Hey, thanks for the video! I have a question maybe you could help So I'm in a foundation course in Canada. Not doing too great and want to be really prepared for my final in 30 days. If I asked my teacher for one now he would tell me it's not fair to give to just one student and would tell me to wait until the last 2 weeks when he will with everyone ( weird I know). Anyways all the practice tests I see are for red seal exams... the CSA one looks intriguing but again I'm only in foundations right now. I struggle most with combination circuits, edison 3 wires and code.
Great video. Very helpful. But with respect to the final calculation regarding the broken neutral, the voltage drop across the 60.3 ohm load is 200.136 V, not 199.793 V. 240 V - 0.332 V - 39.231 V - 200.136 V - 0.332 V = 0 (0.031)
If you use the Kirchoff voltage method yes, it is a slight difference from the ohms law current x resistance method, but we are looking at a very minor difference of about .2% deviation. Both answers can mathematically be proved as correct using different laws.
Respected sir, I like your example may appreciate, Then, solve the problem to move, Line drops are very rare to show as for, "Volts and amphers", May thinking right way to guide these example, Thanking you, Guide,
Interesting to see how removing the neutral actually reduced the voltage drop across some components, even though other components get more than their share.
GREAT VIDEO...IF I UNDERSTAND CORRECTLY, THE FOLLOWING: And that is why you call this an Edison circuit (DC) rather than a Tesla (AC) circuit. I am assuming you are 'freezing' the current in time and looking at the "pseudo-DC" situation during the peak AC flow in order to apply Kirchhoff's law for your analysis.
@@schulerruler great video, most in depth i found so far. I had a broken neutral on the utilities cable into the house and observed basically what you described. But still cant understand one thing, if there is a house ground where the broken neutral was in your diagram (between the two loads), wouldnt it bring the voltage total back to 120v on each line? why is it I could still get >120v on the lines.
@@fobbywillie the whole 240v circuit is from utility transformer, so if the neutral is broken anywhere between utility and the panel utilization, you will read the imbalance in voltages like voltage drops in a 240v series circuit. It isn't the connection to earth at the neutral that stabilizes the 120v, but the connection on the neutral that goes back to the centre tap on the 240v transformer winding.
So in parallel wiring, you just consider the wire resistance and in a series circuit, you add the wire resistance plus the load resistance? It is bewildering to me how the numbers are played with. In both system voltage drop equal voltage input. So what is the problem? I guess the cleverness is in series we add the load's resistance but in parallel which really is a series all by its own self ( mini-series if you take the wire as one load and the 10 A as another load) and the other below circuit just the same. Two separate series circuit that is parallel to each other. The first two mini-series is supplied with 120V each and the second one big series 240 volts for 4 loads. 2 wires and 10 and 2 amp loads. And how come we are not allocating a resistance value to the wire that is in between the two loads in both situations? And I am still to wrap my head around how in the neutral wire 8A goes back in, a free bonus. I don't know how it works. It will a while for me to get a mental picture of why it works like this. I assume there is something going on with the way the second coil is wired as well, maybe?
I didnt answer because sometimes comments slip through the cracks. To answer the original question, I agree with you. My opinion is there is no such thing as a true parallel circuit because as you mentioned, the wires in series will take a voltage drop. We need to understand the laws of parallel though, to move forward. Kind of like a unicorn. Everyone knows what it is, but it doesn't really exist. Many things in electrical theory are just that, theoretical. But we need to understand the theory to move forward with new concepts.
You had your polarity markings on both sides of the neutral as the same and I don’t believe that’s correct. Polarity goes clockwise on top circuit and it goes clockwise on bottom circuit. Which is correct. That would make polarity on the neutral opposite each other top and bottom. Top circuit has 10a returning on neutral to winding and 2a leaving winding to load on the neutral. The unbalanced of 8a as a residual current is due to 2a in one polarity being subtracted from 10a in the opposite polarity. 0v neutral can have currents of different angles flow.
Bob, when I first read your comment I thought you were correct, but here is why it is not: First of all, if you check his work with Kirchoffs voltage law, it works. If you check your work done the way you suggest it is wrong. The reason is that the difference between the two opposing currents was already determined at 2:08 and so there is a net flow of 8 amps in one direction. Therefore polarity will be according to the net flow.
@@spruce_goose5169 How can total voltage drop be greater than source voltage? You are of the understanding that current only flows one way on neutral based on one polarity of that neutral. My understanding is it can flow both ways based on different polarity on neutral conductor. If polarity were opposite on bottom part of neutral for bottom circuit it works.
@@bobbrumley3964 I don't understand what you mean regarding voltage drop being greater than source voltage. At 3:34 he states that the voltage drop is 0.8v. Voltage source is 120v. Can you explain what you mean? The current flow is the result of the NET current. When you say it can flow 'both ways' what you should be thinking is that the DIFFERENCE between the two opposing currents will be applied for a net flow in one direction (cycling 60hz). I also don't know what you mean by 'If polarity were opposite on bottom part of neutral for bottom circuit it works.' What works? Kirchoffs voltage law certainly doesn't (not around the outside).
@@bobbrumley3964 Nevermind I understand what you are saying about voltage drop. You are adding 120.6+0.8+0.2=121.6 and wondering how that can be larger than 120, correct? It's because the 0.8 is opposite polarity so it's subtracted, not added (I guess that's the sticking point here isn't it). If you check Kirchoffs law around the outer perimeter (the 240v) hopefully you can at least see that that doesn't work if you swapped the polarity on the 0.8 drop as you wish? It would give you 119 for the load drop, and 119+118.2+1+0.2= 238.4 (exactly 1.6 volts shy of 240 which is 0.8X2 (because reversing it hits you twice in the count). Just thinking about this situation intuitively: take the good ol analogy of water. If you have a single pipe with competing pressures on either side, with you get distinct currents, or a net flow according to the difference in pressure? (assume pressure equally applied to pipe ends and not giant pipes with jets of pressure that allows for turbulence since that breaks down the analogy).
@@spruce_goose5169 I appreciate this conversation. If I were to switch open my 10a load wouldn’t the voltage drop change across the 2a load according to your understanding. I believe it would and therefore the neutral would in a small way be floating. 120.6 vs 119.6 drop. That is not the way it works.
We have learnt it in the class before but never been clear to me till I watched this 15 minutes video. You are great 👌
How is this guy so good at writing backwards?
I'm guessing it's mirrored
Reviewing for my Journeymans Red seal exam. this has been helpful, thank you.
Hey, thanks for the video! I have a question maybe you could help
So I'm in a foundation course in Canada. Not doing too great and want to be really prepared for my final in 30 days. If I asked my teacher for one now he would tell me it's not fair to give to just one student and would tell me to wait until the last 2 weeks when he will with everyone ( weird I know). Anyways all the practice tests I see are for red seal exams... the CSA one looks intriguing but again I'm only in foundations right now. I struggle most with combination circuits, edison 3 wires and code.
Great video. Very helpful. But with respect to the final calculation regarding the broken neutral, the voltage drop across the 60.3 ohm load is 200.136 V, not 199.793 V.
240 V - 0.332 V - 39.231 V - 200.136 V - 0.332 V = 0 (0.031)
If you use the Kirchoff voltage method yes, it is a slight difference from the ohms law current x resistance method, but we are looking at a very minor difference of about
.2% deviation. Both answers can mathematically be proved as correct using different laws.
Respected sir, I like your example may appreciate,
Then, solve the problem to move,
Line drops are very rare to show as for,
"Volts and amphers",
May thinking right way to guide these example,
Thanking you,
Guide,
Interesting to see how removing the neutral actually reduced the voltage drop across some components, even though other components get more than their share.
beautifully written and explained simplified it for made it understandable
Excellent! Thank you.
Love these videos. Very well done 👍
Thank you for the encouraging words!
GREAT VIDEO...IF I UNDERSTAND CORRECTLY, THE FOLLOWING: And that is why you call this an Edison circuit (DC) rather than a Tesla (AC) circuit.
I am assuming you are 'freezing' the current in time and looking at the "pseudo-DC" situation during the peak AC flow in order to apply Kirchhoff's law for your analysis.
That's correct. Though the theory holds as the current alternates, just the neutral current would change direction with the polarity alternation.
@@schulerruler great video, most in depth i found so far. I had a broken neutral on the utilities cable into the house and observed basically what you described. But still cant understand one thing, if there is a house ground where the broken neutral was in your diagram (between the two loads), wouldnt it bring the voltage total back to 120v on each line? why is it I could still get >120v on the lines.
@@fobbywillie the whole 240v circuit is from utility transformer, so if the neutral is broken anywhere between utility and the panel utilization, you will read the imbalance in voltages like voltage drops in a 240v series circuit. It isn't the connection to earth at the neutral that stabilizes the 120v, but the connection on the neutral that goes back to the centre tap on the 240v transformer winding.
Great presentation.
you're my spirit animal
Haha thanks!
So in parallel wiring, you just consider the wire resistance and in a series circuit, you add the wire resistance plus the load resistance? It is bewildering to me how the numbers are played with. In both system voltage drop equal voltage input. So what is the problem?
I guess the cleverness is in series we add the load's resistance but in parallel which really is a series all by its own self ( mini-series if you take the wire as one load and the 10 A as another load) and the other below circuit just the same. Two separate series circuit that is parallel to each other. The first two mini-series is supplied with 120V each and the second one big series 240 volts for 4 loads. 2 wires and 10 and 2 amp loads.
And how come we are not allocating a resistance value to the wire that is in between the two loads in both situations?
And I am still to wrap my head around how in the neutral wire 8A goes back in, a free bonus.
I don't know how it works. It will a while for me to get a mental picture of why it works like this. I assume there is something going on with the way the second coil is wired as well, maybe?
I didnt answer because sometimes comments slip through the cracks.
To answer the original question, I agree with you. My opinion is there is no such thing as a true parallel circuit because as you mentioned, the wires in series will take a voltage drop.
We need to understand the laws of parallel though, to move forward. Kind of like a unicorn. Everyone knows what it is, but it doesn't really exist.
Many things in electrical theory are just that, theoretical. But we need to understand the theory to move forward with new concepts.
i just subscribed are you realy writeing backwords?
I wish I was that good! It's an inverted image.
@@schulerruler
I was wondering lol
Awesome 👍
Excellent!!!
Thanks!
Thanks you!!!!!!
If the 60ohm is an incandescent 120V bulb... likely poof
You had your polarity markings on both sides of the neutral as the same and I don’t believe that’s correct. Polarity goes clockwise on top circuit and it goes clockwise on bottom circuit. Which is correct. That would make polarity on the neutral opposite each other top and bottom. Top circuit has 10a returning on neutral to winding and 2a leaving winding to load on the neutral. The unbalanced of 8a as a residual current is due to 2a in one polarity being subtracted from 10a in the opposite polarity. 0v neutral can have currents of different angles flow.
Bob, when I first read your comment I thought you were correct, but here is why it is not:
First of all, if you check his work with Kirchoffs voltage law, it works. If you check your work done the way you suggest it is wrong.
The reason is that the difference between the two opposing currents was already determined at 2:08 and so there is a net flow of 8 amps in one direction. Therefore polarity will be according to the net flow.
@@spruce_goose5169 How can total voltage drop be greater than source voltage? You are of the understanding that current only flows one way on neutral based on one polarity of that neutral. My understanding is it can flow both ways based on different polarity on neutral conductor. If polarity were opposite on bottom part of neutral for bottom circuit it works.
@@bobbrumley3964 I don't understand what you mean regarding voltage drop being greater than source voltage. At 3:34 he states that the voltage drop is 0.8v. Voltage source is 120v. Can you explain what you mean?
The current flow is the result of the NET current. When you say it can flow 'both ways' what you should be thinking is that the DIFFERENCE between the two opposing currents will be applied for a net flow in one direction (cycling 60hz).
I also don't know what you mean by 'If polarity were opposite on bottom part of neutral for bottom circuit it works.'
What works? Kirchoffs voltage law certainly doesn't (not around the outside).
@@bobbrumley3964 Nevermind I understand what you are saying about voltage drop. You are adding 120.6+0.8+0.2=121.6 and wondering how that can be larger than 120, correct?
It's because the 0.8 is opposite polarity so it's subtracted, not added (I guess that's the sticking point here isn't it). If you check Kirchoffs law around the outer perimeter (the 240v) hopefully you can at least see that that doesn't work if you swapped the polarity on the 0.8 drop as you wish? It would give you 119 for the load drop, and 119+118.2+1+0.2= 238.4 (exactly 1.6 volts shy of 240 which is 0.8X2 (because reversing it hits you twice in the count).
Just thinking about this situation intuitively: take the good ol analogy of water. If you have a single pipe with competing pressures on either side, with you get distinct currents, or a net flow according to the difference in pressure? (assume pressure equally applied to pipe ends and not giant pipes with jets of pressure that allows for turbulence since that breaks down the analogy).
@@spruce_goose5169 I appreciate this conversation. If I were to switch open my 10a load wouldn’t the voltage drop change across the 2a load according to your understanding. I believe it would and therefore the neutral would in a small way be floating. 120.6 vs 119.6 drop. That is not the way it works.