10 Ohm 100 Watt resistor, inductor circuit

Thread Starter

sab201

Joined Nov 18, 2023
17
Good Day,

I need some assistace with my circuit.

I have a 24 V DC adapter powering a 10 Ohm, 100 Watts resistor connected in series with an inductor. This is the main circuit.

I have a separate cooling circuit with two 12 V DC, 0.3 A fans connected in parallel powered by a separate 12 V DC adapter to cool the 10 Ohm resistor in the main circuit.

Now the current flowing through the main circuit will be 2.4 A. I have two questions.

1. Can I use two 24 V DC, 0.1 Amp fans connected in parallel across the 10 Ohm resistor instead of using a separate 12 V cooling fan circuit??? Will the two high resistance fans taking 0.1 Ampere current be in parallel with a resistor giving out 57.6 Watts and taking 2.5 Amp current. Or should the fans be connected in parallel to the supply instead???

2. I want to install an indicating lamp in series with the main circuit. But since the main circuit draws a 2.4 Amps current, there is no solution I could arrive at but using a current transformer. Any other ideas to make an indicating lamp would be of great assistance.

Thanks.
 

Alec_t

Joined Sep 17, 2013
14,223
Welcome to AAC!
1) Provided the DC resistance of the inductor is much less than 10 Ohms and the supply is rated for at least 2.4A + 2 x 0.1A = 2.6A then yes. Or they could be connected in parallel with the supply.
2) Why do you want the indicator lamp in series with the main circuit? Current transformers work with AC, not DC. You could use a Hall sensor to detect the magnetic field around a wire/inductor due to current flow, or you could wrap a few turns of the supply wire around a reed relay to switch on an indicator lamp.
What is the inductor for?
 

Thread Starter

sab201

Joined Nov 18, 2023
17
Yes you can put the fans across the 24V resistor supply, what do you need the 100W resistor for?
Actually my main task is to create an electromagnet with the inductor so I need a reasonable current flowing through the coil. That is the use of the 100 W resistor in series.

Thanks I will connect it across the resistor instead of connecting across the power supply.
 

Thread Starter

sab201

Joined Nov 18, 2023
17
Welcome to AAC!
1) Provided the DC resistance of the inductor is much less than 10 Ohms and the supply is rated for at least 2.4A + 2 x 0.1A = 2.6A then yes. Or they could be connected in parallel with the supply.
2) Why do you want the indicator lamp in series with the main circuit? Current transformers work with AC, not DC. You could use a Hall sensor to detect the magnetic field around a wire/inductor due to current flow, or you could wrap a few turns of the supply wire around a reed relay to switch on an indicator lamp.
What is the inductor for?
Yes the dc resistance is less than 10 Ohms when measured with a multimeter...

Thanks I will try to look into the option of the reed relay and Hall Sensor.

Actually my main task is to create an electromagnet with the inductor so I need a reasonable current flowing through the coil. That is the use of the 100 W resistor in series.
 

BobTPH

Joined Jun 5, 2013
8,665
Actually my main task is to create an electromagnet with the inductor so I need a reasonable current flowing through the coil. That is the use of the 100 W resistor.
Why not reduce the voltage and get the same current without the resistor? The resistor is simply wasting 57.6W.

If the only supply you have is 24V, power the magnet with a buck converter set to 2.4A.
 

Dodgydave

Joined Jun 22, 2012
11,243
Actually my main task is to create an electromagnet with the inductor so I need a reasonable current flowing through the coil. That is the use of the 100 W resistor in series.

Thanks I will connect it across the resistor instead of connecting across the power supply.
Then why don't you omit the resistor, and use a lower voltage across the inductor to create the magnet, AC will do .
 

Thread Starter

sab201

Joined Nov 18, 2023
17
Why not reduce the voltage and get the same current without the resistor? The resistor is simply wasting 57.6W.

If the only supply you have is 24V, power the magnet with a buck converter set to 2.4A.
Hi, thanks for the idea, but I suppose still the buck converter would consume power and it produces heat to give out the 2.4A.
 

LesJones

Joined Jan 8, 2017
4,173
If the inductance of the coil is high enough you could use half wave rectification and add a flywheel diode in parallel with the coil to help maintain the current in the coil during the half cycle when there was no output from the half wave rectifier. Without knowing the purpose of the coil I don't know if this would be acceptable.
For an indicator light you could have 4 diodes in series with the DC supply which would give a voltage drop of about 2.4 to 2.8 volts you could then connect a red led and current limiting resistor between the ends of the 4 diodes. You could also use a current transformer before the rectifier.

Les.
 

BobTPH

Joined Jun 5, 2013
8,665
Hi, thanks for the idea, but I suppose still the buck converter would consume power and it produces heat to give out the 2.4A.
It will indeed waste some power, maybe 1/2 W as opposed to the 57W wasted by dropping the voltage with a resistor. This is why we use DC to DC converters to reduce voltage instead of resistors.
 

Thread Starter

sab201

Joined Nov 18, 2023
17
It will indeed waste some power, maybe 1/2 W as opposed to the 57W wasted by dropping the voltage with a resistor. This is why we use DC to DC converters to reduce voltage instead of resistors.
That is right. If you are talking about reducing voltage then I could bring the current down to 2.4 A. For that I need to connect a (approximately)2 Ohm resistor that would be a 10 Watts of power wasted since the inductor cannot be connected directly to the voltage source. However from my understanding, reducing the voltage to 5V will reduce the strength of the electromagnet. That is the reason I had to waste 57 Watts.
 

Thread Starter

sab201

Joined Nov 18, 2023
17
If the inductance of the coil is high enough you could use half wave rectification and add a flywheel diode in parallel with the coil to help maintain the current in the coil during the half cycle when there was no output from the half wave rectifier. Without knowing the purpose of the coil I don't know if this would be acceptable.
For an indicator light you could have 4 diodes in series with the DC supply which would give a voltage drop of about 2.4 to 2.8 volts you could then connect a red led and current limiting resistor between the ends of the 4 diodes. You could also use a current transformer before the rectifier.

Les.
The purpose of the coil is to produce a magnetic field similar to a permanent bar magnet. It is wound around an iron core.
 

BobTPH

Joined Jun 5, 2013
8,665
However from my understanding, reducing the voltage to 5V will reduce the strength of the electromagnet. That is the reason I had to waste 57 Watts.
Your understanding is wrong. The magnetic field produced depends on the current only. The voltage across the coil, at any steady current, is determined by Ohm’s law. Measure the resistance of your coil and you can determine the voltage needed to get any current you want.

You cannot maintain the same current at any other voltage.

Why do you think the voltage dropped across a resistor has any effect on your magnet? It does not. Once again, the resistor simply wastes power and has no benefit.
 
However from my understanding, reducing the voltage to 5V will reduce the strength of the electromagnet. That is the reason I had to waste 57 Watts.
EDIT; Bob beat me to the reply.
Your understanding is not correct.
Magnetic force, for a given geometry, is only proportional to ampere-turns.
Of course, all those turns will have resistance, and if a current is passed thru, a voltage will be developed.

The point I am attempting to drive, is that you feed the electromagnet a voltage only as high to allow the required current. Every thing else is wasted.
 

Thread Starter

sab201

Joined Nov 18, 2023
17
Your understanding is wrong. The magnetic field produced depends on the current only. The voltage across the coil, at any steady current, is determined by Ohm’s law. Measure the resistance of your coil and you can determine the voltage needed to get any current you want.

You cannot maintain the same current at any other voltage.

Why do you think the voltage dropped across a resistor has any effect on your magnet? It does not. Once again, the resistor simply wastes power and has no benefit.
Thanks for correcting it. Yes the current is what affects the intensity of the magnetic field of the electromagnet. I have been reading about buck converters and will soon try to use it instead of the high 24 V and avoid wasting power in the 10 Ohm resistor. I am thinking of using this type of buck converter:

https://www.electronicscomp.com/xl4...sQGAOrwCz3ZtTXi6YZ5t9snwGsVj1JCxoCaVUQAvD_BwE

Then for 2.5 Amps to flow through the inductor I will need a 2 Ohm, 10 Watts resistor in series with the electromagnet/inductor. The Power supplied by the buck converter would be 12.5 Watts which is much lesser than the 57 Watts wasted by the resistor.

However I also read the buck converter generates heat which could be dissipating electric power to some extent but that may not be as high as the 57 Watts dissipated by the resistor.
 

BobTPH

Joined Jun 5, 2013
8,665
Then for 2.5 Amps to flow through the inductor I will need a 2 Ohm, 10 Watts resistor in series with the electromagnet/inductor
No. Where did you get that idea? You simply set the current on the back converter to 2.5A and connect the coil directly to the output of the converter. There is no resistor needed.
 
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