| Author | Message | ||
| don (Bottomacher)
Registered Member Username: Bottomacher Post Number: 173 Registered: 12-2000 Posted From: 72.15.86.59 Rating: N/A |
Hi Folks. I need a dimmer for my instrument lights (MC9). Do any of you know if I can use a rheostat from a passenger car (12volt)on a 24 volt system? | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 472 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
Don, Can't give you a definite answer, but consider the following. Rheostats are made with specific resistance values, and wattage ratings. A rheo with too much resistance will give you dim lights in the bright position, too little res will give you bright in the dim, and too low a wattage rating will smoke in a short time. The answer is to read the ratings on the two rheos... G | ||
| Debbie and Joe Cannarozzi (Joe_camper)
Registered Member Username: Joe_camper Post Number: 101 Registered: 10-2006 Posted From: 71.239.202.82 Rating: N/A |
Switch the dash lights to 12 volt along with the rheostat. | ||
| don (Bottomacher)
Registered Member Username: Bottomacher Post Number: 175 Registered: 12-2000 Posted From: 72.15.86.59 Rating: N/A |
Thanks to both of you. A friend gave me one today from his parts shed. Don | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 473 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
Don, 12V bulbs in a 24V bus will blow as soon as the knob is turned to the bright position. Most automotive style rheostats have almost no resistance at one end of the scale, which will allow full system voltage thru them. If the bulbs aren't rated for it, FLASH! G | ||
| don (Bottomacher)
Registered Member Username: Bottomacher Post Number: 176 Registered: 12-2000 Posted From: 72.15.86.59 Rating: N/A |
Thanks, George. I wouldn't use any 12v device with 24v. The rheostat I got is from a 24v MCI. Don | ||
| Debbie and Joe Cannarozzi (Joe_camper)
Registered Member Username: Joe_camper Post Number: 102 Registered: 10-2006 Posted From: 71.239.202.82 Rating: N/A |
All who own Prevo have experienced the pitiful dash lighting. One way I have seen this problem solved is to put a multimeter on the rheostat and turn it up until you get to 13.9-14.1 volts and either mark that point and or put a stop there. Then use all 12 volt bulbs for the dash. WORKS GREAT. | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 476 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
Now we have an explanation which makes sense for the prior post which didn't make sense! All's well. G | ||
| Bruce Henderson (Oonrahnjay)
Registered Member Username: Oonrahnjay Post Number: 243 Registered: 8-2004 Posted From: 72.12.39.78  Rating: N/A |
__. Thank you for that one George. I don't quite understand those electric-ron things but I was pretty sure that a rheostat hooked to a 24v supply and 12v bulbs was burned out bulbs looking for a place to happen. __. Am I right that a rheostat variably controls voltage from a low value to a higher one as the knob is turned (not current/amps)? So, I guess if the rheostat were limited so it couldn't go over about 14v, then things could work OK. | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 477 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
Bruce, This is a HUMOROUS reply, as nobody is offended, and I want to keep it that way, so: Actually, you are wrong, we know right where the burned out bulbs would be! A rheostat, also called a potentiometer, actually varies resistance, so as the resistance increases, the current flow, and the voltage decrease, because the total resistance in the circuit increases. Old man Ohm figured this out well over a hundred years ago. Now, lets get practical. A 12V bulb will be bright at 14V, which is normal charging voltage. As the rheostat is turned toward dim, it increases resistance in the circuit, and the lights get dimmer. In the bright position, automotive rheos have nearly zero ohms. They also have a wattage (capacity) rating, and will smoke if they are run right at their rating in the lower resistance position. (Only a few turns doing all the work.) Without being qualified on Prevost dashboard problems, I would install higher wattage 24V bulbs instead of over-voltage 12V? Heck, we don't know the wattage of either the 12 or 24 volt bulbs. The REAL issue with under-voltage bulbs, is that the rheo is in series with all the bulbs which are in parallel with each other. So, we dial up 14V with say 6 good 12V bulbs, which means 14V drop across the rheo WITH THE CURRENT FLOW OF SIX BULBS! When one bulb burns out, the voltage will increase on the remaining five, because of the reduced current flow across the rheo. Failure of the rest will happen fairly quickly. Now I need a Foster's Bruce, and its all your fault! Regards, mate, George | ||
| Bruce Henderson (Oonrahnjay)
Registered Member Username: Oonrahnjay Post Number: 244 Registered: 8-2004 Posted From: 72.12.39.78 Rating: N/A |
George wrote: " Actually, you are wrong, we know right where the burned out bulbs would be! " __. Yeah, on the end of a wire that's having 24v being fed to it, right???? "A rheostat, also called a potentiometer, actually varies resistance, so as the resistance increases, the current flow, and the voltage decrease, because the total resistance in the circuit increases. Old man Ohm figured this out well over a hundred years ago. " __. Now, am I getting this right? The current doesn't really matter because that's kinda controlled by the amount the lights "draw" (except how it's related to resistance)? But the important part is the vottage? (I know about vottage - that's what Ms. Hirry Rodman Clifford didn't get enought of in her bid to become President again, right? And what her husband didn't get enough of, we'll leave for another discussion.) __. But we can't forget about Old Man Ohm - I know about him, too -- I got one of his meters ... it makes a good paperweight. " Now, lets get practical. A 12V bulb will be bright at 14V, which is normal charging voltage. As the rheostat is turned toward dim, it increases resistance in the circuit, and the lights get dimmer. In the bright position, automotive rheos have nearly zero ohms. They also have a wattage (capacity) rating, and will smoke if they are run right at their rating in the lower resistance position. (Only a few turns doing all the work.) __. And I hear you about the likelihood of bulbs blowing in a 12/24V resistorstat setup, but I'm going to guess that a resistorstat in that 24V setup is going to be making a lot of heat when 14V is coming out of it (when those electric-rons get resisted, they get angry and heat up, right?). If it's the turned down, won't the heat go up even more? I'd think that that would be described -- in technical language -- as "not too good". "The REAL issue with under-voltage bulbs, is that the rheo is in series with all the bulbs which are in parallel with each other. So, we dial up 14V with say 6 good 12V bulbs, which means 14V drop across the rheo WITH THE CURRENT FLOW OF SIX BULBS! When one bulb burns out, the voltage will increase on the remaining five, because of the reduced current flow across the rheo. Failure of the rest will happen fairly quickly. " __. Is this why, like, taillights will "burn out together"? You'll run for 41 months with no problem but then both will go. I'm guessing one burns out (normal aging) and then the remaining bulb is the only resistance (and it's had its normal aging, too), so it's then more likely to go out, even without one of those resistorstats in that circuit? " Now I need a Foster's Bruce, and its all your fault! Regards, mate, George " __. My friend, if we ever meet up (and I hope we do), the first one's on me! BH, NC USA (Maybe I'd do better if I could just *see* those electric-rons. Why do they have to be invisible?) | ||
| Don Evans (Doninwa)
Registered Member Username: Doninwa Post Number: 127 Registered: 1-2007 Posted From: 208.81.157.90  Rating: |
"A rheostat, also called a potentiometer" Actually they are two different things.  A rheostat is a variable resistor, two connections. Can be had to handle considerable wattage. Ignore the periods, just trying to get it to display correctly. Terminal 1______/\/\/\/\/\/\ .........................../\___________Terminal 2 A potentiometer is a fixed resistor with a variable tap, three connections. Usually low wattage. Terminal !1____/\/\/\/\/\/\/\/\_______Terminal 2 ............................/\_______Terminal 3 To confuse the issue more you can wire a pot as a rheostat but not a rheostat as a pot. If the dash lites are a two wire affair and accessible you could group the dash lites as pairs of 12 volt bulbs in series. There is always a way to make a simple system more complex. If I were redesigning the dash lites I would go LED. Good luck Don 4107 | ||
| Frank Allen (Frank66)
Registered Member Username: Frank66 Post Number: 81 Registered: 10-2005 Posted From: 64.12.117.73 Rating: N/A |
i got one from a freightliner truck and put it on my 06, works well Frank Allen 4106 | ||
| Richard Bowyer (Drivingmisslazy)
Registered Member Username: Drivingmisslazy Post Number: 2056 Registered: 1-2001 Posted From: 75.108.85.14 Rating: N/A |
Don, I seriously suspect that you may get some argument on this statement. LOL There is also an adjustable resistor. Usually in the higher wattages which has a band around the resistor which is loosened with a screwdriver to slide it up and down the ceramic shell. Richard | ||
| Don Evans (Doninwa)
Registered Member Username: Doninwa Post Number: 128 Registered: 1-2007 Posted From: 208.81.157.90 Rating: N/A |
Richard, I agree, just another meaningless point of interest about how terms change over time. They are used almost interchangeably these days. Even the suppliers mix them up. But if you are searching for higher wattage units, look for rheostats. Many pots marketed as rheostats these days because putting the third connection on fills the bill for two uses. Yeah, those big ole resistors with the slider and the 'dropping' resisters sold to drop 12V to 6V, when 12V vehicles were 'the new thing', were good for screwing up a lot of gear. Showing our age now. Don 4107 | ||
| Leland Bradley (Lee_bradley)
Registered Member Username: Lee_bradley Post Number: 31 Registered: 11-2006 Posted From: 138.163.106.71 Rating: N/A |
Since light output of the bulbs (incandescent) is a function of wattage, the wattage requirement is probably the same for 12 or 24 volts for the same number of lights. As rheostats are mainly rated by wattage (yes there are voltage limits but shouldn't apply to low voltage systems) then there is probably no difference in the rheostats used in 12 or 24 volt systems. | ||
| Gus Causbie (Gusc)
Registered Member Username: Gusc Post Number: 685 Registered: 11-2005 Posted From: 208.54.200.58 Rating: N/A |
I thought rheostats vary amps and potentiometers vary voltage(Potential)? But what do I know? | ||
| Richard Bowyer (Drivingmisslazy)
Registered Member Username: Drivingmisslazy Post Number: 2057 Registered: 1-2001 Posted From: 75.108.85.14 Rating: N/A |
Gus, both vary resistance. They are variable or adjustable resistors. As the resistance varies so do amps and voltage. As you increase resistance in a circuit the amps decrease and the voltage at the load also decreases. The voltage drop across the resistance increases. Richard | ||
| Gus Causbie (Gusc)
Registered Member Username: Gusc Post Number: 687 Registered: 11-2005 Posted From: 208.54.200.201 Rating: N/A |
Richard, Thanks. Always was confused about that. But, if I=E/R and you increase R to decrease I how will this change E? | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 481 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
Evening Gus, The direct answer to your question, is that it doesn't. Ohm's law has been around for a loooong time, so the terms are a little out of our everyday language. I is essentially induced current (Amps) E was first used to describe electromotivo, (electromotive force) or Volts to us gringos, and R is still resistance. Nowadays, A=V/R may be clearer? So, if you increase the resistance, you decrease the amperage, and the voltage SUPPLIED TO THE CIRCUIT stays the same. Yes the voltage drop across the rheo increases, and the voltage across the load in series with it decreases by the same amount. Does this clear up the mud? Regards, George | ||
| Richard Bowyer (Drivingmisslazy)
Registered Member Username: Drivingmisslazy Post Number: 2058 Registered: 1-2001 Posted From: 75.108.85.14 Rating: N/A |
Duplicate Post deleted. Richard (Message edited by drivingmisslazy on June 29, 2008) | ||
| Richard Bowyer (Drivingmisslazy)
Registered Member Username: Drivingmisslazy Post Number: 2059 Registered: 1-2001 Posted From: 75.108.85.14 Rating: N/A |
"So, if you increase the resistance, you decrease the amperage, and the voltage SUPPLIED TO THE CIRCUIT stays the same." Read this very carefully. At least it took me several readings to understand that George is talking about the INPUT voltage source, such as a 12 volt battery, which does not vary, and this is typical when discussing a circuit such as this. "Yes the voltage drop across the rheo increases, and the voltage across the load in series with it decreases by the same amount." Gus, this is what I stated. As the resistance of the adjustable resistance is increased, the current thru the circuit will decrease, as long as the SOURCE voltage remains the same. With the current thru the load decreasing, the voltage drop across it will decrease in the same amount the voltage drop across the variable resistor increases. Richard | ||
| Len Silva (Lsilva)
Registered Member Username: Lsilva Post Number: 153 Registered: 12-2000 Posted From: 71.99.27.221 Rating: |
Here is my attempt to explain the circuit. FWIW, Len  | ||
| Gus Causbie (Gusc)
Registered Member Username: Gusc Post Number: 688 Registered: 11-2005 Posted From: 208.54.200.164 Rating: |
When I learned Ohms Law he was still alive!! I had never really thought this through before and always assumed the voltage at the bulb stayed the same as source voltage!! If the source is 24v this brings me to one post about using 12v bulbs in a 24v system. If you do that will the 12v bulbs burn at normal brightness at the rheo mid-point and only the high resistance half of the rheo is usable for dimming? | ||
| Richard Bowyer (Drivingmisslazy)
Registered Member Username: Drivingmisslazy Post Number: 2061 Registered: 1-2001 Posted From: 75.108.85.14 Rating: N/A |
Len, I have to respectfully disagree with your drawings. As long as there is current flowing in the circuit there will always be some voltage drop across the light bulb. It might not be enough to illuminate the bulb, but it will be there. Richard | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 484 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
Richard, I hear what you are saying, but in Len's third drawing, he is showing a 12 Volt battery, and 12 Volts drop across the rheostat. Another way to look at it would be to consider the rheostat in this situation, to be an open switch. Both show 12 Volts, (system, or battery voltage) drop across the rheostat, which leaves nothing thru the bulb. Would 1 Volt (theoretical) cause current to flow thru the bulb and not light it? YEP! But we're talking theory here, and I'm going to take Len's side in this one. To try to explain Gus's last question, just double all of the voltage figures in all three examples. Then draw a fourth, with 10 Volts drop on the rheo, which would give 14 Volts drop across the bulbs. BUT, remember A in the previous post! Several bulbs are in parallel, then in series with the rheo. You dial the rheo to give 14 Volts across say eight bulbs, and when one burns out, you have the resistance to limit the current requirement of 8, but only 7 drawing current. This WILL cause the voltage across the remaining seven to rise. Pretty soon, another bulb can't stand say 15 Volts, and it fails. That will put 16+ on the remaining 5, which equals imminent failure. Len, Thanks for the effort in the drawings, it helps the verbal explanations. You're always helpful, and I really appreciate it. KUTGW, George | ||
| Len Silva (Lsilva)
Registered Member Username: Lsilva Post Number: 154 Registered: 12-2000 Posted From: 71.99.27.221 Rating: N/A |
Richard, I agree with you completely. If the rheostat were 50,000 ohms and the lamp were 1 ohm, there would still be a voltage drop across the lamp (good luck measuring it). My post was intended for those in this conversation with little grasp of Ohms law. Len (Message edited by lsilva on July 02, 2008) | ||
| Leland Bradley (Lee_bradley)
Registered Member Username: Lee_bradley Post Number: 32 Registered: 11-2006 Posted From: 138.163.106.72 Rating: |
Georg Simon Ohm was born on March 16 1789 and died on July 6 1854. He was a German physicist. As a high school teacher, Ohm started his research with the recently invented electrochemical cell, invented by Italian Count Alessandro Volta. Using equipment of his own creation, Ohm determined that there is a direct proportionality between the potential difference (voltage) applied across a conductor and the resultant electric current which flows through it -- which we now know as Ohm's law. Had no idea anyone on the board was quite that old but .... | ||
| Richard Bowyer (Drivingmisslazy)
Registered Member Username: Drivingmisslazy Post Number: 2062 Registered: 1-2001 Posted From: 75.108.85.14 Rating: N/A |
Len, that is a great set of drawings and does illustrate the point we are trying to make. As far as George's comment, I do not care if the rheostat is ten billion ohms, if there is current flowing thru the rheostat, then there MUST be current flowing thru the indicator lamp since that is the rheostats return to the other side of the source. Since there is current flowing thru the indicator lamp, then there is voltage drop across it. Lens drawing is not technically correct since he shows 12 volt drop across the rheostat. It should show, to be technically correct, something like 11 volts across the rheostat and 1 volt across the lamp. Richard | ||
| Len Silva (Lsilva)
Registered Member Username: Lsilva Post Number: 155 Registered: 12-2000 Posted From: 71.99.27.221 Rating: N/A |
OK, let's beat it to death. I haven't opened one up in many years but I believe that the typical automotive dash light rheostat actually goes open at the end of it's travel. Len | ||
| Gus Causbie (Gusc)
Registered Member Username: Gusc Post Number: 692 Registered: 11-2005 Posted From: 208.54.200.1 Rating: N/A |
Lee, Thanks for the Ohms lesson, but I was attempting a tad of humor there. However, I am a pretty old guy. | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 486 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
This thing is absolutely beat to death, to the point that the corpse is out of rigor mortis, and is beginning to smell. However, and it did take me several days to realize that I didn't say so the first time, Len's drawings are EXACTLY correct, according to Ohm's law. I DON'T want to get into a contest here, Richard, the corpse stinks enough already, but like it or not, AS DRAWN, with a 12V battery, and 12V across the rheostat, ITS ZERO ACROSS THE BULB PERIOD. If there is 0.001V across the bulb, then there is only 11.999V across the rheostat! That's Ohms law, you don't go to jail when you break it, because you can't. Gus, I got your humor, and I appreciate it, we gotta have some fun, right? Happy Fourth, and now I'm going to go mess with something live! George | ||
| Richard Bowyer (Drivingmisslazy)
Registered Member Username: Drivingmisslazy Post Number: 2063 Registered: 1-2001 Posted From: 75.108.85.14 Rating: N/A |
"AS DRAWN, with a 12V battery, and 12V across the rheostat, ITS ZERO ACROSS THE BULB PERIOD. If there is 0.001V across the bulb, then there is only 11.999V across the rheostat! That's Ohms law, you don't go to jail when you break it, because you can't." George you are exactly correct. Since there will be no current thru the rheostat unless it returns to the negative post of the source. And its return is thru the indicator. That is why I state that there can not be 12 volts across the rheostat unless it returns directly to the negative post of the battery. That is why I indicated that Len's drawing is in error. As long as the indicator light is in the circuit there will be current flowing thru it and there will be a voltage drop across the indicator. Even if it is only .0000001 volt. Richard | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 488 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
Richard, As I said several posts before, I hear you. I still mean it, so lets consider this: What is the difference between a rheostat that is open at the end of its travel, and an open switch? The answer is none. NO current flows in either circuit. Apparently you feel that every rheostat does not go open at the end of its travel, which is neither here nor there. If the rheostat is open at the end of its travel, the voltages will be 12 and zero, whether you think so or not. If the rheostat has 50,000 ohms at the end of its travel, then the voltages will be something like 11.99996, and 0.00004. (I might be off a zero beyond the decimal point, but it doesn't matter, and I'm not going to find the calculator.) If we go up and look at Don Evans' post on the difference between a rheostat and a potentiometer, and connect terminal 1 to the +, and terminals 2 & 3 to the indicator, and the other terminal of the indicator to the -, we have to thank him for going to considerable effort to publish what he did, AND how to flummox the computer to get a drawing past it! The voltage readings in those drawings will be the same, depending upon the setting of the rheostats. Does a DC voltmeter have to see both positive and negative in order to read voltage? Yes, but even one of them can't make current flow thru an open circuit. Now, look CLOSELY at the bottom drawing, and you will see that the arrow is BEYOND the last peak, and definitely doesn't touch it, as it does in the upper two. That's an open circuit if I ever saw one. Intentional or not, the drawings are right. Len's demonstration of Ohm's law is correct, and I guess we will have to agreee to disagree. George | ||
| Richard Bowyer (Drivingmisslazy)
Registered Member Username: Drivingmisslazy Post Number: 2064 Registered: 1-2001 Posted From: 75.108.85.14 Rating: N/A |
"I guess we will have to agreee to disagree." Yep, I agree with that. I personally have never seen a rheostat that is open on one end, but there are a lot of things I have never seen. LOL Perhaps you might explain to us how there would be any voltage drop if the end of the rheostat is open? I always thought that for any voltage to be dropped across any load there had to be a small amount of current flowing thru the load. With the open circuit in the rheostat I do not see how any voltage could be read across it. Not trying to be argumentative, just trying to understand how the circuit would work. Thanks for any help. Richard | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 490 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
Richard, I do appreciate a polite reply, it makes it easy to explain!!! Now that I see what you are getting at, here comes an explanation. Please take a voltmeter and read the voltage across a bus battery. Record it. Then get a switch, a couple of alligator clips, a small light bulb rated the same voltage as the battery, and wire the light and switch up in series to the battery terminals. (The same as Len's drawings.) (We do both agree that an open switch carries no current?) With the switch in the open position, (and a good bulb,) you will find that the voltage reading across the terminals of the switch is the same as the previous reading across the battery terminals. Now turn on the light, and then record the voltage across the battery terminals again. It will be slightly less than before, because now you have put some load on the battery. Then read the voltage across the switch, which will be NEAR zero, depending upon the switch. Then read the voltage across the bulb, which will be VERY close to the battery voltage, minus the drop across the switch, if there is any. Any difference in total readings, which won't be a tenth of a volt, is caused by the resistance in the wires and the connections. After you look at the open switch circuit for a while, where I told you to measure voltage across the switch terminals, you are going to want to hit me, and fortunately I'm out of reach! You will see that all you are doing is measuring the battery voltage again, this time with some more wire, and thru the bulb, and as there is no current flowing, there is no drop! Whew, George | ||
| Richard Bowyer (Drivingmisslazy)
Registered Member Username: Drivingmisslazy Post Number: 2065 Registered: 1-2001 Posted From: 75.108.85.14 Rating: |
Thanks George for the time taken to provide an answer. Since you are assuming there is an open circuit at the high end of the rheostat and you are apparently measuring from the negative terminal of the battery, instead of across the rheostat itself, then I agree with you that there will be a 12 volt reading. Richard | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 492 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
Richard, Thanks for the thanks, AND for continuing to ask like a professional, and a gentleman. I'll bet you know what an inversion valve is! Regards, George | ||
| Richard Bowyer (Drivingmisslazy)
Registered Member Username: Drivingmisslazy Post Number: 2066 Registered: 1-2001 Posted From: 75.108.85.14 Rating: |
I am not really familiar with the mechanical systems so I will pass on that one. LOL Now I have a technical question. Has anyone ever actually seen a rheostat or potentiometer that is open at one end. In other words only two connections instead of three. I personally have worked around electronic and electrical for over 50 years and I do not recall ever seeing one. Richard | ||
| George M. Todd (George_mc6)
Registered Member Username: George_mc6 Post Number: 493 Registered: 8-2006 Posted From: 207.231.75.253 Rating: N/A |
Good morning Richard, Your first sentence gets a WHOLE LOT of points in my book, and a vote! There are cab heater rheostats, and rear window defroster rheos that double as the switch. As you say, in the AC motor control business I don't know if I ever saw one, mostly because I never took an ohmmeter to it? Don's drawing of a potentiometer would always show the maximum resistance, as long as read between terminals 1, and 2&3 jumpered together, and the knob in the high resistance position. I just had a horrible thought. In that situation, does R1 X R2/R1 + R2 come into play? Actually I guess it does, but it doesn't matter, as we still get the resistance value we want when we turn the knob. Regards, George |
