The Engineering Mindset if I break one bulb in the parallel circuit does the A reading for the other bulb go back to full flow? Or is it still split in half even tho the electrons are no longer flowing through the other circuit. In other words would one bulb get brighter the moment you broke the other or would it stay the same brightness. Probably a dumb question but this is a new found interest of mine thanks to you. You do a great job of explaining now I just need to commit all of these formulas to memory. Thanks a ton for the videos
I have a degree in mechanical engineering. Was never terribly interested in the electrical side of things because it was harder for me to grasp, so I never paid much mind to my electrical classes. Not to mention I had a very quiet professor from I think the Shandong region of China with a thick accent. She was a perfectly nice and pleasant person but when you can barely hear or understand the lecture in a big lecture hall you become even more disinterested than you already were. Fast forward to today and a large part of my job is troubleshooting and resolving electrical issues in an industrial environment. Finally having hands on experience and excellent one-on-one instruction from my boss (an electrical engineer) has *sparked* an interest in electrical systems and hobby circuitry. So here i am, refreshing myself to fill the gaps in my knowledge. I can wire up a 3-phase motor complete with relays wired into a PLC, but I couldn't remember the basic principles behind current in series vs parallel circuits. And it turns out mechanical systems are quite lovely when you know how to electrically automate them.
See our new video on how to build mechanical versions of electronic circuits? Watch here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Zv9Q7ih48Uc.html
Computer power section or TV power section will benefit from giving a video on how to operate the voltage and current in the LCD, LED TV's power section.
5:40 when you connect a 2nd 1 ohm lamp, the current increases to 3 amps. Why? We just said current = voltage / ohms. 1.5 volts / 2 ohms total = 0.75, not 3. Is it because current splits in parallel circuits and so each lamp is receiving 1.5 amps which *2 totals 3?
I came up with a possible better formula scheme to calculate total parallel resistance. First: R1. Next: 1/(1/R1 + 1/R2) = 1/(R1*R2 / R1+R2). Next: 1/(1/R1 + 1/R2 + 1/R3) = 1/[(R1*R2 / R1+R2) + 1/R3]. Next: 1/(1/R1 + 1/R2 + 1/R3 + 1/R4) = 1/[(R1*R2 / R1+R2) + (R3*R4 / R3+R4) ]. As the number of resistors in parallel increases, for an odd number of resistors you get multiply divided by sum, x number of pairs times, plus 1/R_odd. When the number of resistors in even, leave off the 1/R_odd. Did I do this right? Let me know!
At 7:00 Why do we not measure/place the multimeter after the merge of the 3 branches to measure the total current? Does the Batteries not just release another round of 1,5A of Current at the start before splitting? Can someone explain this to me?
3:05 Your example is 2 Amps, but the meter is showing 3A. This may confuse some. I apologize if I'm not the first to point this out, but with 318 comments, I wasn't going to read through them all.
Also worth noting that this example provides a figure for total resistance before the video has explained how this figure is derived, which is a little confusing on first viewing. Excellent series though, thanks!
I got a wireman exam coming up nowhere online gives u the refreshing of basic electrical work like you do thank you for these videos god bless you 🙏🏻🙏🏻🙏🏻
As someone who struggles to learn things "just because", I really appreciate your explanation of why we calculate total resistance in a parallel circuit that way. Will definitely help make it stick now I can actually rationalise it. Thank you 👍
Paul: At 3.22 the amp meter shows 3A. Although the math is correct. This error was put into the graphic to see if we are (paying attention) learning anything.
I came here looking for this. I dont know how he got the numbers for the formula either. I suppose the formula itself is important. I got that the voltage is 36
@@andrewangerer1399 I got confused at this part too. He does write in the video description this was an error and that the multimeter should read "2A" not "3A" I'm still confused as to why the resistance is 3 ohms
KCL Kirchhoff's current law. Wonderful explanation. Paul: A little too fast on the excellent graphic and voice over. Perhaps trying to squeeze everything into the 16 minutes.
Up until a few weeks ago my electrical understanding extended to being able to wire a plug. With your series of excellent videos you have expanded my understanding in this area. Thank you very much for sharing your knowledge.
for anyone wondering how he got 3 ohms for total parallel resistance at 3:05.... (use this formula.. this works for exactly with 2 resistors) Rt=(R1∗R2)/(R1+R2) R1 = 6 ohms R2 = 6 ohms (easy math, subsitute and solve!) Rt=(6 ohms * 6 ohms)/(6 ohms+6 ohms) Rt=36 ohms/12 ohms Rt= 3 ohms formual for more than 2 resistors is... Rt = 1/(1/R1 + 1/R2 + 1/R3 + etc....) 😸
I gotta ask, do you know any Canadians? This is literally the next module (the one we're doing tomorrow!) in my electrical course. I've watched a lot of your previous videos but I find it funny on the timing for this. Anyway, I shot you $10/CAD on Paypal so hopefully that's enough to buy yourself a coffee. Thanks for making this series! I'm a very visual learner & I haven't found any electrical series on RU-vid that can hold a candle to yours!
I do not understand the example on 3:23. The illustration circuit shows that current is 3A and resistance is 6 Ohms per resistor how is it that when calculating voltage you are using values that are not in the illustrated circuit?
Thank you for these videos. A few weeks ago I knew NOTHING and just wanted to learn some basic circuitry for low voltage DIY projects for gardening, automation, etc. You helped me grasp all of this and realize electronics aren’t just “magic”. Thanks again! :)
You comment is in the front just below the video it showing on screen not in comment section i didn't understand what you mean but after my calculations and when see wrong answer on screen than understand 🥺🖤
god, I love the way you explain things... sometimes I watch videos from different sources and got more confused than before the watch. It is apparent that you put the work into explaining things in the clearest way possible. Thanks :)
I am an amateur, I can answer as much as I learn from your lesson. Answer 1 - 1.02 Ohm... Answer 2 - R1: 10 Ohm and 0.6 A ; R2: 15 Ohm and 0.4 A ; R3: 4 Ohm and 1.5 A. Correct?......
3:15 I’m not quite understanding where you’re getting 2A and 3 ohms from. But if I base this off the diagram Voltage = Current x Resistance Which means base on the numbers in the diagram, the voltage should be 18V 18V = 3A x 6 ohms If I’m wrong about that then the only other answer I can think of is 36V = 3A x 12 ohms
I have a question. At 5:35 a battery at 1.5 V, and a total resistance of 2, equals 3.0A. At 6:12 a battery at 1.5 V, and a total resistance of 3, equals 2.25A. At 7:07 a battery at 1.5 V, and a total resistance of 3, equals 4.5A. At 7:07, you have the same setup(a total resistance of 3), but this time it equals 4.5A, instead of 2.25A. How can a total resistance of 3, with 2 different circuits, equal different Amperage? I see the equations, and I understand the math, but you can't tell me that the numbers stay the same(total resistance, and total voltage), but then you add a lightbulb, and it magically increases the amperage.
OK, after some more research I found a more intuitive answer… Imagine the following We start with a simple circuit (no branch) and a resistance of 1Ohm and 1.5V. Clearly the total current should be: 1.5V/1 Ohm = 1.5A. Now we add a new branch (lightbulb with 1 Ohm). Instead adding the Ohms, calculate the same way as in the first case (only one branch). As every branch gets the same voltage (as there are consumers before the bulb, the „full“ volt is present on each branch). So „before“ ach lightbulb we have 1.5V and 1 Ohm, thus I = U / R = 1.5 V / 1 Ohm = 1.5A „before/after“ each lightbulb. As every branch has/needs 1.5A, we have a total of 3A. If we would add another branch (3 branches with lightbulb with 1 Ohm) this one would also „need“ 1.5A, resulting in 4.5A in total. No matter how many branches/lightbulbs we add, the voltage on each branch is the same, so is the amps. I like to think about ampere as workers who transport the energy (actually ampere are a bulk of electrons, so maybe not that false?). If we add more branches we need more workers, as the total number of workers have to be split across all branches. The total resistance can be calculated with R = U / I and as every branch adds more ampere the total resistance decreases. Maybe one can think of this as the „effectivity“… If we imagine we have only one branch with 2 Ohms we get some numbers, 100% current „flows“ through this branch. If we add another branch with 1 Ohm, some current of that previously 100% will flow through this branch… The percentage will follow the distribution of the resistances, in this case 33% in the 2 Ohm branch, 66% in the 1 Ohm branch. So the „effectivity“ of the 2 Ohms is only 33%, thus the total resistance decreases (compared to only one branch). But how do we get these numbers 33% and 66%? Let’s make it more complicated to see, if it works with also with „not synthetic“ numbers: 15 Ohm, 50 Ohm, 150 Ohm (3 branches). A first idea could be: add everything to 165 Ohm and then 100/165 = ~0.6 (~60%), 50/165 = ~0,3(~30%) and 10/165 = ~0.06(~6%) (should add up to 100%). No matter what these numbers are a bit misleading as 100 Ohm should lead to far fewer current than 15 Ohm… Thus this cannot be right. So if 100 Ohm leads to less current than 15 Ohm, the percentage of 15 Ohm must be higher than the percentage of 100 Ohm. Lets try 1/100Ohm = 0,01, 1/50 Ohm = 0,02, 1/15 Ohm = 0,066 (from Ohms law with 1 volts). It s seems they already add up to 1 (or 100%) but they don’t (imagine we would remove the 15 Ohms, we would get the same numbers but would definitely not add to 1). Instead we calculate the percentage of these numbers… add them up (1/100 + 1/50 + 1/15) and calculate the percentage if 1/100, 1/50 and 1/15 relative to the sum: (1/15) / (1/100 + 1/50 + 1/15) = 0.6896551724 -> ~68% (1/50) / (1/100 + 1/50 + 1/15) = 0,2068965517 -> ~20% (1/100) / (1/100 + 1/50 + 1/15) = 0,1034482759 -> ~10% these will definitely add up to 1 Now we have the percentages of how much current will flow through each branch (depending on the relative resistance). A small test: we pick 225V, which gives us 225/100+225/50+225/15 = 21,75 Amps in total (each brach would be 225V/100 Ohm = 2,25A, 225/50 = 4,5A and 225/15 = 15A). Actually we can achieve the same thing with our percentages (if we somehow knew the total amps)… 21.75 * 0.6896551724 = 14.9999999997 A 21.75 * 0,2068965517 = 4,4999999995 A 21.75 * 0,1034482759 = 2,2500000008 A The percentages seem to work and give us the „amount“ of current „flowing“ through. We can see that we have 100 Ohms but they are only „applied“ to 10% of the current which is why it is less „effective“. Thus, the total resistance gets lower because the resistance is „applied“ to less current. This thinking only works for a fixed number of branches (or a fixed number of amps)… As you can see the amps increase if more branches are added… then we need workers/amps as we have more branches. Again, I’m not a noob but this way it makes a bit more sense and maybe this helps somehow ;) OLD: I’m a complete noob but according to 11:40 the total resistance is calculated differently in parallel circuits. For 1. 6:12 I get a total resistance of 1/( 1/2 + 1/1 ) = 0.6666666667 Ohm For 2. 7:07 I get 1/( 1/1 + 1/1 + 1/1 ) = 0.3333333333 Ohm If we plug this in I = U / R for 1. I = 1.5V / 0.6666666667 Ohm = 2.25A and for 2. I = 1.5V / 0.3333333333 Ohm = 4.5A It seems counterintuitive but I guess this is how current works?
awesome. Just awesome. You should make an online course for electronics. I would buy it. Also...the soldering kit on amazon is not available. Recommend another
Never a fan of the line "you will never need this part of the explanation". Need is subjective - my need is determined more by being curious. Some people's need will be determined by passing an exam. Etc.
When you're saying with these parallel circuits, are the multimeters just seeing what is being drawn form the battery so the voltage from the battery is 4.5A drawn
gracias mi rey se me aclararon varias dudas de el como funciona la electricidad, incluso sobre la importancia de los circuitos en serie (en mi carrera no los entendí, ni mucho menos su importancia y aplicación)
I just dont understand the need for resistors. Ive seen people connecting multiple led lights on one circuit, with each led light having their own wire connections to add on to that one dc fuse wire on the dc panel?Even adding on a switch or dimmer and it all turns on the lights at full brightness. Pls explain.
If converting parallel resistances into conductance, adding them and converting them back give us total resistance in parallel so why does only adding resistance in parallel not doing the job?
How about we change one of the resistor to inductor, in the inductor video, you said current don't go through inductor first but later all go though it, seems outside law of parallel circuits. 🤔
I really found this video very helpful. I have been through so many videos but this really helps to know much more easier for Series connection and Parallel Connection, and their advantages and disadvantages. I have fixed spots lights in series connection and the first light is very bright and the rest are dull and flickering. Then I have been through this video and connected spot lights in parallel connection. Now it's really cool. I must really thank you, you made my day easy. All the best :)
Your videos are helping me get an intuition for electricity and electronics and I really needed this since I started a job in embedded development and I really wanted to be able to understand the PCB designs and why certain things are done the way they are. Amazing content, I have subscribed and am binging like crazy!
Conventional current flow is def easier to understand. Still can't wrap my head how current flows from the negative, but in a car the negative terminal is connected to the chassis.
1st example: 3 amp on the multimeter and (2) 6ohm resistors and we are told that its obviously 2ix3r=6e. confusing. great illustration but there are a few of these inconsistencies. Lord knows I'm not perfect but..
Corrections are shown in the video description. I try to ensure no errors get published but some get through, unfortunately. There's not a single engineering book you can buy without errors in. You have to pay for the book and pay to see the corrections, but this video is free and the corrections are free too.
Hey brother , can you please make some stuffs for Mechanical Engineering as well. It would be a very great of you man. By the way You just won my heart through these types' of videos Wish you could make some vids on M.E. too.
Thank you for this video, And also I've learnt so much from this channel... I've seen the comments about your mistakes in this video... I want to say, We all are imperfect and make lots of mistakes and learn from that.... Ans no.1). The total resistance of the circuit is 1.01 ohm Ans no.2). Current on R2 is 0.4 Amp Current on R3 is 1.5 Amp Resistance on R3 is 4 Ohm
Great video, thanks. But. I don’t understand how placing two 1.5v batteries in series works (how does it double). If I no current flow can occur from the negative terminal to the positive terminal, then how does connecting the positive terminal to the negative terminal of a different battery work? ...shouldn’t the electron difference between the touching terminals balance out immediately (like shorting a single battery with just a piece of wire)?
Great video and I fully understand the explanation. But what I can't get my head around is how in domestic lighting and plug socket circuits, the voltage supply is 230v (UK) so why is it not blowing light bulbs up when from what I've seen all that's resisting the flow is a mere 10 ohms per bulb. Could you do a video on typical measurements, figures, resistances, etc. in domestic circuits please?
All my questions about WHY that i found no answer are solved in your channel. Visual examples + WHY explanations are top quality content, you make people learn for real
studying for a med entrance exam after several years of not studying physics. Your videos have been the most helpful so far for electricity. Thank you so much, I'm very grateful !
At 3:12 why issit said that the current is 2 A and the total resistance is 3 ohms when the multimeter states that it’s 3A and each of the resistor is 6ohms?
