# NODE VOLTAGE METHOD

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Another alternate method of analysis of multiple loop circuits is called the node voltage method. It is based on finding the voltages at each node in the circuit using Kirchhoff s current law. A node is the junction of two or more current paths.

G**ENERAL STEPS**

The general steps for the node voltage method are as follows:

**Determine the number of nodes****Select one node as a reference. All voltages will be relative to the reference node.****Assign current at each node where the voltage is unknown, except at the reference node. The directions are arbitrary.****Apply Kirchhoff s current law to each node where current are assigned.****Express the current equations in terms of voltages and solve the equations for the unknown node voltages.**

## CIRCUIT FOR DEMONSTRATING NODE VOLTAGE ANALYSIS

We will use Figure-1. to ilustrate the general approach to node voltage analysis.

Figure - 1

First |
Establish the node. In this case there are four nodes, as indicated in the figure. |

Second |
Let's use node B as reference. Think of it as circuit ground. Node voltages C and D are already known to he the source voltages. voltage's. The voltage at node A is the only unknown it is designated as V _{A}. |

Third |
Arbitrarily assign the currents at a node A as indicated in the figure-1 |

Fourth |
The Kirchhoff's current equation at node A is I |

Fifth |
Express the currents in terms of circuit voltages using Ohm's law as follow I I I Substituting these into the current equation, we get (V |

The only unknown is V_{A}; so we can solve the single equation by combining and rearranging terms Once the voltage is know, all branch currents can be calculated.

## EXAMPLE FOR DEMONSTRATING NODE VOLTAGE ANALYSIS

**Find the node voltages in Figure**

**SOLUTION**

The unknown node voltage is V_{A}, as indicated in figure. This is the only unknown voltage. Currents are assigned at node A as shown. The current equation is

I_{1} - I_{2} + I_{3} = 0

Substitution for current using Ohm's law gives equation in terms of voltages

(10 -V_{A})/47 - V_{A}/22 + (5 - V_{A})/82 = 0

Solving for V_{A} yields

10/47 - V_{A}/47 - V_{A}/22 + 5/82 - V_{A}/82 = 0

- V_{A}/47 - V_{A}/22 - V_{A}/82 = - 10/47 - 5/82

(1804V_{A} + 3854V_{A} + 1034V_{A})/84788 = (820 + 235)/3854

6692V_{A} / 84788 = 1055/3854

VA = (1055 X 84788)/(6692 X 3854) = 3.47 V

V_{A} = 3.47 volt