# 48 volts DC at 3 amps, resistor or rheostat?

Discussion in 'The Projects Forum' started by aac_dave, Jun 3, 2014.

1. ### aac_dave Thread Starter New Member

Jun 3, 2014
2
0
Forgive my limited knowledge about circuits but here goes...

I need to apply 48 volts at 3 amps to a load. As I understand

I = V/R
so
3 = 48/16
...meaning a 16 ohm resistor/rheostat

This is probably overly simplistic - there will likely be other sources of resistance like the wires themselves and I'm going to add a switch also. There may even be some resistance in the power source...don't know yet.

But I think wattage is also a challenge. As I understand:

P = I*I*R or P = V*R
so
144 = 3*3*16 or 144 = 3*48
...meaning at least a 150 watt resistor/rheostat? Seems like a lot. Is my reasoning correct?

Assuming some of the above is correct - should I go with a rheostat over a resistor? I'm thinking there's going to be enough variability in the circuit to need the rheostat to regulate the current to 3 amps.

The problem I'm having is finding some of these parts based on my assumptions above.

 150 watt/16 ohms resistor assuming all else works out perfectly or
 150 watt/16 ohm rheostat?
 48 volts. I'm thinking I'm going to use 4 12 volt car batteries in serial to achieve the 48 volts. There's also some 48 volt/3 amp battery chargers with AC input which would be a lot easier, BUT, I think the problem is that these devices actually try to be smarter about charging - like varying voltage, etc, based on the state of the battery they are charging and I'm not sure how the charger would react to the load. But I'd sure like to be able to use something like this so that I don't keep draining and then having to recharge the car batteries - and it would be cheaper than 4 car batteries. I'm sure I'd have to cut open some output wires so I have my leads for my load, but I'd be OK with doing that.

I think this is it, thanks for any suggestions that come my way.

2. ### alfacliff Well-Known Member

Dec 13, 2013
2,449
432
you dont apply current to a load, the load pulls current when you apply voltage. the current drawn is controled by the load. apply 48 volts to your load and measure the current. your load might be rated at 3 amps, but under what conditions? a pure resistive load will pull the rated current, but nearly anything else will pull what ever it pulls up to the rated load.

3. ### R!f@@ AAC Fanatic!

Apr 2, 2009
9,647
1,111
Mind telling more about your LOAD

4. ### crutschow Expert

Mar 14, 2008
22,524
6,602
Yes I = V / R but R is the load, not an external resistor.

5. ### aac_dave Thread Starter New Member

Jun 3, 2014
2
0
Thanks everyone. Sorry, as you can tell, I'm still learning. I thought I understood this simple exercise, but now I'm feeling sort of on the spot

Unfortunately, I don't know the current the load will pull and it will probably vary. But, I will have some control over how conductive it is.

And I'll have to be sure of is that the load is conductive enough to pull at least 3 amps of current.

But once it pulls more than that, a rheostat will be helpful in restricting the current to 3 amps?

So for example if the load has a resistance of 8 ohms.

I = V/R
so
6 = 48/8

Now, if I want to restrict the current to 3 amps, is it this:

I = V/(R1*R2)
or this:
I=V/(R1+R2)

Assuming the former, then a resistor of 2 ohms or the later resulting in 8 ohms.

I'm lost on the wattage though...And probably in many other ways

thanks again for the help.

6. ### #12 Expert

Nov 30, 2010
18,076
9,686
Eh...nobody holds a grudge against beginners.

It's R1 + R2

and power is I times E
or I squared times R
or E squared over R
For each individual part.

A real problem here is that low ohm rheostats are expensive!
Much cheaper to experiment or calculate what you need and use a single resistor.
Low ohm, high watt resistors are pretty cheap at (usually) less than \$1 each.

7. ### Alberto Active Member

Nov 7, 2008
169
36
If the device you have is rated @ 48 volts with current @ 3 Amps then you have to worry about to the power supplier. Is my power supplier capable to deliver 3 Amps @ 48 volt ?

This is the question you should answer! And here the power is what you need.

As it has been said before your device will pull the current it will need and very likely 3 amps is the maximum required, placing a resistor in series you will limit the current and conseguently dropping the voltage and this could demage the device.

A good thing you can do is to post more detail of the device you have in mind (data sheet will help a lot) so that we could be more useful in helping you.

Cheers

Alberto

8. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,984
1,844
Nope... the second form is the correct one. Resistors in series add. Resistors in parallel, well... that's for another day as it is a little complex.

The 16 ohm, 144 watt resistor is correct in that it will absolutely limit the current to 3 amps, even into a dead short circuit. In fact, it only delivers that 3 amps into a short. Any other resistance in the line will yeild a lower current.

9. ### BobTPH Senior Member

Jun 5, 2013
1,810
474
Does anyone else smell HHO in here?

Bob

10. ### #12 Expert

Nov 30, 2010
18,076
9,686
I wasn't paying attention to the context. I just caught a question in post #5

11. ### R!f@@ AAC Fanatic!

Apr 2, 2009
9,647
1,111
This is going no where.
If OP does not tell about his load, how can we tell what resistance to use to limit I to 3A.

Any ways to limit current a resistor of appropriate ohms and power should be connected in series if the load is drawing more than 3Amps