Mesh analysis -- Where does the negative sign come from?

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SUMMARY

The discussion focuses on the derivation of the equation IA + IB = -IS in mesh analysis, specifically addressing the directionality of current sources. Participants clarify that the negative sign arises because IA and IB flow in the opposite direction to the current source IS, which is defined by the orientation of the arrows in the circuit diagram. The correct formulation for the node is A + B + S = 0, reinforcing the importance of consistent sign conventions in circuit analysis to avoid common errors.

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ace8888
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Hi,

I couldn't figure out from the equation that relate the current source IS to IA and IB. If both IA and IB is coming from the same direction and the positive terminal has been predicted. Wouldnt that be IA + IB = IS?
The solution is IA + IB = -IS. Is the negative comes from the current source actual direction is the opposite of what was predicted, which was positive at the top and negative at the bottom ?

thanks
 

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Look at the arrows. They define the +ve direction. Since the arrows for Ia and Ib are in the opposite direction to ls that means they have the opposite sign.
 
CWatters said:
Look at the arrows. They define the +ve direction. Since the arrows for Ia and Ib are in the opposite direction to ls that means they have the opposite sign.
Ia and Ib moving in opposite of Is so that should be Ia + Ib - Is = 0. where do they get the Ia + Ib = -Is from?
 
Could you please share the resource of this calculation?
 
ace8888 said:
Ia and Ib moving in opposite of Is so that should be Ia + Ib - Is = 0.
No, it most certainly should not. Look at the top node. They all enter, none leave, so the proper equation for that node is A + B + S = 0, or A + B = -S
 
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ace8888 said:
Ia and Ib moving in opposite of Is so that should be Ia + Ib - Is = 0. where do they get the Ia + Ib = -Is from?

See reply by phinds.
 
ace8888 said:
If both IA and IB is coming from the same direction and the positive terminal has been predicted.

I think that might be your mistake.

No values are given for Is or Vs, so either could be -ve. The + and - marks next to Vs don't really "predict" which side of the current source is +ve and -ve. They simply define what a +ve or -ve value of Vis actually means. Likewise for the direction arrow for Is.

When solving circuits you don't always know at the outset which direction a current actually flows or which side of a component is +ve or -ve. What you do is mark up the circuit with voltage and current labels and add arrows or +/- signs to define "what you mean by a +ve" values. Later when you solve the loop equations you might find that some values turn out negative. The most common mistake to make is a sign error.

Just for interest... Depending on the value of Is and other components in your circuit the current Ia and Ia could both be +ve, both be -ve, or one of each.
 
phinds said:
.. the proper equation for that node is A + B + S = 0, or A + B = -S

Just to add that I find I make fewer sign errors if I write node equations in the form A + B + S = 0 rather than A + B = -S.
 
CWatters said:
Just to add that I find I make fewer sign errors if I write node equations in the form A + B + S = 0 rather than A + B = -S.
I agree completely. I was simply making the point that the equation he asked about could be trivially derived from what you and I both consider the more correct form.
 

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