∑iin=∑iout
“>∑iin=∑iout: sum of current going into a node = sum of current going out of the same node
Now we add a resistor in series to change the current
So now A2 and A4 are in series so the resistors add together to 100Ω
A3: 10A * ( A24 / ( A24 + A3) ) = 10A * 100Ω / 150Ω = 6.6667A
A2 and A4 would have the same current flowing through both of them since they are in series.
= 10A * A3 / (A24 + A3 ) = 10A * 50Ω / ( 100Ω + 50Ω ) = 3.333A for A2 and A4.
10A = A2 + A3 = 3.3333A + 6.6667A