Mesh Analysis Question: Why I3-I2?

[influx]

http://photouploads.com/images/untitleda.png

Basically, in the last step, why have they done I3 - I2 rather than the other way round?

Thanks

PS: I have not embedded the image because its size is too large..

 

[gneill]

Note the indicated polarity of the voltage VE across that resistor (indicated by the arrow next to the resistor). What direction should you assume for the net current flow to result in that polarity?

 

[influx]

Note the indicated polarity of the voltage VE across that resistor (indicated by the arrow next to the resistor). What direction should you assume for the net current flow to result in that polarity?

Well V_e is positive so I am assuming the direction the arrow is pointing in is correct. However, I'm not sure how to progress from here.

 

[gneill]

Well V_e is positive so I am assuming the direction the arrow is pointing in is correct. However, I'm not sure how to progress from here.

The arrow indicates assumed polarity of the voltage. If you were to do a measurement, you'd place the "+" lead of the voltmeter at the arrow tip end of the resistor and the "-" lead at the other end in order to measure according to the defined assumption. For the arrow tip to be at a higher potential with respect to the base end, what direction should the current flow through the resistor?

Note that these arrows and currents are all assumed directions... actual calculated or measured values may turn out to be negative, indicating that the assumption(s) happened to be incorrect. Incorrect assumptions of polarity or direction will NOT lead to incorrect answers! The math, if carried out correctly, will automatically "correct" for this.

 

[influx]

The arrow indicates assumed polarity of the voltage. If you were to do a measurement, you'd place the "+" lead of the voltmeter at the arrow tip end of the resistor and the "-" lead at the other end in order to measure according to the defined assumption. For the arrow tip to be at a higher potential with respect to the base end, what direction should the current flow through the resistor?

Note that these arrows and currents are all assumed directions... actual calculated or measured values may turn out to be negative, indicating that the assumption(s) happened to be incorrect. Incorrect assumptions of polarity or direction will NOT lead to incorrect answers! The math, if carried out correctly, will automatically "correct" for this.

Doesn't voltage always flow from a high potential to a low potential? So the voltage would have to go from positive to negative?

 

[gneill]

Doesn't voltage always flow from a high potential to a low potential? So the voltage would have to go from positive to negative?

Current flows from high (more positive) potential to low potential, yes. But this has nothing to do with assumptions or definitions made about things for purposes of analyzing a circuit. After all, the reason one analyzes a circuit is to determine how things are actually working. Often a voltage polarity indicated on a circuit diagram is really a definition of how to interpret a given potential difference (how to orient your voltmeter to read the value as intended by the person who drew the diagram).

Just marking down an assumed direction for a current or potential gradient on the diagram does not make it so. It's just a placeholder for an assumption that may be a definition of some quantity (how it's to be interpreted) or a 'guess' at how things might be, to be used in setting up the mathematics of analysis in a consistent fashion.

 

[influx]

Current flows from high (more positive) potential to low potential, yes. But this has nothing to do with assumptions or definitions made about things for purposes of analyzing a circuit. After all, the reason one analyzes a circuit is to determine how things are actually working. Often a voltage polarity indicated on a circuit diagram is really a definition of how to interpret a given potential difference (how to orient your voltmeter to read the value as intended by the person who drew the diagram).

Just marking down an assumed direction for a current or potential gradient on the diagram does not make it so. It's just a placeholder for an assumption that may be a definition of some quantity (how it's to be interpreted) or a 'guess' at how things might be, to be used in setting up the mathematics of analysis in a consistent fashion.

I understand that the assumed direction is not always correct depending on the answer you get. However, I still don't understand why its I3 - I2 rather than I2 - I3

 

[gneill]

I understand that the assumed direction is not always correct depending on the answer you get. However, I still don't understand why its I3 - I2 rather than I2 - I3

For voltage specifications, the pointy bit of the arrow defines the assumed more positive potential. So:

https://www.physicsforums.com/attachment.php?attachmentid=66326&stc=1&d=1391627964

What direction does that imply for the assumed direction of the net current for the resistor in question? Which of the two mesh currents flowing through that resistor is defined to be in the same direction?

 

[influx]

For voltage specifications, the pointy bit of the arrow defines the assumed more positive potential.

Ah I see. I didn't know that.

What direction does that imply for the assumed direction of the net current for the resistor in question? Which of the two mesh currents flowing through that resistor is defined to be in the same direction?

Well I assumed I₂ is flowing clockwise and my answer was positive meaning it does indeed flow clockwise. I also assumed I₃ flows clockwise and once again my answer was positive meaning I₃ also flows clockwise.

However, if you trace the path followed by each current, I₂ enters on the side of negative potential whilst I₃ enters on the side of positive potential.

So since clockwise is positive, shouldn't it be I₂-I₃? (I₃ is flowing in the opposite direction to the arrow head)

 

[gneill]

Clockwise or counterclockwise directions don't imply a sign directly. What counts is how the given current contributes to the potential drop as defined by the arrow on the diagram.

For the moment, disregard the actual values determined for the mesh currents. The values obtained (positive or negative) depend upon the initial assumptions for current directions, assumptions made before the math was even written. Those assumptions form the basis of the analysis mathematics, and you cannot change them afterwards else the answers would then become incorrect; the sign of the values applies strictly to the original direction assumptions.

The assumption regarding the polarity of the V_E voltage implies that the net current through the 10Ω resistor must be flowing from left to right. That's the only way a potential drop of the indicated polarity can occur across it.

Current I₃ is defined as flowing through that resistor in that same direction. So its contribution to the net current must be positive. On the other hand, I₂ flows in the opposite direction, so its contribution must be negative. We are talking algebraically here, and NOT about the actual values you happened to calculate for them. Thus, the net current flowing in the assumed direction is I₃ - I₂.

 

[influx]

Clockwise or counterclockwise directions don't imply a sign directly. What counts is how the given current contributes to the potential drop as defined by the arrow on the diagram.

For the moment, disregard the actual values determined for the mesh currents. The values obtained (positive or negative) depend upon the initial assumptions for current directions, assumptions made before the math was even written. Those assumptions form the basis of the analysis mathematics, and you cannot change them afterwards else the answers would then become incorrect; the sign of the values applies strictly to the original direction assumptions.

The assumption regarding the polarity of the V_E voltage implies that the net current through the 10Ω resistor must be flowing from left to right. That's the only way a potential drop of the indicated polarity can occur across it.

Current I₃ is defined as flowing through that resistor in that same direction. So its contribution to the net current must be positive. On the other hand, I₂ flows in the opposite direction, so its contribution must be negative. We are talking algebraically here, and NOT about the actual values you happened to calculate for them. Thus, the net current flowing in the assumed direction is I₃ - I₂.

Regarding the part in bold, you are saying the current's direction is always opposite to the direction of the voltage drop?

Thanks

 

[gneill]

Regarding the part in bold, you are saying the current's direction is always opposite to the direction of the voltage drop?

Thanks

Your text or notes should have a description of how potential drop across a resistor occurs in the direction of the current flow. The end of a resistor where the current enters is at a higher potential than the end where it leaves.

 

[influx]

Your text or notes should have a description of how potential drop across a resistor occurs in the direction of the current flow. The end of a resistor where the current enters is at a higher potential than the end where it leaves.

Yeah that is what I thought but that appears to contradict what you said:

The assumption regarding the polarity of the V_E voltage implies that the net current through the 10Ω resistor must be flowing from left to right. That's the only way a potential drop of the indicated polarity can occur across it.

According to the above, the net current is flowing from left to right but the voltage drop is from right to left...

 

[gneill]

 

[influx]

:facepalm:

I finally get it!

The current always flows from positive to negative (left to right or high to low potential). This current flow is in the same direction as I3 and opposite to I2, hence why its I3 - I2!

Thanks a lot for all your help :) !