Here you have two more examples:
First
For this circuit from KCL
I3 =
I1 +
I2
(Va - 4V)/6Ω =
(10V - Va)/10Ω +
(6V - Va)/4Ω
And after solve this I get
Va -> 190/31 = 6.12903226V
Second
And again form KCL
(-I1) + (-I2) + (-I3) = 0
Or
- I1 - I2 - I3 = 0
So know all current flow outward (away) from the node.
So this assume that voltage at node Va should be at higher potential.
And current flow from + to - , I assume that if current entering into a node I give him " +"
and current that come out form the node I give "-".
But you can choose whatever you wont but you must be consistent in your choice
So I can write:
-(Va - 10V)/10Ω -
(Va - 6V)/4Ω -
(Va - 4V)/6Ω = 0
And again the answer is exactly the same
Va -> 190/31 = 6.12903226V
And maybe another example for this diagram I choose
So I can write
I5 = I4 + I3
I1 = I3 - I2
(V1 - E1)/R1 = E1/R2 + (E1 - E2)/R3 (1)
I1 = (E1 - E2)/R3 - (E2 - V2)/R4 (2)
And after solving I get
E1=0.5V and
E2=0.5V
But I can choose that all current flow out from the node
Then
-(E1 - V1)/R1 - E1/R2 - (E1-E2)/R3 = 0
-(E2 -E1)/R3 - (E2 - V2)/R4 - I1 = 0
And again after solve this I get the same result.