This method is mainly based on Kirchhoff's Current Law (LCL). This method uses the analysis of the different nodes of the network. We have already defined a node. Every junction point in a network, where two or more branches meet is called a node. One of the nodes is assumed as reference node whose potential is assumed to be zero. It is also called zero potential node or datum node. At other nodes the different voltages are to be measured with respect to this reference node. The reference node should be given a number zero and then the equations are to be written for all other nodes by applying KCL. The advantage of this method lies in the fact that we get (n - 1) equations to solve if there are 'n' nodes. This reduces calculation work.

Consider the following network shown in the Fig. 1.

Let voltage at node 1 and node 2 be V

_{1}and V_{2}. Mark various branch currents as shown in the Fig.2. Now analyse ech node using KCL. independently. Now applying KCL at node 1,

I

_{1}- I_{3}- I_{4}= 0 ...........(1)At node, I

_{2}+ I_{4}- I_{5}= 0 ..........(2) The currents in these equations can be expressed interms of node voltage as,

I

_{1}- (V_{1}/R_{1})- (V_{1}-V_{2}/ R_{2}) = 0 ...........(3)and I

_{2}+ (V_{1}-V_{2}/R_{2}) - (V_{2}/R_{3})= 0 ..............(4) As I

_{1}and I_{2}are known, we get two equations (3) and (4) with the two unknowns V_{1}and V_{2}. Solving these equations simultaneously, the node voltages V_{1}and V_{2 }can be determined. Once V_{1}and V_{2 }are known, current through any branche of the network can be determined. If there exists a voltage source in any of the branches as shown in the Fig.3 then that must be considered while writing the equation for the current through that branch.Fig. 2 |

Now V

_{1}is at higher potential with respect to base, forcing current downwards. While polarity of V_{x}is such that it tries to force current upwards. So in such a case equation of current becomes,I = (V

_{1 }- V_{x})/R ......... I leaving the node If the direction of current I is assumed entering the node then it is assumed that V

_{x }is more than V_{1 }and hence equation for current I becomes,I = (V

_{x }- V_{1})/R ........... I entering the node__1.1 Points to Remember for Nodal Analysis__

1. While assuming branch currents, make sure that each unknown branch current is considered at least once.

2. Convert the voltage source present into their equivalent current sources for node analysis, wherever possible.

3. Follow the same sign conversion, currents entering at node are be considered positive, while currents leaving the node are to be considered as negative.

4. As far as possible, select the directions of various branch currents leaving the respective nodes.

__1.2 Supernode__

Consider a circuit shown in the Fig. 4. In this circuit, the nodes labelled V

_{2 }and V_{3}are connected directly through a voltage source, without any circuit element. The region surrounding a voltage source which connects the two nodes directly is called supernode.Fig. 4 Region of supernode |

In such a case, the nodes in supernodes region can be analysed separately and the relation between such node voltages and a source voltage connecting them can be separately obtained. In the circuit shown in the Fig.4 we can write,

V

_{2 }= V_{3}+ V_{x} In addition to this equation, apply KCL to all the nodes assuming different branch currents at the nodes. The current through voltage source, connecting supernodes must be expressed interms of node voltages, using these KCL. equations. Then the resulting equations and supernode equations are to be solved simultaneously to obtain the required unknown.

Applying KVL to the loop,

- V

_{x}- V_{3 }+ V_{2 }= 0 V

_{2 }= V_{x }+ V_{3} Thus the relationship between supernode voltages can be obtained using KVL.

Such equation can be written by inspection also. For the Fig. 6 shown, the equation is,

V

_{3 }= V_{2 }+ V_{x}__1.3 Steps for the Node Analysis__

Step 1 : Choose the nodes and node voltages to be obtained.

Step 2 : Choose the currents preferably leaving the node at each branch connected to each node.

Step 3 : Apply KCL at each node with proper sign convection.

Step 4 : If there are supernodes, obtain the equations directly interms of voltages which are directly connected through voltage source.

Step 5 : Obtain the equation for the each branch current interms of node voltages and substitute in the equations obtained in step 3.

Step 6 : Solve all the equations obtained in step 4 and step 5 simultaneously to obtain the required node voltages.

**Key Point**: If there are many number of branches in parallel in a network then node method is advantageous for the network analysis.

**Example 1**: Find the current through each resistor of the circuit shown in the Fig. 7, using node analysis

Fig. 7 |

At node 1, -I

_{1}-I_{2}-I_{3}= 0 -(V

_{1}-15 /1) - (V_{1}/1) + (V_{1}-V_{2 }/0.5)= 0 -V

_{1}+ 15 - V_{1}- 2V_{1}+ 2V_{2}= 0 4V

_{1}- 2V_{2}= 15 .........(1)At node 2, I

_{3}- I_{4}- I_{5}= 0 (V

_{1}-V_{2 }/0.5) - (V_{2}/2) - (V_{2}-20 /1) = 0 2V

_{1 }- 2V_{2 }- 0.5V_{2}- V_{2}+ 20 = 0 2V

_{1 }- 3.5 V_{2 }= -20 .............(2) Multiplying equations (2) by 2 and subtracting from equation (1) we get,

5V

_{2}= 55 V

_{2 }= 11 Vand V

_{1 }= 9.25 V Hence the various currents are,

I

_{1}= (V_{1}- 5)/1 = 9.25 - 15 = -5.75 A i.e. 5.75 A**↑** I

_{2 }= V_{1}/1 = 9.25 A I

_{3 }= (V_{1}-V_{2})/0.5 = -3.5 A i.e. 3.5 A**←** I

_{4}= V_{2}/2 = 5.5 A I

_{5 }= (V_{2 }- 20)/1 = (11-20)/1 = -9 A i.e. 9 A**↑** The various branch currents are shown. Applying KCL at various nodes.

Node 1 : 9 - I

_{1 }- I_{2 }- I_{3 }= 0 .........(1)Node 2 : I

_{3 }- I_{4 }+ 4_{ }= 0 .........(2)Node 3 : I

_{2 }- 4 - I_{5 }= 0 ..........(3)**Key Point**: Nodes V

_{1 }and V

_{2 }from supernode region and nodes V

_{1 }and V

_{2 }from super node region.

Super node : V

_{1 }- 10 = V_{3 }i.e. V_{1 }- V_{3 }= 10 .........(4)Super node : V

_{1 }+ 6 = i.e. V_{1 }- V_{2 }= -6 ........(5)From equation (2), I

_{3 }= I_{4 }- 4 and from equation (3), I_{2 }= I_{4 }+ 4Using in equation (1), 9- I

_{1 }- I_{5 }- 4 - I_{4 }+ 4 = 0i.e. I

_{1 }+ I_{4 }+ I_{5 }= 9 .........(6)But I

_{1 }= V_{1}/4, I_{4 }= V_{2}/10, I_{5 }= V_{3}/5 (V

_{1}/4) + (V_{2}/10)+ (V_{3}/5) = 9 0.25 V

_{1 }+ 0.1 V_{2}+ 0.2 V_{3}= 9 ..........(7) Solving equations (4), (5) and (7) simultaneously,

V

_{1}= 18.909 V, V_{2}= 24.909 V, V_{3}= 8.909 V
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