The Trainer #81: When Voltage Drop Is A GOOD Thing!
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- Опубліковано 4 вер 2018
- With all the preaching I do on voltage drop testing to locate electrical problems, you would think that voltage drop is a bad thing. And it is when it’s not supposed to be there!
There are times, though, where voltage drop is designed in. And that’s the topic for this edition of The Trainer!
Watch the June Trainer on reading a wiring diagram • The Trainer #78: How ...
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06:43 That feeling you get when you finally understand a problem with a blower motor fan going out on setting 5 (high) and only on 5. . . after like 6 years, and you no longer even own that vehicle. Your videos are just awesome. I'm going to binge-watch them all.
Good basic information. Sometimes these circuits can get real complicated, but with a good wiring and component diagram your on your way to knowing the circuit and fixing the problem.
Good information. Truthfully, electrical was never my strong point. Your explanation helps! Thanks for sharing! ~Matthew
Great training as always . Thanks brother : )
Snap Off - Great refresher on ohm's law which every electrical technician understands. Not sure if advanced electrical theory was necessary here, but to simplify that equation...you missed the entire point of my comment. Basic rule #1 - keep it simple until there is no simple solution.
Thank you for these great tips. Very good and logical !
Thanks Pete great Tip
developing wire diagram muscles thank you Pete!
chevy is on another level
Engine running at idle reading at 14V, then load it headlights A/C on, voltage drops fast from 14V to 10V in 3minustes... Voltage drops is absolutely good!
Have you fixed the issue? If you have, can you please share with us what it was?
KVL Kirchhoff's voltage law. A good primer.
Having an issue trying to find a parasitic draw on a 2004 Jeep Grand Cherokee Special Edition 4.0L Straight 6 4WD engine pretty much loaded....I have an inexpensive multimeter and I want to use your method to check voltage drop across fuses....So would I have my black lead on COM and my red lead on V? and my dial set to 200m on DCV side? Does this method work on all fuses? What about relays or JK fuses? Also trying to figure out how long it takes for it to go to sleep....Any help would be greatly appreciated!
Ed, first - did you confirm that you do, indeed, have a parasitic drain? I would recommend you use a quality amp clamp and measure the drain overall first. As for how long it takes for the modules to "go to sleep", it depends on OEM and model. I wouldn't expect your vehicle to take more than a few hours, tops. The method shown works on any fuse that you can access both sides of. On some, you'll need to remove the clear plastic cover on top to gain that access. But before you do, get a copy of the Power Distribution diagram and identify the first fuses in line from the battery.
sir I don't get what causes the resistors value to change by adding or subtracting them in a circuit
The value of the individual resistors doesn't change, but the total resistance of the circuit does. In an electrical circuit, all available voltage will be used to overcome the resistance(s) in the circuit, and in proportion to those resistances. Focus on the blower motor resistor, made up of three resistors. Let's just use your number of 3 ohms each. We also have to take into account the resistance of the blower motor (which is a little harder since it changes a bit when it's running). To keep things simple, let's say it's also 3 ohms. So all the resistance(s) in the circuit are the same - which means all will demand an equal share of the applied voltage (12.0 v in our example).
In LO, the circuit path goes through all 3 resistors in the resistor assemble and the blower motor. That's a total of 12 ohms of resistance. Using Ohm's Law, we can determine that current flow through the circuit is 12v/12 ohms = 1 amp. Now, if one resistor is 3 ohms and we know now current is 1 amp, the voltage demand on each is 3 ohms time 1 amp = 3 volts.
Continuing in LO - the first resistor uses 3v of the 12v applied, leaving 9v to travel to the next resistor in series. It, too, uses 3v so now only 6v is left. The third resistor takes its share, 3v, leaving only 3v remaining for the blower motor.
In M1, there are two resistors in the circuit path in the resistor assembly and in M2 there is only one - plus the blower motor - to consider. Does that clarify it for you?
inrush current on high speed
wouldn't there be 6 volts to the motor if you moved the switch to m2 tan wire if the resistors value is 3 ohms each and 9 volts to the motor at the light blue wire
What is your rationale? Let's learn something together!
@@MotorAgeMagazine Pete stated that R1-R3+blower motor all had the same resistance values and created a three volt drop each, thus leaving 3 volts available at the blower supply. When he then switched to R2 and R3 only (M1 speed), he said there would be 4 volts available. Being a constant fixed resistor, if R2=3v drop, and R3=3v drop, 6 volts would be available to blower, M2 would leave 9 volts, and relay engaged would allow full system voltage to blower. Pete divided the 12 volt supply by the number of resistors in the second example to come up the 4 volts available. The resistor values are fixed, being coiled resistance wire, and can not change. seekintruth1977 is correct.
@@patrickmccarel Based on what you're saying, the drop across the R2 resistor - all by itself in the circuit - would still be only 3v?
@@MotorAgeMagazine Pete stated..." In this case each resistor in the circuit will drop 3 volts with a constant 12 volt supply". The actual values of the resistor are inconsequential, since we are discussing voltage drop. If R1+R2+R3 leaves 3 volts available to the blower, and each one drops 3 volts, then switching to M1 using only R2+R3 (3v+3v) would leave 6 volts available, not 4.
@@patrickmccarel But isn't it true that all available voltage will be shared proportionally across the individual circuit resistances? If R2, R3, and the blower motor are equal in resistance, shouldn't the voltage drop across them also be equal - or now 4 volts?