Solve a node connected to several sources

Count connections, not drawing positions

All points joined by an ideal wire are one node, even if the drawing bends or uses a repeated net label. Choose ground and mark unknown node voltages. For each resistor from node x to node y, current leaving x is $(V_x-V_y)/R$. This expression supplies both magnitude and direction without guessing the eventual current flow.

Consider one unknown node x connected to 5 V through 10 kΩ, to ground through 20 kΩ, and to an ideal 2 V reference through 10 kΩ. KCL, using outward currents, is $(x-5)/10000+x/20000+(x-2)/10000=0$. Multiplying by 20000 gives $2x-10+x+2x-4=0$, hence x = 2.8 V.

Interpret the negative branch

Current toward the 5 V node is −0.22 mA, so 0.22 mA actually enters x from that source. Current to ground is 0.14 mA. Current toward the 2 V reference is 0.08 mA. The two outward currents total 0.22 mA. The 2 V reference must absorb current in this example; a real reference that cannot sink current would not hold the assumed voltage.

The result lies between the fixed potentials because all conductances are positive. You can also see x as a conductance-weighted average: sum of each fixed voltage divided by its resistance, divided by total conductance.

Extend carefully

For two unknown nodes, write two KCL equations and solve them together. Do not use a series-resistor simplification across a node that has a third connection. Before a simulator run, create a connection list with component name, first node, second node, and value. A misplaced wire changes the equations even when the drawing looks familiar. Check residual currents against zero after solving; rounding is allowed, a missing branch is not.