### Nodal Analysis of Electric Circuits
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.
The steps in the nodal analysis method are:
- Count the number of principal nodes or junctions in the circuit. Call this number
*n*.
(A **principal 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
*N*_{1}, *N*_{2}, . . . , *N*_{n} and draw
them on the circuit diagram. Call the voltages at these nodes *V*_{1}, *V*_{2},
. . . , *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 *G*_{11}, *G*_{12}, . . . , *G*_{mm} and
*I*_{1}, *I*_{2}, . . . , *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 *V*_{1},
*V*_{2}, . . . , *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
*V*_{1} the voltage at node 1,
*V*_{3} the voltage at node 3, and remember that *V*_{2} = 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:
*V*_{1}=68.2 volts and *V*_{3}=27.3 volts
**Example 2: **Use nodal analysis to find the voltage at each node of this circuit.
**Solution:** Click here for solution.
**Example 3: **Use nodal analysis to find the voltage at each node of this circuit.
**Solution:** Click here for solution.
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