Voltage is defined as existing between two nodes.
Even when we refer to the voltage at a specific node,that voltage is always expressed relative to some other node, typically our ground node. Ground is always zero volts. When expressing current using node voltages, we first find the voltage difference between the two nodes. voltage difference = V_{1}  V_{2} current = V/R = (voltage difference)/resistance In the example shown: i_{A} = (V_{1}  V_{2})/R or i_{B} = (V_{2}  V_{1})/R Note that i_{A} = i_{B} 

Using this information, now we can generate a KCL equation at a node: Consider this circuit segment. Generate a KCL equation at Node 1. (V_{1}  V_{2})/R_{2} + (V_{1}  V_{3})/R_{3} + (V_{1}  V_{4})/R_{4} + (V_{1}  V_{5})/R_{5} = 0 Note that the above KCL equation has only node voltages as unknowns. The equation is equivalent to: Σ I_{leaving} = 0I_{2} + I_{3} + I_{4} + I_{5} = 0 When doing Nodal Analysis, always define currents as leaving the node. 
We will apply Nodal Analysis on this circuit: We will generate enough independent equations for the number of unknown node voltages. This is done by writing a KCL equation at each unknown node. All parameters are node voltages when doing Nodal Analysis. 

Here is the same circuit labeled: Step 1: Assign one node to be ground. You need a reference node so that all other nodes are expressed relative to the reference node. Note that V_{3} is defined as ground. You can choose any node to be ground if a ground node does not already exist. Step 2: Count the number of unknown node voltages. In this case we have three nodes but one is ground/zero. Therefore we have two unknown nodes V_{1} and V_{2}. Step 3: Write a KCL equation at each unknown node. Always let currents be leaving the node and express your equations in terms of node voltages  NOT CURRENTS. If you follow these steps you should generate as many equations as you have unknowns: KCL at V_{1}: 2 + (V_{1}  V_{3})/3 + (V_{1}  V_{2})/4 + 7 = 0 But V_{3} = 0, therefore 2 + V_{1}/3 + (V_{1}  V_{2})/4 + 7 = 0
KCL at V_{2}:7 + (V_{2}  V_{1})/4 + (V_{2}  V_{3})/6 = 0 7 + (V_{2}  V_{1})/4 + V_{2}/6 = 0
We now have two equations and two unknowns. All terms are in Amps. 
Here is a circuit with the nodes labeled and ground shown: Note there are real benefits to having voltage sources in the problem. By inspection we can get the following information: V_{2}  V_{4} = 18V Since V_{4} = 0 (ground is always zero) Then V_{2} = 18V Also by inspection we see that V_{1}  V_{3} = 6V We have four nodes, but we already know V_{4} and V_{2} Also we already have one equation. Currently we have two unknowns and already one equation, so we only need one more equation. We will apply KCL at one of our unknown nodes Let's try doing KCL at V_{3}: 2A + (V_{3}  V_{2})/3 + I_{through voltage source} = 0 But we do not know I_{through voltage source}. Therefore we need to draw a supernode. 

Here we show the circuit with a supernode: A supernode is a closed path enclosing part of the circuit. As with a node,
In this particular circuit, all currents are shown as leaving the supernode.
Therefore we can say: 
A mesh current is the current going clockwise within a mesh.
Here we have two meshes and therefore two mesh currents, I_{1} and I_{2}.
When expressing voltage using mesh currents, we use Ohm's Law which states V=IR if current is going from + to . In this case the current in the 6 Ohm going from + to  is (I_{1}  I_{2}) V = 6(I_{1}  I_{2}) 

Again we use Ohm's Law to calculate V. In this case the current in the 6 Ohm going from + to  is (I_{2}  I_{1}) V = 6(I_{2}  I_{1}) This is exactly the opposite to the equation in the previous box. This is because V is defined with an opposite polarity. Both are correct! 

Now let's do a problem using Mesh Analysis.
Our first step is to generate a KVL equation in each mesh:Let's generate a KVL loop equation in mesh 1. 18 + 2I_{1} + 6(I_{1}  I_{2}) = 0 Note that the above KVL equation has only mesh currents as unknowns. The equation is equivalent to: ΣV_{drops}  ΣV_{rises} = 0The equation reduces to: 8I_{1}  6I_{2} = 18 Let's generate a KVL loop equation in mesh 2. 3I_{2} + 2 + 6(I_{2}  I_{1}) = 0 As before the equation is equivalent to: ΣV_{drops}  ΣV_{rises} = 0The equation reduces to: 9I_{2}  6I_{1} = 2 We now have two equations and two unkonwns. Solving for I_{1} and I_{2} requires algebra. When doing Mesh Analysis, always define 
We will apply Mesh Analysis on this circuit using the Short Cut Method: We will generate enough independent equations for the number of unknown mesh currents. This is done by writing a KVL loop equation in each mesh where the mesh current is unknown. All parameters are mesh currents when doing Mesh Analysis. 

Here is the same circuit with all mesh currents labeled:
Step 1: Label all mesh currents and show mesh currents going clockwise. Step 2: Count the number of unknown mesh currents. In this case we have three unknown mesh currents, I_{1}, I_{2}, & I_{3}. Step 3: Write a KVL equation in each mesh. Express your equations in terms of mesh currents  NOT VOLTAGES. The Short Cut Method is used to quickly/easily generate a KVL loop equation for a mesh. To use the Short Cut Method to generate a KVL equation

We will apply mesh analysis to this circuit: When a current source is present, it will give you information about mesh currents by inspection. When applying KVL to the circuit, you can temporarily remove the current sources to 'see' which meshes don't have current sources. It is in these meshes where you will write a KVL equation. Note that sometimes there will not be enough meshes to generate the required number of independent equations. In this case you will have to resort to using a loop instead of just a mesh. 

Here is the circuit labeled:Step 1: Looking at the current sources what information by inspection can be acquired about the mesh currents? I_{3} = 7A I_{2}  I_{1} = 2A Step 2: Count the number of unknowns and the number of equations to determine how many additional equations are needed. We have two equations and three mesh currents. Therefore we need one more equation. Step 3: Determine in which meshes it is possible to write a KVL in equation. This can be determined by temporarily removing current sources to see where closed paths exist. This is shown in the box below. 

Step 4: Looking at the circuit on the right, there are no meshes where KVL can be done. All the meshes have an opencircuit in them. Therefore we have to resort to doing KVL in a loop to generate the the independent equation that we need. There is only one loop here. 4 + 2(I_{1}I_{3}) + 4(I_{2}I_{3}) + 6I_{2} = 0 We now have three equations and three unknowns. Therefore all mesh currents can be found. 