I don't understand several things in the potentiometric voltmeter:

1- if the AVS is providing 0V while the main voltage source is providing > 0v
would the null detector move? would current pass through the AVS?

2-how does providing equal voltage stops current from flowing through the detector?

I appreciate your help.

 

1. In this case, 0 really is the absence of voltage. So less than 0 is not meaningful, unless you intend to say "a negative voltage".

2. It takes a difference in potential to move electrons (current). Two point that have the identical voltage won't cause current flow if a conductor is placed between them. That is why the meter shows a null, meaning no current.

 

This is how i think current should flow(as seen in the attachment).....my main issue here is lack of understanding of how things go when we have two voltage sources.

1- Dear "beenthere", I assumed that as long as there is a voltage/(attraction for electrons) current should flow toward that source. In this case we have two voltage sources which are attracting electrons and i assumed that both would drive electrons toward their respective sources.

2- Now lets assume that both leads were put on the circuit under test but without any voltage from the voltmeter voltage source, would we have a parallel circuit or is it that current can't pass through voltage sources the way it would pass through a resistor.

Thanks your work!

 

For both 1 & 2 - there must be a difference in potential between two points to drive current. The "actual" voltage present on the two points is not significant; it is the voltage difference that will make electrons move from the more negative point to the one that is more positive.

So, if the null meter has identical voltages on either side of it, there will be no current through the meter.

 

I completely understand that potential difference is required.....but I thought that the polarity of the AVS was supposed to be the reverse of what is shown......

What is making me stuck here is the idea that the AVS drops its voltage on R2 as the main VS does.

 

I completely understand that potential difference is required.....but I thought that the polarity of the AVS was supposed to be the reverse of what is shown......

What is making me stuck here is the idea that the AVS drops its voltage on R2 as the main VS does.

It will ONLY if the voltages on the two sides of the AVS are NOT equal. When they are equal, there is no potential difference to cause current flow, thus no voltage drop.
On edit. Depending upon whether the voltage on the right is more positive or more negative than the voltage on the left, there may be current either pulled from or sent to the left node.

 

If no current flows through the null detector doesn't it mean that there is no current through R2 as well?

 

There will always be current through R2 from the 24volt source via R1. The only way to eliminate ALL current through R2 would be to short R2 with the theoretical 0 ohm resistor. In operation, the current through the null meter will upset the calculated value of the R1-R2 node voltage by some small amount. When the measuring voltage source matches the voltage at the R1-R2 node, to reduce the current through the Null Detector, the actual value of the node voltage, with effectively no external loading, is matched.

I guess I'm having a hard time understanding exactly what you don't yet understand about balancing the circuit. Maybe that is the key! Think of it as a simple balance scale. When the weights are equal on both sides, the pointer is in the middle of its scale, just as it should be if no weights are applied at all.

If R1 and R2 are equal value, the node voltage should be 12 volts. If the adjustable power supply is set to 12 volts, the null meter will have the same voltage on both sides so no current will flow through the meter, therefore the currents through R1 and R2 are unchanged from the value they would have without the meter connected. If the variable voltage is set lower than 12 volts, more current will flow through R1, then the meter, to the lower voltage. On the other hand, if the variable source is set higher than 12 volts, more current will flow through the meter and mix with the current from R1 and then on to R2 .

 

I really appreciate your explanation. The thing is....I am trying to make intuitive sense of the matter, which at the moment needs to be worked on alot .

 

Another analogy I just thought of would be two buckets full of water connected with a hose at the bottom of each. The level of the water in each bucket would equate to the voltage on either side of the null indicator. If the left bucket is higher than the left bucket, water will flow through the connecting hose toward the lower bucket. (WATER FLOW WOULD BE IN THE OPPOSITE DIRECTION OF ELECTRON FLOW IN AN ELECTRICAL CIRCUIT) If the water level (potential energy) was the same in each bucket, there would be no water flow through the connecting hose.

 

great analogy.