Voltage drop / Ohms law / DC equations

[xformer]

Can someone shed some light on why this is?

http://www.youtube.com/watch?v=7pjLaGzEGGM&feature=youtu.be

Does the Voltage Drop Formula hold true in all instances or only for non looped circuits?

 

[Gary Glaenzer]

Voltage drop / Ohms law / DC equations

Can someone shed some light on why this is?

Can someone shed some light in what this guy is trying to demonstrate ?

Thanks

Gary

 

[GrayHair]

Can't see well or concentrate right now (allergies) but some things stand out:

1. His resistance and voltage appear unrelated since he measures from different points; i.e. different circuits.
2. Left and right circuits are equal and fed in parallel. Right now, I'd say that adding the two 1K "jumpers" during voltage measurement would have no effect on the circuit.

Would have to try to draw it out to analyze it and at the moment, that is beyond me.

 

[ATSman]

He is adding resistors, not jumper wires to connect the 2 parallel circuits which turns out to be a more complex elec circuit than he is portraying. My guess it that the resistance of these "jumpers" is so high (minimal current flow) compared to the resistance of the individual parallel circuits that the measured voltage would not change.
He needs to use jumper wires (o ohms) to prove his point and ohm's law. Also, I only see the left side circuit fed from the battery. He does not show how the right side circuit is being fed or where he has the meter connected.
Maybe GAR can chime in and explain it further.

 

[gar]

151027-1528 EDT

ATSman:

I have no idea what this guy was trying to prove. There were no clear definitions, equations, or statement of the problem. No circuit diagram.

The guy has some heresay information and is clearly ignorant of circuit analysis, and does not know how to define the problem he wants to address. His presentation skills are poor. He is hyper and one should not hand hold a video camera for this presentation.

He is a "snake oil" salesman.

.

 

[GoldDigger]

I think I see exactly what he is doing and why he is confused. He has two identical load circuits, both connected to the same supply. The two points that he is bridging are therefore at the same voltage, so bridging them has no effect on system currents and voltages.
His problem is that when he measures the lower resistance back to the supply he does not take into account that that resistance is carrying twice the current! If the right side circuit had not included any loads he would have seen the voltage drop go down by 1/2.

 

[GrayHair]

I think I see exactly what he is doing and why he is confused. He has two identical load circuits, both connected to the same supply. The two points that he is bridging are therefore at the same voltage, so bridging them has no effect on system currents and voltages.
His problem is that when he measures the lower resistance back to the supply he does not take into account that that resistance is carrying twice the current! If the right side circuit had not included any loads he would have seen the voltage drop go down by 1/2.

Agree with GoldDigger. The guy hasn't done (or maybe doesn;t know how to do) a circuit analysis.