1. Method 2c: Nodal Analysis
Find voltage at a node: nodal voltage
Express branch currents in terms of nodal voltages
Use KCL to solve the nodal voltages

2. Nodal Analysis: Cont.
Find nodal voltages: Voltage between two nodes and use one of them as a reference node.
Example, going from point A to B
A is the reference node, calculate VB C
For a resistor crossed in the same direction as the current, the change in voltage is - iR.
For a resistor crossed in the opposite direction of the current, the change in voltage is + iR.
If a source of emf is crossed in the same direction as the emf (- to +), the change in voltage is +.
If a source of emf is crossed in the opposite direction of the emf, the change in voltage is -.
VBA=VB-VA=e1-i1R1+i2R2-e2
VCA=VC-VA=e1-i1R1

3. Nodal Analysis: Cont.
Express branch currents in terms of nodal voltages C D
VCA=VC-VA=e1-i1R1
VBC=VB-VC=i2R2-e2
i=(Vc-VD)/R3
i1=(VC-VA-e1)/R1
i2=(VB-VC-e2)/R2
i1+i2=i

4. Example
Consider three lossy voltage sources in parallel. Note that there are actually 2 nodes, but the bottom one has been grounded to 0.0 V, so there is only one unknown nodal voltage.
Step 1: Assignment of Nodal Voltage Assign a variable name to all nodes with unknown voltages.
Step 2: Apply KCL to Each Node Assume that all currents flow away from the node. Apply KCL.

5. Detailed Example Find currents in each branches
Step 1: simply the circuit
Step 2: identify nodes and label them accordingly
Step 3: Write the node equation.
(V1 - 9)/5 + V1/6 + (V1 + 9)/12 = 0
Here I suppress the "k" for each resistor; there is no problem as long as all resistances are written in the same units.
V1 : 7/3 V  2.33 V
i1=-1.33mA, i2=0.39mA, i3=0.94mA V1 i1 i2 i3

6. Nodal analysis with current sources
branch current is the same as the current provided by current source.

7. Example with mixed sources: nodal analysis
Identify nodes and label accordingly. Here I have grounded the central node.
Write the nodal equations
There are three unknowns so three equations will be needed. Derive the first equation by considering the grounded node.
V1/2 + V2/4 + V3/4 = 0 2V1 +V2 + V3 = 0
The next two equations may be written quickly by inspection of the voltage sources connecting pairs of nodes.
V1 - V3 = 10 V3 - V2 = 12
1.5 A
3.5A
V1 = 8 V V2 = -14 V V3 = -2 V
Finally, to find the current through the 12 V course, apply KCL at the top node. Since the 8 ohm resistor has 12 V across it; 1.5 A flows through it toward the node.
The top 4 ohm resistor has 14 V across it; thus 3.5  flows through it toward the node.
Accounting for the 2 A source, application of KCL yields 3 A flowing away from the node through the 12 V source.