Misunderstanding about the choices of current values in KCL

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SUMMARY

The discussion centers on the application of Kirchhoff's Current Law (KCL) in analyzing current flow at specific nodes in an electrical circuit. The equation presented, 0 = ID2 + I2, is confirmed as correct, indicating that the sum of currents entering and leaving node V2 equals zero. The participants clarify that the negative signs in the equation arise from the defined directions of the currents, with ID2 and I2 being opposite in direction. The importance of consistent sign conventions when applying KCL is emphasized to avoid errors in circuit analysis.

PREREQUISITES
  • Understanding of Kirchhoff's Current Law (KCL)
  • Familiarity with circuit analysis and node voltage concepts
  • Knowledge of current direction conventions in electrical engineering
  • Basic proficiency in interpreting circuit diagrams
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  • Study the application of Kirchhoff's Voltage Law (KVL) in circuit analysis
  • Learn about node voltage analysis techniques in electrical circuits
  • Explore the impact of current direction on circuit behavior
  • Review common mistakes in applying KCL and how to avoid them
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Electrical engineering students, circuit designers, and anyone involved in analyzing electrical circuits using Kirchhoff's laws.

kookamlok
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Moved from general forum. Homework template is missing.

By KCL,
-ID2 + (-I2) = 0

0 = ID2 + I2

Q1: Do you think this equation is correct? (i assume it is correct because it came from my professor's notes)

Q2: Which node do you think this equation is used for making this equation with values currents which is going in and out to the selected partcular node? (I would guess node at V2 but why the input current will be 0 according to 0 = ID2 + I2?)

Q3: why will my professor choose this node? (to be honest, i have no clue. I will ask my professor later but want to see your opinion.)
 

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Hi kookamlok. :welcome:

These equations can arise in cases such as where you have been provided with the labelled figure in white, and asked to find something in terms of the current coloured in blue, I2.

You can see that I2 is the current ID2 but opposite in direction, so you can write
I2 = –ID2

Alternatively, you could consider the point in the wire near where you see the voltage label V2, and sum the currents into (or out of) that point.

There is a current –ID2 directed in from the left, and a current –I2 directed up from the ground. KCL says these must sum to zero. (Otherwise there would have to be current spilling out of the wire, or something equally preposterous).
 
kookamlok said:
I would guess node at V2..

It must be node V2 because that's the only node that involves only ID2 and I2. If you applied KCL at node V1 it would feature ID1.

but why the input current will be 0 according to 0 = ID2 + I2?)

Not sure what you mean by input current but the zero arises because KCL states that the sum of currents going into (or out of) a node equals zero. It so happens that in this case ID2 and I2 have been defined in opposite directions so you can rearrange your equation to ..

ID2 = -I2

When applying KCL you must first decide if current entering or leaving a node is going to be defined as positive (+ve). This can be an arbitrary decision but you must be consistent. It looks like your professor decided that current entering a node would be +ve when he wrote..

-ID2 + (-I2) = 0

Both ID2 and I2 are -ve in his equation because on the drawing they are shown leaving node V2. That only makes sense if he defined current entering a node as +ve.

Here is KCL for node V1...

(+ID1) + (+ID2) + (-V1/1k) = 0

I have tried to be very explicit with the signs because this is the greatest source of error when applying KCL.
ID1 is +ve because it's entering node V1.
Likewise for ID2.
The last term -V1/1k is negative because I have made the reasonable assumption that V1 stands for the voltage across the 1K resistor and it's polarity is as shown by the arrow that I have added in this version of your drawing. This implies the current in the 1k resistor is flowing downwards out of V1 so it's -ve. In some cases you have to add your own arrows to a drawing to define what you mean by a positive voltage.

t2.PNG
 

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