Use the node-voltage method to find the branch currents i1, i2, and i3 in the circuit in (Figure 1) if v1 = 40 V and v3 = 74 V.?

This ckt has 2 nodes. Va above the 500Ω and Vb above the 400Ω = 74V
KCL at V1:
(V1+40)/5000 + V1/500 + (V1-74)/1000 + 0.01 = 0 Multiply by 5000
V1+40+10V1+5V1 -370 +50 = 0
16V1 = 370 - 90 = 280
V1 = 280/16 = 35/2 = 17.5 V
i2 = 17.5/500 = 35mA
-40 - 5000*i1 = V1 = 35/2
i1 = -(35/2 + 80/2)/5000 = -23/2 = -11.5mA
The current right to left in the 1kΩ = (74-V1)/1000 = 113/2 mA
The current down into the 4kΩ = 74/4 = 37/2 = 18.5mA 
The current right to left from V3 = 37/2 - 10 = 17/2 mA
Therefore the current up out of V3 = 130/2 mA = 65mA = -i3
i1 = -11.5mA
i2 =  35mA
i3 =  -65mA...Show more