True, but as a teacher I have to consider my students and what their interests and abilities are. I teach mechanics, and mechanics need a mechanical illustration. Since no one has actually ever seen voltage or current, I feel pretty confident in using the example I did. It's all observation of effects - not an observation of reality. Voltage is pressure anyway, and it doesn't move. Voltage pushes Amperage thru Resistance. It's all what you care to believe, and as long as it makes sense, use it.
Daniel, That is very consise teachingt. I needed a refresher to make sure I was interpreting the voltage drop across the ingnition switch on an old goldwing I am restoring. I am having a charing system problem and all the trouble shooting instructions result in good components. Then as I decided to look further I noticed the switched voltage at the regulator was about 1 volt low. Now its on to the search for a replacement. Thanks for the refresher!!!
Great video! It was very informative and helped me understand the material, I'm currently studying this in my AP Physics B class and this helped me understand voltage drop well.
It simply makes every voltage reading a voltage DROP reading. So, anytime you ever read voltage just push the button to load the circuit. If the wire is a sensor input, you can isolate it, put voltage on one end and read and load at the other. Several techs use it regularly on computer circuits - but they're pretty knowledgeable about their systems. I'm posting a video tomorrow about fuel level systems and you can simply read voltage and when you load the circuit, the gauge moves. Cool, huh?
Thanks for the reply that was one of the better explanations I've come across. I've spent too much time pondering over this problem. I think I now need to find out what factors determine the speed of the charges. Perhaps that will help me understand, although I thought it was just the source voltage and load resistance that determined the speed.
Hi Daniel, I work for a premium German automotive company and your videos are some of the most informative and concise I have ever seen. Your methods used in training work very well. I have some questions regarding the voltage drop method, is there any examples of circuits where this method does not/can not find faults or an example where this method should not be used? Is there any scenario where the 25ohm resistance in the leads could cause damage to control units? Also, is it possible to order these leads in Asia?
+Steven Webb Thanks. Glad to help. Use whatever you can and need. That's why I created it. if you're not aware of my book and LOADpro leads, contact Mary at Electronic Specialties (maryk@esitest.com) and tell her you're an educator. There's and educational price list. No meter company *really* gets the education market - we're (I'm) trying to focus on the training aspect of the business rather than the sales. Also - look this over: th-cam.com/video/DfMZxWXb8M8/w-d-xo.html (I don't make a ton of money - less than a couple of hundred dollars a month). I've been in the classroom about 27 years, and I'm all about simple yet effective. Call me if you have problems or questions.
excellent video and train bought your tool works great just have to know how to use it and understand elect. have power probe and lots of other testers it is still good but yours works 100%
Hi everyone, I'm a proffesional tech who uses Dan's methods, and they work! His way of troubleshooting has made me a better tech. Check out his website, you won't be sorry. Keep on putting out more videos Dan, this is a great idea.
You can't do a voltage drop on an open circuit. Current must be flowing for there to be a "drop". The drop is what the resistor is consuming and to consume there must be flow. The voltage across an open will be full voltage because all you're really doing is reading at the battery posts. The wires are connected to the battery, right? So, if you read across an open, the wires are extensions of the battery terminals.
Thanks for the reply.I am a newbie at this .I understand ohms law .I have always wanted to learn how to diagnose faults (electrical) on my own vehicles but never had any training only read from books .The mechanical side i can do pretty good,changing brakes,timin belts ,head gaskets etc but this side really interests me .Just a little worried that i don't fry my ECU if i test the wrong way with the loadpro lol as i would love to get this and do some voltage drop testing on my car.
Daniel, Just want to commend you on the way you explain and draw out what you are explaining. Your awesome!!! Thanks for your time, please teach more. Generators, UPS's, Pulse Wit Mod, Rectifier, scr. Also how do you determine or calculate what component you will put a circuit. Say like, when do I put a resistor, where do I put a resistor, where and why would I put a capacitor after the resistor. Hope I make sense. But thanks a lot for you vids.
@Daniel Sullivan. I remember when I tried to watch your video when I was first starting to learn about electricity and skipped it after the first min. because everything was so fast. I've learned more since then (about a year ago) and was able to keep up this time. It would have been very helpful THEN if I would have just paused the video after every one of your points and wrote that point down or just memorized it and then moved on to the next one.
Thanks, its understandable now. Some of the other videos are misleading. I'll find the resistance of my coils with a meter when I can pick one up from the store. I get it now. 24 volt power supply can work with a 4 ohm load to give 6 amps to the circuit. From the other 12vdc transformers, the coils must have been about 100 Ohms or less drawing more than their 0.1 amp rating.
Yes - but the videos don't pay the mortgage, so I've had to focus on training. I'm in North Dakota now and will be in Wyoming in 2 weeks. I have several planned - but timing is everything. Thanks for the comments though.
can you use the loadpro the same way on computer controlled cars ,say for checking for faults in wiring for lights ,fuel pump circuits etc without damaging the ecu .I am in the UK and thinking of getting one and your book to learn trouble shooting .Could you please let me know what you can test with this on modern computer controlled cars.Thanks Shaun.
What does jump across 2 resistors mean? Around 12:37 on the video, how did you read (A VOLT coming into the bulb)? Sorry for all the questions, but I'm a beginner when it comes to electrical and i'm trying to learn everything about this subject as much as I can. :)
Sir, Tnank you so much, i can use a 50 watts 57 ohm resistor to keep cool going. our domestic line voltage is around 250 to 260 i am facing frequent burn outs of electricals like bulbs etc, thanks once again.
Hey Dan, I used your loadpro leads on a 5v signal wire and a ground wire going to a camshaft position sensor. When I loaded the circuit, the wire dropped to .118v. I took apart the tape in the wiring harness to inspect the wire for a lot of corrosion but I was not able to find any corrosion. This wire goes into the PCM, is it possible for the PCM driver to have too much resistance in it? I'm suspecting that the PCM driver is bad because I could not see any fault in the wire. Am I right, am I thinking correctly about this? Thanks for your time! By the way, the problem came back when I moved the black lead to a good ground, which menas that the problem was in the 5v signal wire.
+Roger Sahagun It's too late to reply you on your particular problem. Hope you already solved it. I also experienced the same thing with some sensor in car. When loading the reference circuit using a test lamp drawing 100ma even, it was loading it and dropping to some value less than 5V. But when I connected the actual sensor and test using a voltmeter, it was sustaining at 5V constant. The reason for such case is the design of reference circuit inside the PCM. They designed the reference circuit with such impedance so that it may work when connect to the certain low current drawing sensor but will drop down if you will draw more current when testing with a loading device like you was using or a test lamp. Regards,
ur videos r 2 good. but i have 1 doubt that how electrons travels even after voltage drop i,e there energy is losed so how they complete there further path in circuit
You're welcome. If you're into car stereo, you can easily understand series and parallel resistor combinations using speakers. Two 8Ω speakers in series are 16Ω, but if you put them in parallel with two others in series, the total is 8Ω. And, the secret to scrambled eggs is low heat and whipped cream...
Amazing video. The only thing I don't understand is at the end... When you do a voltage drop on an incomplete circuit, shouldn't the meter read 0V? Because the circuit is open, and current can't flow. Not only that, but in volts mode, a DMM has over 1mil resistance, so current can't flow through that and bypass the open part of the circuit.
@earnamint Voltage drop is the only way to really, really know what's happening in a circuit. If we assume (correctly 99.9% of the time) that there's only one load per circuit, then nothing in the circuit other than the load should drop voltage. If the voltage drop across (ACROSS means (+) of the load to the (-) of the load) is full voltage then there can't be a wire problem - it has to be the load. If the voltage is less than full voltage, then the only cause must be added resistance.
I am confused about the voltage drops in a series circuit equaling the supplied voltage - for example, a wire has some small resistance, so isn't it technically a resistor, relatively speaking? and if that is so, if I connect a wire from a car batter directly from positive to negative terminal, would there be zero volts at the negative terminal? thanks
No - but it shouldn't be needed since the terminals on larger systems are exposed. That permits a regular voltage drop test without needing the LOADpro.
Hi Daniel, can you make a video solving this problem: which size copper conductor is required to supply a remote head rated at 7 A located 6.5 m from a 6-V emergency lighting unit. Will greatly appreciated.
Hi Daniel, thanks so much for your videos, they are very helpful. I'm just beginning electronics and this has clarified a concept that I've struggled with. I have one question that I haven't been able to work out elsewhere, if you don't mind. In your example with two resistors and one bulbs in a circuit, what factor or characteristic determines that the resistors each drop 5.5V and the bulb drops 1V? Or to put it another way, what would have to change for the resistors to each drop 2V, and the bulb to drop 8V? Thanks, Max.
First - why are you a Meanie? Second, this is where the math can help - because V÷Ω=A and V=AxΩ. Soooo - if A is constant in a series circuit - you multiply the A x Ω to get voltage drop. I REALLY want the formula to be written as VD=AxΩ, because that's what you're really doing. (1) What is the source voltage - V? (2) What is TOTAL resistance? --- Total Ω = R1 + R2 + R3 + etc..... (3) 1 and 2 will give you TOTAL amps (4) NOW - use that TOTAL amperage -A- with EACH resistance in the VD=A x Ω formula. OUTCOME: ALL Voltage Drops MUST add up to source voltage; each voltage drop will be proportional to its resistance. Try this - 24V. R1 = 6Ω, R2 = 4Ω, R3 = 2Ω. Let me know what you get.
Short answer is, screen names are hard. Long answer, it was a nickname I used when I used to play starcraft in highschool in the early 2000s with three friends, who used Eenie, Miney and Moe. Had I spent less time griefing people in starcraft I probably could have learned this stuff in class. As for your scenario, I got (1) 24 volts, (2) 12 ohms, (3): 2 amps, (4a) 12 VD on R1, (4b) 8 VD on R2, (4c) 4 VD on R3. 4a+4b+4c= 24 volts of drop across the circuit. I hope I've got that right. So I guess the proportion of voltage drop across each component could be described as a product of its resistance as a proportion of total circuit resistance?
Well, as long as you're not a "big meanie"... On to the problem: Perfect! Note that VD=AxΩ clearly states that the more resistance (Ω), the higher the drop. In mechanical terms, the harder it is for the current to flow, the more energy (volts) is consumed getting through - voltage drop (sort of) equals heat in resistors. A perfectly clean 0Ω circuit has 0V drop - while an open circuit (∞Ω) has a "full" drop. In a normal circuit with only one load component, that one load will consume 100% of voltage because that one load IS total Ω.
Awesome, thanks for the help. Your vids have provided a clarity on concepts I've been scratching my head over for a bit now. I'm definitely going to check out the rest of them while I learn more about practical electronics. Thanks again.
Meanie010 You're welcome. Try to visualize the current flow - it helps to "see" a mechanical picture in your head. Feel free to ask more questions. BTW - what I said about the voltage drop across the sole load is VERY crucial in understanding parallel circuits.
Hmmm. You're correct about wires having resistance (Ω), but it's hopefully very low. If you do as you suggest with a wire and a battery, technically you're correct, but because the resistance is nothing, amperage will be HUGE, and the wire will melt in milliseconds. Mathematically, the voltage drop is V=AΩ. If there's 0.0 Ω, then V=0.0. This is why any resistance in the wiring will cause heat. Rust will never blow a fuse - it will melt the fuse panel though.
The speed at which they're moving is different than the energy they possess. The assumption is that the stress on them from the like charges is what pushes them, but they're being pulled at the same time. Think of it like a car in a carwash. The car moves at a constant speed through the wash, but as it moves the dirt gradually washes away. It is vague - you'll just have to keep thinking about it.
What was the amperage sent into the circuit along with the 12.45 volts? Nobody covers this and it wo9uld help so we can determine how to predict amperage in the circuit given a certain amperage feed into the circuit along with the voltage. I need to step 24 volts @ 6 amps down to 12 volts @ 1 amp. But nobody's covering the topic of bumping down amperage.
Are those leads for sale? I ran into this problem at a friends house. Neutral in the transformer came loose in a storm and turned his house into a big series circuit. But you couldnt tell what was wrong till you turned something in the house on. The local utility company told him the problem was in his house because they tested the voltage with nothing in the house actually operating. Everything electronic in his house was fried. But the utilities paid to have it all replaced.
I understand what a voltage drop is and how to do calculations with them. However, I am struggling to understand something. If there is a load in a circuit then there is a voltage drop across that load. I don't understand how the current can be the same through this circuit when there is a voltage drop across the load. Current is the movement of electrons so how is it that the electrons after the load can be moving at the same speed (current being Q/t) if they have less energy (due to the drop).
wow this is one of the best vids on youtube i've seen. i was hoping you would explain why you read full voltage across an open (after the resistors) in a circuit, and you did! it's still a little fuzzy for me, and i guess that is because i don't quite understand the function of the meter. does it not close the circuit? is it not providing a load, and just measuring potential difference on each side of the open? that's the part that gets me. awesome vid, again.
Hi there, just a bit confused about what voltage drop is. My understanding is that voltage potential, similar to if you have a ball held in the air, has a certain potential energy and as it falls that potential energy transforms into kinetic energy at a uniform rate. My question is, what about voltage potential allows it to drop all of it's potential in the resistor. Why is the potential not lost uniformly, and how does it lose all it's potential through the resistor when after the resistor it still has "distance from the ground" which you would think would mean remaining "potential" or voltage. It would seem that the voltage's potential energy should drop uniformly based on the current relative position of the electrons in the cirucuit, assuming the gravity analogy makes sense. Any clarification would be appreciated! Thanks, B.
Braxton Ryback Hmmm - wow. Well, I'm not sure why you'r thinking it *doesn't* dissipate uniformly -- because it actually does. If you break open a 100Ω wirewound ceramic resistor (of which I have about a billion) and you unwrap the wire, the voltage in the middle will be half of source. As for the ground path, it *will* drop voltage if it has resistance. The V=AxΩ version of the Ohm's Law formula calculates voltage drop. So, if the wire has essentially 0Ω, then 1,000,000A x 0.0Ω = 0.0V. To me, the best explanation is that voltage drop is the energy that is transformed to another form as current passes through the resistance. Corrosion heats up when current flows through it, which robs energy from the component that subsequently operates poorly - because of a lack of adequate energy. Think it over and ask again...
I don't see the "ball in the air" example. I can't make it fit. The electrons are energized going into the resistance; they dissipate the energy in the resistance. The energy is converted into something else - heat, light, sound, etc. I prefer water buckets in a bucket line at a fire. Buckets have potential (to put out the fire). More water in each bucket is more potential. Speed of the buckets is amperage. The number of buckets going towards the fire is equal to the number going away (amps on positive and negative). Total volume of water over time (V x A) is wattage...
Thanks - the concepts are pretty easy, and as long as you focus on them before trying to worry about the math, it seems to be a bit easier to comprehend. I think "AP" should mean advanced PERFORMANCE - so keep working and let me know if you have any questions. Good luck.
The LOADpro can't "fry" the ECU. The maximum load is 40mA per volt and that means less than a 1/4 amp at 5V and 1/2 amp at 8V. There's no voltage on a sensor input so you wouldn't be using it. It's very safe. CAT gave it a part number. You're good. Thanks.
Be very careful, and make sure the leads on the resistors are covered. Personally, I'd use a 100-200W resistor to avoid any real cooling problems. This will work as long as the voltage (258VAC) remains essentially constant. Don't leave anything exposed. Where are you?
My calculations say you need a 57Ω resistor in series with the 200W load. V÷Ω=A. 258V ÷ 0.77A = 335Ω. 220V ÷ 335Ω= 0.65A. 258V - 220V = 38V. 38V ÷ .65A = 57Ω. 335Ω is the assumed resistance of your load, under load, so its current draw at 220V would be 0.65A. If you put the 57Ω resistor in series with your load, the resistor will drop 38V, leaving 220V for the original load. Kirchoff's Law. Assumes the load is resistive. Caution - the resistor will get hot. 38V x 0.65A = 25W.
But the socket was open with preceding resistors (1 and 2), it's just as you said, the resistor aren't doing anything because there is no flow. I guess it would be more suitable to ask what the meter does, does the meter not effectively close an otherwise open circuit?
Daniel Sullivan Its a wire to ecu, but I think its normal because I have another car thats the same and measures the same ohms. I was trying to figure out why I get link error when I try to to scan the honda civic 2005. I dont know exactly why I would get this error.
Daniel Sullivan I bought the powerprobe 4 to check if it had a bad wire on the obd2 port, but the eect3000 did not work on some pins on the obd2 port. I think I heard you talk about this before. As far as im concerned, im returning this powerprobe. Im going to try out the LoadPro instead.
Everything you said is technically correct - but don't forget I work in an environment where these tests aren't always easy to do. Theory is one thing, practice is another.
I know what voltage drop is but it took me a second to understand what you were getting at first I was totally confused it was explained to me in a different way but then the last few statements you made I felt like the little bulb in your test circuit the light in my head came back on! glad it did too! Thanks for clarifying
I would measure the voltage across the bulb, with it connected to the circuit. If the reading is showing less than source/battery voltage. I will know there a fault in the circuit wiring.
As long as there aren't any data transmission wires --- if so, those need to be twisted. As long as no copper is touching copper and you don't overfill the conduit, it shouldn't be problematic.
@7:30, you talk about a "circuit", but I don't see a circuit, or at least not one that looks "circular". Rather, the connection goes from a power source, through some switches, through a load, then out to ground. Can you please explain how this is a circuit? (It seems to me that this is a one way process, rather than being circular.) Thank you.
I thought voltage goes from Negative to Positive... Since atoms tend to want to be neutral. So whenever something is positively charged it draws electrons?
Thanks I guess? I think the point was made without a correction. BTW. Been using your method for a few months now as a second year HET. Increased my efficiency quite a bit. Only had one time where it didn't work. A light break/rear marker light had a bunch of corrosion on one of the pins in the jump harness connector and for some reason when loading the circuit with a resistor it didn't drop the voltage but on a 5 LED light it drops and cuts the light out. Was a one off thing and pretty crazy. btw, have a question. Had a higher amp diode for a branch of a smart sense alternator module. When testing the diode it read .5v one way and not the other meaning its good but when i connected it and tested the output voltage it read .01v. Why is that? The Diode works and it ended up being something different but i don't know why when i stuck my positive lead on the end of the connector and hooked my neg to a good ground it read .01v.
You can't have two true opposing theories. That's a contradiction. Either A) electrical current is going from negative to positive or B) positive to negative. It is proven to go in on direction... Or else diodes would not work properly. Now if you are referring to A/C that's Alternating. Not a switching of electron flow to positive to negative/negative to positive. I'm not trying to argue.
I don't have a perfect answer - it's really all conjecture on one level. You're correct that voltage and resistance determine amperage (Q/t) - remember that the measure of wattage (power) is a function of Volts and Amps. A higher voltage system means the electrons have a higher energy level to start with, so they're inclined to move more quickly as a result. It's one of those things I can visualize in my mind, but have difficulty explaining. Think very small and try to think mechanically.
Very good video and well explained examples or current flow and how circuit parts cause voltage drops in DC circuits.... But I must disagree with your statement about voltage drop in the wires of a DC circuit. I have worked with many DC to AC power inverters and have done lots of tests using heavy loads , drawing lots of amps from a DC power source to the inverter to run a electric heater for example. Now if the wires that connect to the DC power are too small/thin and you attempt to draw lots a amps through a small gage conductive wire, that will cause a fair amount of voltage to drop of DC current to the inverter causing the heater not to run or run very poorly and not fully operate. Now after increasing the size of the DC wires and also connecting more the one DC wire from the battery power source to the inverter provides a much larger/wider path for DC amps to flow with much less drop in DC volts when powering heavy loads and in some cases allow the heavy load circuit to run longer while drawing lots of amps to operate. As the heavy load runs the DC volts will slowly drop in the battery until it gets too low to run your load. I find that using very large gage wires and connecting more then 1 wire of Pos/neg connections will make the DC volts to drop much slower then say using a small AC wire. Also having more then 1 battery connected to the inverter also provides less voltage drops of DC power from the battery to the inverter while running a heavy load. Keep in mind that this is only for DC circuits using battery power sources, a power supply that runs off AC power will provide a constant voltage and amps of DC power regardless of wire size. But when using DC power only not from a AC power source supplying continues power of voltage. I found that I could not run heaters , cookers, motors , or anything that draws lots of amps to run for very long before my inverters would start screaming "low Voltage" , within 5 min the voltage was too low to keep anything running. by adding thicker wire and more of them allowed a far less drop in DC volts and I found I could run heavy loads up to 1/2 hour , sometimes longer depending on the charge in the batteries.....
It's not complicated - thicker wires have a lower resistance than thinner ones. The smaller wires you're using, the higher their resistance. That's all it is.
I really don't understand why voltage is dropped. If V = IR, then wouldn't a circuit need MORE voltage to push the current of electrons at a point of high resistance (a resistor)?
Voltage doesn't flow. Amperage flows - voltage drops. Voltage is present - amperage moves. Any absence of voltage anywhere means it's somewhere else - which is likely wrong.
I guess I should have worded that different,I'm getting 3 volts on my multimeter on the ground side of 1 injector and I've isolated the ground to rule out something else interfering with it.There's no short,no open and actually not enough resistance.
Right, there is no voltage drop. What I don't see is why there exists a potential difference between the two terminals of the light bulb, it's like the resistance is being neglected. So my question is: What if you replaced a resistor with a material high in resistivity, such as a wooden stick?. Would we still be able to see a potential difference of 12V? Elaborating further, can you just remove the resistors and still see 12 V? (air as "insulator")
ALSO --------- this calculation (57Ω) is ONLY good for the one situation you described. The calculation and resistor will be completely different in ANY OTHER application. This is very critical.
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The more and more I watch your videos the easier it gets thanks Dan.
True, but as a teacher I have to consider my students and what their interests and abilities are. I teach mechanics, and mechanics need a mechanical illustration. Since no one has actually ever seen voltage or current, I feel pretty confident in using the example I did. It's all observation of effects - not an observation of reality. Voltage is pressure anyway, and it doesn't move. Voltage pushes Amperage thru Resistance. It's all what you care to believe, and as long as it makes sense, use it.
Finally helpful information. You explained it WAY better then the CAT web based training I've been taking at school.
Great video! Voltage drop is key to electrical diagnostics, and you did an awesome job!
Awesome video(s). Your videos have helped me a lot with electrical fundamentals!
Very brisk, concise and clear presentation. Great job!
You answer my question in voltage drop in about 38seconds! Thank you sir!!!
Daniel, That is very consise teachingt. I needed a refresher to make sure I was interpreting the voltage drop across the ingnition switch on an old goldwing I am restoring. I am having a charing system problem and all the trouble shooting instructions result in good components. Then as I decided to look further I noticed the switched voltage at the regulator was about 1 volt low. Now its on to the search for a replacement. Thanks for the refresher!!!
very helpful look forward to future uploads!
Great video! It was very informative and helped me understand the material, I'm currently studying this in my AP Physics B class and this helped me understand voltage drop well.
It simply makes every voltage reading a voltage DROP reading. So, anytime you ever read voltage just push the button to load the circuit. If the wire is a sensor input, you can isolate it, put voltage on one end and read and load at the other. Several techs use it regularly on computer circuits - but they're pretty knowledgeable about their systems. I'm posting a video tomorrow about fuel level systems and you can simply read voltage and when you load the circuit, the gauge moves. Cool, huh?
Daniel once again a wonderful video i had from u!!!!!
Daniel once again a wonderful video i had from u
Great video, thanks.
badass explanation thank you
Thanks for the reply that was one of the better explanations I've come across. I've spent too much time pondering over this problem. I think I now need to find out what factors determine the speed of the charges. Perhaps that will help me understand, although I thought it was just the source voltage and load resistance that determined the speed.
Brilliant, thank you.
Good stuff here - I just found your loadpro and troubleshooting book on Amazon and ordered mine! Thanks!
just took my ASE L1 test and failed it. i forgot how to do voltage drops i wish i would've watched this before.. thanks for posting. will subscribe
Thanks...great video!!!
Hi Daniel, I work for a premium German automotive company and your videos are some of the most informative and concise I have ever seen. Your methods used in training work very well. I have some questions regarding the voltage drop method, is there any examples of circuits where this method does not/can not find faults or an example where this method should not be used? Is there any scenario where the 25ohm resistance in the leads could cause damage to control units?
Also, is it possible to order these leads in Asia?
Brilliant. George C.
Great Video. I am going to show my students at school this so it will help them understand a voltage drop better.
+Steven Webb Thanks. Glad to help. Use whatever you can and need. That's why I created it. if you're not aware of my book and LOADpro leads, contact Mary at Electronic Specialties (maryk@esitest.com) and tell her you're an educator. There's and educational price list. No meter company *really* gets the education market - we're (I'm) trying to focus on the training aspect of the business rather than the sales. Also - look this over: th-cam.com/video/DfMZxWXb8M8/w-d-xo.html (I don't make a ton of money - less than a couple of hundred dollars a month). I've been in the classroom about 27 years, and I'm all about simple yet effective. Call me if you have problems or questions.
+Daniel Sullivan I am going to tell some other Automotive instructors about you video's.
excellent video and train bought your tool works great just have to know how to use it and understand elect. have power probe and lots of other testers it is still good but yours works 100%
Hi everyone, I'm a proffesional tech who uses Dan's methods, and they work! His way of troubleshooting has made me a better tech. Check out his website, you won't be sorry. Keep on putting out more videos Dan, this is a great idea.
I was goofed up...until I saw this video! Thank You!
I loved this video ur are awesome man
You can't do a voltage drop on an open circuit. Current must be flowing for there to be a "drop". The drop is what the resistor is consuming and to consume there must be flow. The voltage across an open will be full voltage because all you're really doing is reading at the battery posts. The wires are connected to the battery, right? So, if you read across an open, the wires are extensions of the battery terminals.
Wow! Excellent!
great vid..thank you
Thank you SO much, Sir!
Thanks for the reply.I am a newbie at this .I understand ohms law .I have always wanted to learn how to diagnose faults (electrical) on my own vehicles but never had any training only read from books .The mechanical side i can do pretty good,changing brakes,timin belts ,head gaskets etc but this side really interests me .Just a little worried that i don't fry my ECU if i test the wrong way with the loadpro lol as i would love to get this and do some voltage drop testing on my car.
Thanks for the help
Daniel, Just want to commend you on the way you explain and draw out what you are explaining. Your awesome!!! Thanks for your time, please teach more. Generators, UPS's, Pulse Wit Mod, Rectifier, scr. Also how do you determine or calculate what component you will put a circuit. Say like, when do I put a resistor, where do I put a resistor, where and why would I put a capacitor after the resistor. Hope I make sense. But thanks a lot for you vids.
Great video thanks :)
Wow - teaching electricity and poetry at the same time. Cool.
Excellent
thanks.that helped!!
@Daniel Sullivan. I remember when I tried to watch your video when I was first starting to learn about electricity and skipped it after the first min. because everything was so fast. I've learned more since then (about a year ago) and was able to keep up this time. It would have been very helpful THEN if I would have just paused the video after every one of your points and wrote that point down or just memorized it and then moved on to the next one.
wow! best for my review...like it
Thanks, its understandable now. Some of the other videos are misleading. I'll find the resistance of my coils with a meter when I can pick one up from the store. I get it now. 24 volt power supply can work with a 4 ohm load to give 6 amps to the circuit. From the other 12vdc transformers, the coils must have been about 100 Ohms or less drawing more than their 0.1 amp rating.
Yes - but the videos don't pay the mortgage, so I've had to focus on training. I'm in North Dakota now and will be in Wyoming in 2 weeks. I have several planned - but timing is everything. Thanks for the comments though.
can you use the loadpro the same way on computer controlled cars ,say for checking for faults in wiring for lights ,fuel pump circuits etc without damaging the ecu .I am in the UK and thinking of getting one and your book to learn trouble shooting .Could you please let me know what you can test with this on modern computer controlled cars.Thanks Shaun.
i loved the video
i loved the video...
hi dan very good videos, pm trying to watch all,I have a question ,what is a crossed circuit? it is bad or is normal?
What does jump across 2 resistors mean?
Around 12:37 on the video, how did you read (A VOLT coming into the
bulb)? Sorry for all the questions, but I'm a beginner when it comes to
electrical and i'm trying to learn everything about this subject as much as I
can. :)
Hi dan have you got any new videos planned?? I and many others really enjoy and learn a lot from them.
Hey dan how do you get resistors do you buy them or you get them online
Sir,
Tnank you so much, i can use a 50 watts 57 ohm resistor to keep cool going. our domestic line voltage is around 250 to 260 i am facing frequent burn outs of electricals like bulbs etc, thanks once again.
Thanks Dan. I am assuming you refer to a regular voltage drop test using a carbon pile load tester for the larger cables.
No - if you just test the voltage drop using the actual component when current is flowing you're good.
What is the amperage required by the load?
How come in electrical distribution in a building wires must be upsized for long runs due to voltage drop?
The resistance of a material is proportional to its length
Hey Dan, I used your loadpro leads on a 5v signal wire and a ground wire going to a camshaft position sensor. When I loaded the circuit, the wire dropped to .118v. I took apart the tape in the wiring harness to inspect the wire for a lot of corrosion but I was not able to find any corrosion. This wire goes into the PCM, is it possible for the PCM driver to have too much resistance in it? I'm suspecting that the PCM driver is bad because I could not see any fault in the wire. Am I right, am I thinking correctly about this? Thanks for your time! By the way, the problem came back when I moved the black lead to a good ground, which menas that the problem was in the 5v signal wire.
+Roger Sahagun It's too late to reply you on your particular problem. Hope you already solved it. I also experienced the same thing with some sensor in car. When loading the reference circuit using a test lamp drawing 100ma even, it was loading it and dropping to some value less than 5V. But when I connected the actual sensor and test using a voltmeter, it was sustaining at 5V constant. The reason for such case is the design of reference circuit inside the PCM. They designed the reference circuit with such impedance so that it may work when connect to the certain low current drawing sensor but will drop down if you will draw more current when testing with a loading device like you was using or a test lamp. Regards,
ur videos r 2 good. but i have 1 doubt that how electrons travels even after voltage drop i,e there energy is losed so how they complete there further path in circuit
What' kind of voltage is that but thanks you helped me electrical conductivity
You're welcome. If you're into car stereo, you can easily understand series and parallel resistor combinations using speakers. Two 8Ω speakers in series are 16Ω, but if you put them in parallel with two others in series, the total is 8Ω. And, the secret to scrambled eggs is low heat and whipped cream...
Amazing video. The only thing I don't understand is at the end...
When you do a voltage drop on an incomplete circuit, shouldn't the meter read 0V?
Because the circuit is open, and current can't flow. Not only that, but in volts mode, a DMM has over 1mil resistance, so current can't flow through that and bypass the open part of the circuit.
@earnamint Voltage drop is the only way to really, really know what's happening in a circuit. If we assume (correctly 99.9% of the time) that there's only one load per circuit, then nothing in the circuit other than the load should drop voltage. If the voltage drop across (ACROSS means (+) of the load to the (-) of the load) is full voltage then there can't be a wire problem - it has to be the load. If the voltage is less than full voltage, then the only cause must be added resistance.
I am confused about the voltage drops in a series circuit equaling the supplied voltage - for example, a wire has some small resistance, so isn't it technically a resistor, relatively speaking? and if that is so, if I connect a wire from a car batter directly from positive to negative terminal, would there be zero volts at the negative terminal? thanks
What is the purple wire used for on the training board?
Dan can the LOADpro be used effectively on larger cables i.e. starter and/or alternator cables ranging from 2/0 to 4/0.
No - but it shouldn't be needed since the terminals on larger systems are exposed. That permits a regular voltage drop test without needing the LOADpro.
Hi Daniel, can you make a video solving this problem: which size copper conductor is required to supply a remote head rated at 7 A located 6.5 m from a 6-V emergency lighting unit. Will greatly appreciated.
Search online for a "wire size calculator". I can use a chart, but I can't do the math. My chart says 12AWG.
Hi Daniel, thanks so much for your videos, they are very helpful. I'm just beginning electronics and this has clarified a concept that I've struggled with. I have one question that I haven't been able to work out elsewhere, if you don't mind. In your example with two resistors and one bulbs in a circuit, what factor or characteristic determines that the resistors each drop 5.5V and the bulb drops 1V? Or to put it another way, what would have to change for the resistors to each drop 2V, and the bulb to drop 8V? Thanks, Max.
First - why are you a Meanie? Second, this is where the math can help - because V÷Ω=A and V=AxΩ. Soooo - if A is constant in a series circuit - you multiply the A x Ω to get voltage drop. I REALLY want the formula to be written as VD=AxΩ, because that's what you're really doing. (1) What is the source voltage - V? (2) What is TOTAL resistance? --- Total Ω = R1 + R2 + R3 + etc..... (3) 1 and 2 will give you TOTAL amps (4) NOW - use that TOTAL amperage -A- with EACH resistance in the VD=A x Ω formula. OUTCOME: ALL Voltage Drops MUST add up to source voltage; each voltage drop will be proportional to its resistance. Try this - 24V. R1 = 6Ω, R2 = 4Ω, R3 = 2Ω. Let me know what you get.
Short answer is, screen names are hard. Long answer, it was a nickname I used when I used to play starcraft in highschool in the early 2000s with three friends, who used Eenie, Miney and Moe. Had I spent less time griefing people in starcraft I probably could have learned this stuff in class. As for your scenario, I got (1) 24 volts, (2) 12 ohms, (3): 2 amps, (4a) 12 VD on R1, (4b) 8 VD on R2, (4c) 4 VD on R3. 4a+4b+4c= 24 volts of drop across the circuit. I hope I've got that right. So I guess the proportion of voltage drop across each component could be described as a product of its resistance as a proportion of total circuit resistance?
Well, as long as you're not a "big meanie"... On to the problem: Perfect! Note that VD=AxΩ clearly states that the more resistance (Ω), the higher the drop. In mechanical terms, the harder it is for the current to flow, the more energy (volts) is consumed getting through - voltage drop (sort of) equals heat in resistors. A perfectly clean 0Ω circuit has 0V drop - while an open circuit (∞Ω) has a "full" drop. In a normal circuit with only one load component, that one load will consume 100% of voltage because that one load IS total Ω.
Awesome, thanks for the help. Your vids have provided a clarity on concepts I've been scratching my head over for a bit now. I'm definitely going to check out the rest of them while I learn more about practical electronics. Thanks again.
Meanie010 You're welcome. Try to visualize the current flow - it helps to "see" a mechanical picture in your head. Feel free to ask more questions. BTW - what I said about the voltage drop across the sole load is VERY crucial in understanding parallel circuits.
Hmmm. You're correct about wires having resistance (Ω), but it's hopefully very low. If you do as you suggest with a wire and a battery, technically you're correct, but because the resistance is nothing, amperage will be HUGE, and the wire will melt in milliseconds. Mathematically, the voltage drop is V=AΩ. If there's 0.0 Ω, then V=0.0. This is why any resistance in the wiring will cause heat. Rust will never blow a fuse - it will melt the fuse panel though.
The speed at which they're moving is different than the energy they possess. The assumption is that the stress on them from the like charges is what pushes them, but they're being pulled at the same time. Think of it like a car in a carwash. The car moves at a constant speed through the wash, but as it moves the dirt gradually washes away. It is vague - you'll just have to keep thinking about it.
thanks sir. i have a if the voltage drops increases then the current flow will decrease?
What was the amperage sent into the circuit along with the 12.45 volts? Nobody covers this and it wo9uld help so we can determine how to predict amperage in the circuit given a certain amperage feed into the circuit along with the voltage. I need to step 24 volts @ 6 amps down to 12 volts @ 1 amp. But nobody's covering the topic of bumping down amperage.
Are those leads for sale? I ran into this problem at a friends house. Neutral in the transformer came loose in a storm and turned his house into a big series circuit. But you couldnt tell what was wrong till you turned something in the house on. The local utility company told him the problem was in his house because they tested the voltage with nothing in the house actually operating. Everything electronic in his house was fried. But the utilities paid to have it all replaced.
I understand what a voltage drop is and how to do calculations with them. However, I am struggling to understand something. If there is a load in a circuit then there is a voltage drop across that load. I don't understand how the current can be the same through this circuit when there is a voltage drop across the load. Current is the movement of electrons so how is it that the electrons after the load can be moving at the same speed (current being Q/t) if they have less energy (due to the drop).
thanks ,,
wow this is one of the best vids on youtube i've seen. i was hoping you would explain why you read full voltage across an open (after the resistors) in a circuit, and you did! it's still a little fuzzy for me, and i guess that is because i don't quite understand the function of the meter. does it not close the circuit? is it not providing a load, and just measuring potential difference on each side of the open? that's the part that gets me.
awesome vid, again.
Hi there, just a bit confused about what voltage drop is. My understanding is that voltage potential, similar to if you have a ball held in the air, has a certain potential energy and as it falls that potential energy transforms into kinetic energy at a uniform rate. My question is, what about voltage potential allows it to drop all of it's potential in the resistor. Why is the potential not lost uniformly, and how does it lose all it's potential through the resistor when after the resistor it still has "distance from the ground" which you would think would mean remaining "potential" or voltage. It would seem that the voltage's potential energy should drop uniformly based on the current relative position of the electrons in the cirucuit, assuming the gravity analogy makes sense. Any clarification would be appreciated! Thanks, B.
Braxton Ryback Hmmm - wow. Well, I'm not sure why you'r thinking it *doesn't* dissipate uniformly -- because it actually does. If you break open a 100Ω wirewound ceramic resistor (of which I have about a billion) and you unwrap the wire, the voltage in the middle will be half of source. As for the ground path, it *will* drop voltage if it has resistance. The V=AxΩ version of the Ohm's Law formula calculates voltage drop. So, if the wire has essentially 0Ω, then 1,000,000A x 0.0Ω = 0.0V. To me, the best explanation is that voltage drop is the energy that is transformed to another form as current passes through the resistance. Corrosion heats up when current flows through it, which robs energy from the component that subsequently operates poorly - because of a lack of adequate energy. Think it over and ask again...
I don't see the "ball in the air" example. I can't make it fit. The electrons are energized going into the resistance; they dissipate the energy in the resistance. The energy is converted into something else - heat, light, sound, etc. I prefer water buckets in a bucket line at a fire. Buckets have potential (to put out the fire). More water in each bucket is more potential. Speed of the buckets is amperage. The number of buckets going towards the fire is equal to the number going away (amps on positive and negative). Total volume of water over time (V x A) is wattage...
Thanks - the concepts are pretty easy, and as long as you focus on them before trying to worry about the math, it seems to be a bit easier to comprehend. I think "AP" should mean advanced PERFORMANCE - so keep working and let me know if you have any questions. Good luck.
The LOADpro can't "fry" the ECU. The maximum load is 40mA per volt and that means less than a 1/4 amp at 5V and 1/2 amp at 8V. There's no voltage on a sensor input so you wouldn't be using it. It's very safe. CAT gave it a part number. You're good. Thanks.
Be very careful, and make sure the leads on the resistors are covered. Personally, I'd use a 100-200W resistor to avoid any real cooling problems. This will work as long as the voltage (258VAC) remains essentially constant. Don't leave anything exposed. Where are you?
My calculations say you need a 57Ω resistor in series with the 200W load. V÷Ω=A. 258V ÷ 0.77A = 335Ω. 220V ÷ 335Ω= 0.65A. 258V - 220V = 38V. 38V ÷ .65A = 57Ω. 335Ω is the assumed resistance of your load, under load, so its current draw at 220V would be 0.65A. If you put the 57Ω resistor in series with your load, the resistor will drop 38V, leaving 220V for the original load. Kirchoff's Law. Assumes the load is resistive. Caution - the resistor will get hot. 38V x 0.65A = 25W.
What does jump across 2 resistors mean?
But the socket was open with preceding resistors (1 and 2), it's just as you said, the resistor aren't doing anything because there is no flow. I guess it would be more suitable to ask what the meter does, does the meter not effectively close an otherwise open circuit?
can a resistor cause a voltage drop?
Is that the circuitry for real life circuit
If a ground measures 2645 ohms, what does that mean when measuring an obd2 port pin?
Either corroded or a pull-up resistor. Was it the wire or inside the ECM?
Daniel Sullivan Its a wire to ecu, but I think its normal because I have another car thats the same and measures the same ohms. I was trying to figure out why I get link error when I try to to scan the honda civic 2005. I dont know exactly why I would get this error.
Daniel Sullivan I bought the powerprobe 4 to check if it had a bad wire on the obd2 port, but the eect3000 did not work on some pins on the obd2 port. I think I heard you talk about this before. As far as im concerned, im returning this powerprobe. Im going to try out the LoadPro instead.
You could use a current or resistant test to find the fault in the circuit.
Everything you said is technically correct - but don't forget I work in an environment where these tests aren't always easy to do. Theory is one thing, practice is another.
I know what voltage drop is but it took me a second to understand what you were getting at first I was totally confused it was explained to me in a different way but then the last few statements you made I felt like the little bulb in your test circuit the light in my head came back on! glad it did too! Thanks for clarifying
sir, amperage required by load is .77 when measured at 258 volts with 200 watt bulb
I would measure the voltage across the bulb, with it connected to the circuit. If the reading is showing less than source/battery voltage. I will know there a fault in the circuit wiring.
Don't forget the book:FET: fundamental electrical troubleshooting
As long as there aren't any data transmission wires --- if so, those need to be twisted. As long as no copper is touching copper and you don't overfill the conduit, it shouldn't be problematic.
@7:30, you talk about a "circuit", but I don't see a circuit, or at least not one that looks "circular". Rather, the connection goes from a power source, through some switches, through a load, then out to ground. Can you please explain how this is a circuit? (It seems to me that this is a one way process, rather than being circular.) Thank you.
w0mbles Sorry for the delay - the "circle" is from battery (+) to battery (-). The ground return path is the part you probably are not visualizing.
I thought voltage goes from Negative to Positive... Since atoms tend to want to be neutral. So whenever something is positively charged it draws electrons?
Current flow direction is completely useless in the real World. If you want to "argue" with someone, the electron moves from (-) to (+).
There are two theories on electrical current flow. In the end, both are true. Which is why its mute point.
Moot point.
Thanks I guess? I think the point was made without a correction.
BTW. Been using your method for a few months now as a second year HET. Increased my efficiency quite a bit. Only had one time where it didn't work. A light break/rear marker light had a bunch of corrosion on one of the pins in the jump harness connector and for some reason when loading the circuit with a resistor it didn't drop the voltage but on a 5 LED light it drops and cuts the light out. Was a one off thing and pretty crazy.
btw, have a question. Had a higher amp diode for a branch of a smart sense alternator module. When testing the diode it read .5v one way and not the other meaning its good but when i connected it and tested the output voltage it read .01v. Why is that? The Diode works and it ended up being something different but i don't know why when i stuck my positive lead on the end of the connector and hooked my neg to a good ground it read .01v.
You can't have two true opposing theories. That's a contradiction. Either A) electrical current is going from negative to positive or B) positive to negative. It is proven to go in on direction... Or else diodes would not work properly. Now if you are referring to A/C that's Alternating. Not a switching of electron flow to positive to negative/negative to positive. I'm not trying to argue.
Sir, can you please help me dropping 250 volts AC to 220 Volts AC at 200 watts load wahat resistor to be used WITH SOME THEORY
I don't have a perfect answer - it's really all conjecture on one level. You're correct that voltage and resistance determine amperage (Q/t) - remember that the measure of wattage (power) is a function of Volts and Amps. A higher voltage system means the electrons have a higher energy level to start with, so they're inclined to move more quickly as a result. It's one of those things I can visualize in my mind, but have difficulty explaining. Think very small and try to think mechanically.
Very good video and well explained examples or current flow and how circuit parts cause voltage drops in DC circuits.... But I must disagree with your statement about voltage drop in the wires of a DC circuit. I have worked with many DC to AC power inverters and have done lots of tests using heavy loads , drawing lots of amps from a DC power source to the inverter to run a electric heater for example. Now if the wires that connect to the DC power are too small/thin and you attempt to draw lots a amps through a small gage conductive wire, that will cause a fair amount of voltage to drop of DC current to the inverter causing the heater not to run or run very poorly and not fully operate. Now after increasing the size of the DC wires and also connecting more the one DC wire from the battery power source to the inverter provides a much larger/wider path for DC amps to flow with much less drop in DC volts when powering heavy loads and in some cases allow the heavy load circuit to run longer while drawing lots of amps to operate. As the heavy load runs the DC volts will slowly drop in the battery until it gets too low to run your load. I find that using very large gage wires and connecting more then 1 wire of Pos/neg connections will make the DC volts to drop much slower then say using a small AC wire. Also having more then 1 battery connected to the inverter also provides less voltage drops of DC power from the battery to the inverter while running a heavy load. Keep in mind that this is only for DC circuits using battery power sources, a power supply that runs off AC power will provide a constant voltage and amps of DC power regardless of wire size. But when using DC power only not from a AC power source supplying continues power of voltage. I found that I could not run heaters , cookers, motors , or anything that draws lots of amps to run for very long before my inverters would start screaming "low Voltage" , within 5 min the voltage was too low to keep anything running. by adding thicker wire and more of them allowed a far less drop in DC volts and I found I could run heavy loads up to 1/2 hour , sometimes longer depending on the charge in the batteries.....
It's not complicated - thicker wires have a lower resistance than thinner ones. The smaller wires you're using, the higher their resistance. That's all it is.
I really don't understand why voltage is dropped. If V = IR, then wouldn't a circuit need MORE voltage to push the current of electrons at a point of high resistance (a resistor)?
So,if I have a bad coil on a fuel injector it could let 3 volts flow through to ground.
Voltage doesn't flow. Amperage flows - voltage drops. Voltage is present - amperage moves. Any absence of voltage anywhere means it's somewhere else - which is likely wrong.
I guess I should have worded that different,I'm getting 3 volts on my multimeter on the ground side of 1 injector and I've isolated the ground to rule out something else interfering with it.There's no short,no open and actually not enough resistance.
It's electrically impossible for there to be voltage on ground without there being a problem on ground.
Right, there is no voltage drop. What I don't see is why there exists a potential difference between the two terminals of the light bulb, it's like the resistance is being neglected. So my question is: What if you replaced a resistor with a material high in resistivity, such as a wooden stick?. Would we still be able to see a potential difference of 12V? Elaborating further, can you just remove the resistors and still see 12 V? (air as "insulator")
ALSO --------- this calculation (57Ω) is ONLY good for the one situation you described. The calculation and resistor will be completely different in ANY OTHER application. This is very critical.
Fucking awesome!!!
YES secret to scrambled eggs
Only if current is flowing through it...