Jump to content

Electromagnets and voltage


battousai

Recommended Posts

How does an increase in voltage with a constant DC current affect an electromagnet? Is the increased voltage just transformed into heat, or extra strength in the electromagnet?

 

For example, at start there is an electromagnet with a current of 5A at 6V.

The current is shut off, and a new supply of 5A at 30V is applied to the same electromagnet.

 

How does the electromagnet react?

Link to comment
Share on other sites

It's not homework; it's my own curiosity. What I mean by constant, is that in the example, the two different supplies provide the same current of 5A. I'll clarify my example.

 

Stage 1. There is an electromagnet with a DC current powering it. The current is a constant 5A, and the voltage is a constant 6V.

 

Stage 2. The electromagnet is disconnected from the power source.

 

Stage 3. The electromagnet is connected to a second DC power supply. The supply provides a constant current of 5A, and a constant voltage of 30V.

 

Now, will the electromagnet be stronger in Stage 3, as opposed to Stage 1. Will more heat be produced in Stage 3 than Stage 1? Is there a significant voltage drop in an electromagnet? Will this drop be significantly larger in Stage 3 than Stage 1?

Link to comment
Share on other sites

It's the same electromagnet. Would it not have the same resistance? If you need a number, let's just say it's 1 ohm.

 

V=IR

 

You can't dictate that the current and resistance remain constant and have a different voltage.

Link to comment
Share on other sites

Ok, I think I see where this is going. I thought with a power supply, you could set both the voltage and current magnitudes. For example, on an AC adapter it says there's a DC output of 7.5V, 5.7A, but that's with respect to the resistance of the machine it's designed to power.

 

In my electromagnet scenario, it would be :

 

Stage 1. The power supply is set to 6V, and the electromagnet has a resistance of 1 ohm, therefore the current is 6A.

 

Stage 2. The power supply is set to 30V, and thus the current is 30A.

 

I'm sorry for my mistake. Thank you for helping me solve this.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.