# Question about electric motors and speed **URL:** \ **Category:** Factual Questions\ **Created:** [August 25, 2004, 3:21pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451 "2004-08-25T15:21:48Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [August 26, 2004, 2:17pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/21 "2004-08-26T14:17:29Z")
Just buy a smaller pulley for the motor. Or a variable pulley that you can set for the speed that gives the air flow you want and then leave alone. Electronics are great but fancy electronic gadgets aren’t necessarily the solution to all problems. ---
**Author:** ![caymus28](https://avatars.discourse-cdn.com/v4/letter/c/7bcc69/32.png) [@caymus28](https://boards.straightdope.com/u/caymus28)\ **Post date:** [August 26, 2004, 2:24pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/22 "2004-08-26T14:24:49Z")
Thanks for all the replies. As far as it being induction motor or not…it looks [http://www.electricmotorwarehouse.com/Furnace\_Motors.htmlike](http://www.electricmotorwarehouse.com/Furnace_Motors.htmlike) this], if that helps. So from what I gather, a variac would be a safe, albeit, expensive option, and a potentiometer would be an unsafe, inexpensive option? I was really looking for a cheap option, along the lines of adding a dimmer switch or the like, but if is unsafe, it is not worth it (I sometimes leave the fan on for long periods while I’m away when trying to dry things (paint, drywall, flooded basements…)). So what’s the cheapest, safe way of doing what I would like? (I think a Variac is too expensive for this purpose…) ---
**Author:** ![caymus28](https://avatars.discourse-cdn.com/v4/letter/c/7bcc69/32.png) [@caymus28](https://boards.straightdope.com/u/caymus28)\ **Post date:** [August 26, 2004, 2:31pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/23 "2004-08-26T14:31:16Z")
> [@David Simmons](#): > > Just buy a smaller pulley for the motor. Or a variable pulley that you can set for the speed that gives the air flow you want and then leave alone. Electronics are great but fancy electronic gadgets aren’t necessarily the solution to all problems. The pulley on the current motor seems to be permanently attached to the shaft- I’ve never worked on a motor, but it seems that it would be too much of a hassle to try to replace it. ---
**Author:** ![awldune](https://avatars.discourse-cdn.com/v4/letter/a/c77e96/32.png) [@awldune](https://boards.straightdope.com/u/awldune)\ **Post date:** [August 26, 2004, 2:47pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/24 "2004-08-26T14:47:24Z")
What’s the amperage on your blower? A quick google indicates that it’s not uncommon for dimmers to be rated for 600W, so a dimmer might be OK if your blower pulls 5A or less. ---
**Author:** ![caymus28](https://avatars.discourse-cdn.com/v4/letter/c/7bcc69/32.png) [@caymus28](https://boards.straightdope.com/u/caymus28)\ **Post date:** [August 26, 2004, 2:58pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/25 "2004-08-26T14:58:26Z")
i’ll check the amperage on the motor when I get to the job site later today, and I’ll post all the relvant info…thanks again! ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [August 26, 2004, 3:24pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/26 "2004-08-26T15:24:10Z")
> [@catsix](#): > > …a current will not exist in the secondary unless it is placed under load (there must be a complete path for current to flow). This is true of an ideal transformer, but in the real world, ideal transformers do not exist. If you examine an equivalent circuit schematic model for a real-world transformer such as [this](http://www.midcom-inc.com/tech/tn17.htm), you’ll note a capacitance in parallel with the secondary (among other loss paths), which is due to various intrawinding stray capacitances. A small leakage current flows through this capacitance. At 60 Hz, the capacitive reactance of this shunt capacitance is large, and usually negligible, but at higher frequencies it becomes more significant. ---
**Author:** ![Dag\_Otto](https://avatars.discourse-cdn.com/v4/letter/d/439d5e/32.png) [@Dag\_Otto](https://boards.straightdope.com/u/Dag_Otto)\ **Post date:** [August 26, 2004, 3:28pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/27 "2004-08-26T15:28:22Z")
To reiterate the best advice you’ve received on this subject - just add a damper to the inlet or outlet of the fan to restrict the flow. It really is that simple. ---
**Author:** ![catsix](https://avatars.discourse-cdn.com/v4/letter/c/e19b73/32.png) [@catsix](https://boards.straightdope.com/u/catsix)\ **Post date:** [August 26, 2004, 3:42pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/28 "2004-08-26T15:42:17Z")
> [@](#): > > **Q.E.D** said: > A small leakage current flows through this capacitance. At 60 Hz, the capacitive reactance of this shunt capacitance is large, and usually negligible, but at higher frequencies it becomes more significant. And since we were discussing, in this thread, devices plugged into 60 Hz AC supplies, why would I include a detailed explanation of the negligible leakage current? > [@](#): > > **Demostylus** said: > Look, catsix, I’m sure you mean well, but any potentiometer you’re going to find in an electrical supply store isn’t going to handle the voltage or the current, and it’s dangerous advice to suggest that unqualified people should fiddle around with mains voltage wiring. Really? A 600W potentiometer won’t handle a voltage of 115V and a current of 5 Amps? Last time I did the calculations, that meant a 575 W, which is below the pot’s rating. And it’s ‘dangerous’ to fiddle around with the wiring in a device that’s unplugged or that the circuit breaker is turned off? Since when? Someone with knowledge enough to handle a schematic and the ability to read a data sheet should be able to handle such a thing. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [August 26, 2004, 4:17pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/29 "2004-08-26T16:17:43Z")
> [@catsix](#): > > And since we were discussing, in this thread, devices plugged into 60 Hz AC supplies, why would I include a detailed explanation of the negligible leakage current? Because what you said was wrong, that’s why. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [August 26, 2004, 4:29pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/30 "2004-08-26T16:29:34Z")
> [@catsix](#): > > And it’s ‘dangerous’ to fiddle around with the wiring in a device that’s unplugged or that the circuit breaker is turned off? Since when? It’s not that it’s dangerous to work on de-energized circuits–it isn’t, unless there are capacitors present. It’s what might happen _after_ you plug it back in and restore power. If you don’t know what you’re doing the result can be anything from nothing to you or someone else getting electrocuted and your house burning down. Given this, being dismissive of warnings to that effect is highly irresponsible. ---
**Author:** ![catsix](https://avatars.discourse-cdn.com/v4/letter/c/e19b73/32.png) [@catsix](https://boards.straightdope.com/u/catsix)\ **Post date:** [August 26, 2004, 4:33pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/31 "2004-08-26T16:33:34Z")
> [@](#): > > **Q.E.D.** said: > Because what you said was wrong, that’s why. None of what I said was wrong. > [@](#): > > If you don’t know what you’re doing the result can be anything from nothing to you or someone else getting electrocuted and your house burning down. Given this, being dismissive of warnings to that effect is highly irresponsible. Anyone with the ability to read and follow a schematic and a data sheet should be perfectly capable of adding a component to a circuit without burning down the house. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [August 26, 2004, 4:38pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/32 "2004-08-26T16:38:57Z")
> [@catsix](#): > > None of what I said was wrong. Oh, really? > [@catsix](#): > > …a current will not exist in the secondary unless it is placed under load. …is wrong for the reasons I stated above. > [@catsix](#): > > Anyone with the ability to read and follow a schematic and a data sheet should be perfectly capable of adding a component to a circuit without burning down the house. You do not know that the OP has those skills. He may, he may not. To assume that he does is, again, irresponsible. ---
**Author:** ![catsix](https://avatars.discourse-cdn.com/v4/letter/c/e19b73/32.png) [@catsix](https://boards.straightdope.com/u/catsix)\ **Post date:** [August 26, 2004, 4:52pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/33 "2004-08-26T16:52:08Z")
> [@](#): > > **Q.E.D.** said: > …is wrong for the reasons I stated above. Nope, it’s 100% technically correct. Current cannot flow through an incomplete circuit, or are you attempting to say that it can? > [@](#): > > You do not know that the OP has those skills. He may, he may not. To assume that he does is, again, irresponsible. I’ve specifically said that it is a reasonable thing for someone who can read a schematic and a data sheet. It’s up to the OP to decide whether or not he can read a schematic and a data sheet. By the way, current does not flow through capacitors because electrons do not travel through them, because the dielectric does not conduct current. When one plate is charged negatively, electrons in that plate cause electrons in the dielectric to move toward the other plate, which forces electrons in that plate to move and a current will then flow on the other side of the capacitor. But to say that current ‘flows through a capacitor’ is blatantly wrong. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [August 26, 2004, 5:08pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/34 "2004-08-26T17:08:38Z")
> [@caymus28](#): > > The pulley on the current motor seems to be permanently attached to the shaft- I doubt it. It it’s stuck on and hard to get off **Dag Otto’s** reiteration of **casdave’s** suggestion about the shutter is probably the way to go. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [August 26, 2004, 5:11pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/35 "2004-08-26T17:11:06Z")
> [@catsix](#): > > But to say that current ‘flows through a capacitor’ is blatantly wrong. That’s not correct. While it’s true that individual electrons do not generally cross the dielectric (there is a leakage current, since no dielectric is perfect, but we’ll ignore that), for all practical purposes, the current can be said to be flowing through the capacitor. There are several equations for calculating this current, which you’re free to look up for yourself. However, even if we stipulate there there is zero electron current flow through a capacitor, you said “_a current will not exist in the secondary unless it is placed under load._”, which is still wrong. Period. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [August 26, 2004, 5:24pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/36 "2004-08-26T17:24:54Z")
> [@Q.E.D.](#): > > That’s not correct. While it’s true that individual electrons do not generally cross the dielectric (there is a leakage current, since no dielectric is perfect, but we’ll ignore that), for all practical purposes, the current can be said to be flowing through the capacitor. There are several equations for calculating this current, which you’re free to look up for yourself. However, even if we stipulate there there is zero electron current flow through a capacitor, you said “_a current will not exist in the secondary unless it is placed under load._”, which is still wrong. Period. While we’re arguing theoretical fine points, don’t forget the secondary and primary current that flows as a result of radiation from the transformer. And by the way, **caymus28** , if to can borrow a puller to get that pulley off, there are matched sets of stepped pulleys so that you can have a range of blower speeds. Of course you can also have various sizes of shutters. ---
**Author:** ![catsix](https://avatars.discourse-cdn.com/v4/letter/c/e19b73/32.png) [@catsix](https://boards.straightdope.com/u/catsix)\ **Post date:** [August 26, 2004, 6:14pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/37 "2004-08-26T18:14:13Z")
> [@](#): > > **Q.E.D.** said: > That’s not correct. While it’s true that individual electrons do not generally cross the dielectric (there is a leakage current, since no dielectric is perfect, but we’ll ignore that), for all practical purposes, the current can be said to be flowing through the capacitor. Actually, it is correct. Current does not flow through a capacitor, because electrons do not pass through the dielectric, aside from the fact that no dielectric is perfect. You chastise me for not stating what is technically correct, and then you say ‘for all practical purposes’? You’re wrong on this point. While current does appear on the “other side” of a capacitor, it’s not the same current that was applied, and it definitely does not pass through. So, you’d be getting that one ticked against you on my circuits exams. > [@](#): > > However, even if we stipulate there there is zero electron current flow through a capacitor, you said “a current will not exist in the secondary unless it is placed under load.”, which is still wrong. Period. Convenient then that you ignored the rest of that sentence? _but a current will not exist in the secondary unless it is placed under load ( **there must be a complete path for current to flow** )._ Now that I have highlighted it for you, can you accept that I specifically stated a closed path must exist for current flow? Or are you going to haggle over the fact that you can push electrons to or away from the ends of a conductor not in a closed loop, which is not considered to be ‘current flow’? You left off a very important part of what I said so that you could call me wrong. Of course, I’ve had students who do this also, so perhaps you would fit in with them. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [August 26, 2004, 6:20pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/38 "2004-08-26T18:20:23Z")
> [@catsix](#): > > Convenient then that you ignored the rest of that sentence? \*but a current will not exist in the secondary unless it is placed under load ( **there must be a complete path for current to flow** ).[/i So, if I connect a large capacitor across the secondary terminls, you’re going to say there is still zero current flow? Mind telling what my ammeter is reading when I connect it in series with this capacitive load? ---
**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\ **Post date:** [August 26, 2004, 6:45pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/39 "2004-08-26T18:45:03Z")
> [@catsix](#): > > Current does not flow through a capacitor, because electrons do not pass through the dielectric, aside from the fact that no dielectric is perfect. How did we get from “speed control of an AC motor” to a theoretical discussion on capacitors? For the record, you’re both correct, but **Q.E.D.** is (at least) looking at it from a practical and useful perspective. Yes, it is true an electron can’t tunnel through the dielectric of a perfect capacitor. So from the _physical perspective_, **catsix** is correct. But for _circuit analysis purposes_, the physical perspective of a component should rarely (if ever) be used. Instead, we must use lumped & distributive parameter models based on h-parameters, I-V relationships, transfer functions, etc. So from a _circuit analysis perspective_, current _ **does** _ flow trough a capacitor, and the magnitude & phase of the current on the capacitor’s two leads are _always_ identical (just like a resistor or inductor). The time-domain I-V relationship for an ideal capacitor, as I’m sure you’re all aware of, is **i = C dv/dt**. ---
**Author:** ![catsix](https://avatars.discourse-cdn.com/v4/letter/c/e19b73/32.png) [@catsix](https://boards.straightdope.com/u/catsix)\ **Post date:** [August 26, 2004, 8:07pm UTC](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451/40 "2004-08-26T20:07:18Z")
> [@](#): > > **Q.E.D.** said: > So, if I connect a large capacitor across the secondary terminls, you’re going to say there is still zero current flow? Mind telling what my ammeter is reading when I connect it in series with this capacitive **load**? Because of capacitive reactance, which by the way is measured in Ohms, you are placing a load condition on the circuit at that point in time and you will get a current out of it. If you’d like me to explain how capacitive reactance leads to apparent power but not true power because there is not a true resistance, sit in on my circuits class next week. > [@](#): > > **Crafter\_Man** said: > Yes, it is true an electron can’t tunnel through the dielectric of a perfect capacitor. So from the physical perspective, catsix is correct. But for circuit analysis purposes, the physical perspective of a component should rarely (if ever) be used. Instead, we must use lumped & distributive parameter models based on h-parameters, I-V relationships, transfer functions, etc. So from a circuit analysis perspective, current does flow trough a capacitor, and the magnitude & phase of the current on the capacitor’s two leads are always identical (just like a resistor or inductor). Which is why every one of my college circuits students is taught that the current they measure on the two leads of a capacitor are equal in magnitude, direction, and phase, it is still not technically correct to say that a current ‘flows through’ a capacitor. > [@](#): > > The time-domain I-V relationship for an ideal capacitor, as I’m sure you’re all aware of, is i = C dv/dt. Yup. First semester circuits course. Not something currently covered in my circuits courses because my students are not learning calculus based circuits and would have a heart attack if they saw dv/dt in any formula, but the books on my reference shelf definitely cover it, and I tend to give them problems like that when they think the approximations aren’t good enough for them. [Previous page](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451.md?page=1) [Next page](https://boards.straightdope.com/t/question-about-electric-motors-and-speed/261451.md?page=3)