# Variable resistor question

Discussion in 'General Electronics Discussion' started by John Spender, Jan 11, 2015.

1. ### John Spender

11
1
Jan 10, 2015
I have an older variable resistor - potentiometer I think they were called - and stamped on the side is "35 ohm; .845 amp."

I understand the 35 ohm is the max. resistance, but need help understanding the .845 amp. I know amp means current, and it probably has to do with ohm's law, but doesn't one need to know the voltage before a current can be specified, or are they assuming a certain voltage as a given. Thank you.

2. ### KrisBlueNZSadly passed away in 2015

8,393
1,267
Nov 28, 2011
Yes, that's right, the current that flows will depend on the voltage across, and the resistance. That figure of 0.845A must be a maximum rating; you're not supposed to pass more than that amount of current through it, or you might damage it.

That's a strange and very specific figure to specify; generally, resistors and resistive components are specified for maximum allowable power dissipation, and in round numbers - 0.5W, 1W, etc. So there may be something else going on here.

These devices are called potentiometers or "pots". They usually have three terminals - two ends and the wiper. Some have only two terminals, and these may also be called rheostats.

3. ### duke37

5,267
727
Jan 9, 2011
The 'pot' will be made of resistance wire which can stand 0.845A before being damaged. The voltage across the pot will be determined by how much of the wire is in action. Thus the voltage will be near zero when little wire is used and 35V when all the wire is used (at 0.845A).
The maximum power will be 25W, a nice round number.

4. ### John Spender

11
1
Jan 10, 2015

OK thanks KrisBlueNZ and duke37 I'm working on a coffee roasting gizmo; thus my questions.

1) How did you figure 35V when the wiper is at full end of the resistance wire? Does that mean this pot must not be used for any voltage exceeding 35V regardless of Amps?

2) If I have a length of resistance wire (nichrome) I know I can test it with my ohmmeter and tell the resistance in Ohms. Is there any way to tell how many
volts or amps it can be used for by other measuring devices, or must one be specified to calculate the other? I'm thinking of the heating element in this thing, but
would like to understand the general principle behind it.

Thanks again

5. ### KrisBlueNZSadly passed away in 2015

8,393
1,267
Nov 28, 2011
Trevor's calculations were written slightly wrong. When the rheostat is at maximum resistance, 35Ω, and is passing the maximum allowable current, 0.845A, the voltage across it will be:

V = I × R
= 0.845 × 35
= 29.575V

The amount of power the rheostat will dissipate will be:

P = V × I
= 29.575 * 0.845
= 25W.

So that explains the strange, "not very round" number in the rating - 0.845A corresponds to 25W when the rheostat is at maximum resistance.

Heating effect is proportional to power, and when the full length of the rheostat's wire is in the circuit, it can dissipate 25W. The heat that's generated will be evenly distributed over the whole length of the wire, and will be absorbed by the other material (bakelite or whatever it is) in the rheostat.

When the rheostat is at a lower resistance, the power dissipation at the maximum rated current will be lower. For example at mid-position, R = 17.5Ω so:

V = I × R
= 0.845 × 17.5
= 14.7875V

and the power dissipation will be:

P = V × I
= 14.7875 × 0.845
= 12.5W

So the rheostat will dissipate half as much power. That makes sense, since only half of the wire in the rheostat will be in the circuit, and the heat will be dissipated in an area only half the size compared to the first example. So only half as much heat energy can be dissipated before parts of the rheostat will get too hot and be damaged.

So specifying the rheostat for 0.845A maximum current actually makes a lot of sense. No matter what position it's at, as long as the current is less than 0.845A, the heat dissipated by the rheostat will be manageable.

Those two formulas, Ohm's Law and the Power Law, are all you need to calculate the relationships between voltage, current, resistance, and power for anything "ohmic"; that is, anything whose resistance is more or less constant over a reasonable range of voltage and current. Resistance wire is not exactly like that, because as it gets hotter (due to power being dissipated), its resistance (per distance) increases. But they should be enough to get you started.

Ohm's Law can be expressed in three ways:

I = V / R
R = V / I
V = I × R

The Power Law can also be expressed in three ways:

P = V × I
V = P / I
I = P / V

And the two can be combined in various way to eliminate one quantity or another. Common arrangements are:

P = V2 / R
P = I2 × R

6. ### duke37

5,267
727
Jan 9, 2011
Sorry about the typo, Kris is right as always. I should have put 30V not 35V.

This is the maximum voltage which should be across the pot when it is at full resistance. You can use it at higher voltages as long as the rest of the voltage is dropped across the load. The current is the critical parameter, putting 30V across it at half resistance will soon see its end.

You can measure the resistance of wire with a meter but this does not say what the maximum power it can take, this will depend on the way the heat is dissipated. Fan heaters will soon burn out the heater if the fan stops. Nichrome or Kanthal can run up to 1000degC and stainless perhaps 5 or 600degC. The higher the temperature, the shorter the life. The wire will oxidise particularly down the grain boundaries and so become brittle. Advice: never move a storage heater. Wire from an electric fire will be scrap.

hevans1944 and KrisBlueNZ like this.
7. ### John Spender

11
1
Jan 10, 2015
You guys are great. Thank you for the time to write quality answers. As I work with young people, I get to teach them what I'm learning from others (you) so thank you.

Now, getting more into the heart of the project, the small fan motor is connected across incoming house voltage (110 VAC) but is actually a low voltage DC motor. To
accomplish this they put a rectifier (4 diodes) which makes DC (I think) and put it in series with a nichrome coil. The primary heater coil is straight 110, so this
smaller secondary coil is - as Duke said above - dropping the voltage across the load. Since it's a small coil, it probably only provides a small percentage of the total heat
needed, but does double duty.

So here's the need. I want to be able to run the fan with no, heat at times. Easy to insert a switch and cut the primary heater coil off. But can't do without the secondary coil
or the motor/fan won't work. Some suggest using a transformer to supply the motor with what it needs. Could do that. But having lots of wire wounds pot, fixed ceramic
resistors of all sizes, I'm wondering if I couldn't substitute one of them? If It was big enough, there would be no appreciable heat. But how do I calculate all this as
I don't want to burn out the motor experimenting. Or is my thinking here faulty. Maybe a starting point is to measure the resistance of the secondary coil? But then . . .

8. ### duke37

5,267
727
Jan 9, 2011
The size of the resistor will not affect the heat generated but a bigger resistor will run cooler. The current into the motor can be controlled with a capacitor in one of the mains supply lines. The motor will not be isolated so if not insulated properly there will be risk of shock.

The resistance of the motor should be low and, when run from a DC supply, the current will be determined by the torque and the subsequent back EMF.

Using DC you should measure the voltage and current at the speed and load you need.

1,114
157
Aug 13, 2011
Using a resistor in place of the heater coil only distributes the same heat differently. If you use a large enough resistor/rheostat, the specific heat will be reduced so it's only warm to the touch but the total heat dissipated will be the same.

The use of part of the heater to limit current to the motor is a technique commonly seen in blow dryers but if you want to run the fan cold I'd suggest a separate power supply path for the fan.

10. ### John Spender

11
1
Jan 10, 2015
OK thanks for those posts. Based on all the above, I have a tentative conclusion and a couple questions:

Conclusion: Seems like trying to use a resistor (fixed or variable) in place of the heater coil is not a good idea. I'll have to use a transformer power supply.

Questions: The statement "Using DC you should measure the voltage and current at the speed and load you need." Does this mean I get the thing running at normal speed
and then measure the voltage across the motor terminals? Can I take this reading as the operating voltage of the motor? Then to check the current, can I put
a digital VOM in series with one supply lead and read current?

I'm intrigued by the statement "The current into the motor can be controlled with a capacitor in one of the mains supply lines." I didn't know capacitors could be
used this way. Will that work on either AC or DC? Thanks to all again . . .

11. ### BobK

7,660
1,675
Jan 5, 2010
Using a capacitor to control current only works with AC. With AC, a capacitor looks like a resistor that varies with frequency, but, unlike the resistor it will not dissipate any power and therefore not get hot.

Bob

12. ### John Spender

11
1
Jan 10, 2015
Is there anyway to determine what a good starting value would be? Are we talking large electrolytics or just small foil caps?

13. ### BobK

7,660
1,675
Jan 5, 2010
It depends on the voltage and current needed for the motor, do you know these?

Bob

14. ### duke37

5,267
727
Jan 9, 2011
The capacitor will need to be sized to pass the correct current I suggested that you measure the current required and you seemed to understand this. The reactance is 1/(2*pi*f*C).

The capacitor should be of the type (X or Y?) intended to be connected across the mains so that if it fails, it will fail open circuit. It most certainly NOT be an electrolytic. My guess is that it will be 1 to 3μF. This technique is used to change 120V toob radios to 240V valve radios.

15. ### John Spender

11
1
Jan 10, 2015
Bob, Duke;
Here is the imprint from the motor label: 20.5 VOC 1.78 A MAX. I am not familiar with the expression VOC.

The language above implied the capacitor was in series with one of the supply lines; you say across the mains?

If the capacitor will not dissipate power (as mentioned above) what use does it serve? Someone said it is used to "control"
the power. But capacitors are fixed; not variable No? Sorry to seem dense but I'm trying to "get it."

16. ### duke37

5,267
727
Jan 9, 2011
When connected to AC, a capacitor will pass current. This is out of phase with the applied voltage and the energy going into the capacitor is returned to the supply so no net power is consumed. The current level is determined by the reactance (see #14) and the voltage across it. In your case, you have a motor in series with the capacitor which will take a little of the voltage but most of the voltage will be dropped across the capacitor.

The capacitor will have a difficult job to do so the type of capacitor is critical and the type which can be connected across the mains should be used. Here it will be in series with the supply to the rectifier and motor.

The radio modifications I mentioned use about 3μF capacitors to control the current in 0.1A valve heaters so I would expect that the capacitor will be too large at an amp or so to be used in your application. Measurement of the motor current when under load is required. A transformer may well be smaller and cheaper.

17. ### John Spender

11
1
Jan 10, 2015
Thanks Duke; I'm going to look through my stuff to find a transformer with about 20 V output. My understanding is that if it's too small,
it will get warm/hot, but if it's bigger than needed it won't burn out the circuit. By the way, if you get a moment, can you define
"reactance" for me. I see the formula above, but not sure what it means. Thank you

18. ### BobK

7,660
1,675
Jan 5, 2010
Reactance is a generalization of resistance which applies to AC circuits having capacitors and inductors (as well as resistors). It is actually expressed as a complex number because the phase of current and voltage are off by 90 degrees in capacitors and inductors. The phase differs in opposite directions in inductors and capacitors, which leads to the bizarre ability of the reactance of a capacitor cancelling the reactance of an inductor. Which is how LC filters work. At some specific frequency, their reactance of a combined capacitor and inductor will be 0.

Bob

19. ### duke37

5,267
727
Jan 9, 2011
Wikipedia has a page on reactance but I will try to explain a little.

Resistance - R
A resistor needs a voltage across it in order for it to pass current either AC or DC. Ohms law describes this V = I * R.

Reactance - Xc and Xl
A capacitor will pass AC current but not DC. The equation is similar to Ohms law. V = I * Xc.
Inductive reactance is the opposite of capacitive reactance and, if combined you get a tuned circuit.

Impedance - Z
If you connect a resistor and capacitor in series, the combined effect cannot be calculated as the sum of the two because the effect of phase.
The impedance is Z = sqrt (R*R + X*X)
Ohms law can be modified again so V = I * Z

In your case, the capacitor will drop most of the voltage so the effective resistance of the motor can be ignored and the current I will be V / Xc

20. ### John Spender

11
1
Jan 10, 2015
Just a quick note to state my sincere thanks to those who have taken the time to write and explain these fascinating principles.
After I work on the project for awhile, I'll probably have more questions and it's nice to have somewhere to go with them, but that will
be fodder for a new thread.  