Kirchoff’s Current Law. Solve node P2

View attachment Scan Feb 28, 2019.pdf

What bothers me here is that node P₂ has all the sense arrows pointing into the node. I would have figured at least one would leaving the node. This is pertaining to exercise #4. I mean actually it works out. But should’nt one arrow point leaving the node?

Hi,

Remember that the choice of current directions is arbitrary. You can choose whatever direction that you like as long as you stay consistent with it throughout your calculation.

The algebraic summation of voltages that play out at the end leaves you with the actual direction of currents.

Remember that every node equation in the circuit considers every current in the circuit, directly or indirectly. At the end, actual current directionsituation are obtained.

Akanimo said:

Hi,

Remember that the choice of current directions is arbitrary. You can choose whatever direction that you like as long as you stay consistent with it throughout your calculation.

The algebraic summation of voltages that play out at the end leaves you with the actual direction of currents.

Remember that every node equation in the circuit considers every current in the circuit, directly or indirectly. At the end, actual current directionsituation are obtained.

Click to expand...

Yes, I guess its this arbitrary current direction that confuses me. I mean what you say is right because the all currents do = 0. I have to look at this again and wrap my head around it. Thanks for the swift reply.

LeatherNeck.

Equating to zero is what helps us determine the actual directions of current salary they flow in different paths in the circuit. Logically, equating to zero tells you that some of the current are actually -ve (flowing in directions opposite to the assumed). However, which of the currents are negative?

Take for instance that two nodes, V1 and V2, connect a resistance R. If it is determined that:

1) V1>V2, then V1-V2=I*R
2) V1<V2, then V1-V2=-I*R

To understand quickly what's happening, take a simple circuit with 5 nodes and indicated resistance values. Assign an arbitrary direction of currents and solve for node voltages and branch currents. When you are done, assign a different direction of currents arbitrarily and solve again. Assign another current directions arbitrarily and solve again. Compare solutions and try to find matching equations in the solution steps. This should make it clear what the concept does.

I hope this helps.

LeatherNeck said:

View attachment 151405

What bothers me here is that node P₂ has all the sense arrows pointing into the node. I would have figured at least one would leaving the node. This is pertaining to exercise #4. I mean actually it works out. But should’nt one arrow point leaving the node?

Click to expand...

You are given 4 assumed current directions into a node and three finite currents. Using the mathematical convention, current into a node is assumed to be negative and vice-versa for current leaving a node. Setting up the equations:



Since we assumed the current into the node when we set up the equation, the value of i2 = 5 - 7j is the value of i2 into the node. If you want the value of p2 from the node, negate the value and get i2 = -5 +7j . It all depends on the assumed direction of current.

Another point: You always hear folks say "current flow" . That is scientific slang. Current flow literally means "charge flow flow", which is redundant and ridiculous. Charge does not flow twice. They should say "current exists", "current is present", or sometimes just plain current.

Ratch