you!

Register to Reply
Guru
India - Member - New Member

Join Date: Sep 2009
Location: Temporarily at Ashburn, VA
Posts: 2744
Good Answers: 164
#5
In reply to #4

Re: Basic Understanding of Electromagnetism

01/18/2011 9:48 PM

Hello, and thanks. You seem to be busy and aren't coming on CR4 often....we miss you

__________________
Nothing worthwhile can ever be taught, it can only be learnt.
Register to Reply Off Topic (Score 5)
Guru

Join Date: Oct 2009
Posts: 2061
Good Answers: 169
#6
In reply to #5

Re: Basic Understanding of Electromagnetism

01/19/2011 12:13 AM

Ya! I was on tour frm 25th December till 17th January. I don't have access to net while I am on tour. Only when I am back in my HQ I have access to net and hence the moment I am back at my HQ, I devour the CR4 contents. I too miss all of you during such periods of my absence.

Register to Reply Off Topic (Score 5)
Register to Reply 6 comments

Good Answers:

These comments received enough positive votes to make them "good answers".
Copy to Clipboard

Users who posted comments:

Anonymous Poster (1); electricalexpert65 (2); kvsridhar (2); TonyS (1)

Previous in Forum: testing mothods and procedures   Next in Forum: Reactance Changes with Fault Currents

Advertisement