"Week Five"

Nodal analysis 
    
    In this method, we set up and solve a system of equations in which the unknowns are the voltages at the principal nodes of the circuit. From these nodal voltages the currents in the various branches of the circuit are easily determined.

In analyzing a circuit using Kirchhoff's circuit laws, one can either do nodal analysis using Kirchhoff's current law (KCL) or mesh analysis using Kirchhoff's voltage law (KVL). Nodal analysis writes an equation at each electrical node, requiring that the branch currents incident at a node must sum to zero. The branch currents are written in terms of the circuit node voltages. As a consequence, each branch constitutive relation must give current as a function of voltage; an admittance representation. For instance, for a resistor, I_branch = V_branch * G, where G (=1/R) is the admittance (conductance) of the resistor.

The steps in the nodal analysis method are:

• Count the number of principal nodes or junctions in the circuit. Call this number n. (Aprincipal node or junction is a point where 3 or more branches join. We will indicate them in a circuit diagram with a red dot. Note that if a branch contains no voltage sources or loads then that entire branch can be considered to be one node.)
• Number the nodes N₁, N₂, . . . , N_n and draw them on the circuit diagram. Call the voltages at these nodes V₁, V₂, . . . , V_n, respectively.
• Choose one of the nodes to be the reference node or ground and assign it a voltage of zero.
• For each node except the reference node write down Kirchoff's Current Law in the form"the algebraic sum of the currents flowing out of a node equals zero". (By algebraic sum we mean that a current flowing into a node is to be considered a negative current flowing out of the node.)
  
  For example, for the node to the right KCL yields the equation:

  I_a + I_b + I_c = 0

  
  
  Express the current in each branch in terms of the nodal voltages at each end of the branch using Ohm's Law (I = V / R). Here are some examples:
  
  
  
  The current downward out of node 1 depends on the voltage difference V1 - V3 and the resistance in the branch.
  
  
  
  
  In this case the voltage difference across the resistance is V1 - V2 minus the voltage across the voltage source. Thus the downward current is as shown.
  
  
  
  
  In this case the voltage difference across the resistance must be 100 volts greater than the difference V1 - V2. Thus the downward current is as shown.

• 
  The result, after simplification, is a system of m linear equations in the m unknown nodal voltages (where m is one less than the number of nodes; m = n - 1). The equations are of this form:

  

  where G₁₁, G₁₂, . . . , G_mm and I₁, I₂, . . . , I_m are constants.
  
  Alternatively, the system of equations can be gotten (already in simplified form) by using the inspection method.
• Solve the system of equations for the m node voltages V₁, V₂, . . . , V_m using Gaussian elimination or some other method.

Example 1: Use nodal analysis to find the voltage at each node of this circuit. 
Solution:

• Note that the "pair of nodes" at the bottom is actually 1 extended node. Thus the number of nodes is 3.
  
  
• We will number the nodes as shown to the right.
• We will choose node 2 as the reference node and assign it a voltage of zero.
• Write down Kirchoff's Current Law for each node. Call V₁ the voltage at node 1, V₃ the voltage at node 3, and remember that V₂ = 0. The result is the following system of equations:

  

  The first equation results from KCL applied at node 1 and the second equation results from KCL applied at node 3. Collecting terms this becomes:

  

  This form for the system of equations could have been gotten immediately by using the inspection method.
• Solving the system of equations using Gaussian elimination or some other method gives the following voltages:

  V₁=68.2 volts and V₃=27.3 volts

Basic case

Current through resistor R₁: (V₁ - V_S) / R₁The only unknown voltage in this circuit is V₁. There are three connections to this node and consequently three currents to consider. The direction of the currents in calculations is chosen to be away from the node.

1. Current through resistor R₂: V₁ / R₂
2. Current through current source I_S: -I_S

With Kirchhoff's current law, we get:

This equation can be solved in respect to V₁:

Super Node

In this circuit, we initially have two unknown voltages, V₁ and V₂. The voltage at V₃ is already known to be V_B because the other terminal of the voltage source is at ground potential.

The current going through voltage source V_A cannot be directly calculated. Therefore we can not write the current equations for either V₁ or V₂. However, we know that the same current leaving node V₂ must enter node V₁. Even though the nodes can not be individually solved, we know that the combined current of these two nodes is zero. This combining of the two nodes is called the supernode technique, and it requires one additional equation: V₁ = V₂ + V_A.

The complete set of equations for this circuit is:

By substituting V₁ to the first equation and solving in respect to V2, we get: