Under this method the following procedure is adopted:

1. Assume the voltages of the different independent nodes.

2. Write the equations for each mode as per Kirchhoffs current law.

3. Solve the above equations to get the node voltages.

4. Calculate the branch currents from the values of node voltages.

Let us consider the circuit shown in the Fig. 42 L and M are the two independent nodes ; M can be taken as the reference node. Let the voltage of node L (with respect to M) be VL.

Using Kirchhoff’s law, we get

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

Ohm’s law given I_{1} = V_{1} / R_{1} = ( E_{1 }– V_{L}) / R_{1}

I_{2} = V_{2} / R_{2} = (E_{2} – V_{L} / R_{2}

I_{3} = V_{3} / R_{3} = V_{L} / R_{3}

E_{1 }– V_{L} / R_{1} + E_{2 }– V_{L }/ R_{2} = V_{L} / R_{3}

Rearranging the terms , we get

V_{L} [ 1 / R_{1} + 1 / R_{2} + 1/ R_{3} ] – E_{1} / R_{1} – E_{2} / R_{2} = 0

It may be noted that the above nodal equation contains the following terms :

(i) The node voltage multiplied by the sum of all conductance’s connected to that anode. This term is positive.

(ii) The node voltage at the other end of each branch (connected to this node) multiplied by the conductance of branch. These terms are negative.

– In this method of solving a network the number of equations required for the solution is one less than the number of independent nodes in the network.

– In general the nodal analysis yields similar solutions.

– The nodal method is very suitable for computer work.

**Example 10**. For the circuit shown in Fig 43 find the currents through the resistances R_{3} and R_{4}

**Solution.** Refer Fig. 43.

Let L, M and N =independent nodes, and

V_{L} and V_{M} = voltages of nodes L and M with respect to node N .

The nodal equations for the nodes L and Mare :

V_{L} [ 1 / R_{1} + 1/ R_{2} + 1/ R_{3} ] – E_{1} / R_{1} – V_{M} / R_{2} = 0

V_{M} [ 1 / R_{2} + 1 / R_{4} + 1 / R_{5}] – E_{2} / R_{5} – V_{L} / R_{2} = 0

Substituting the values in (i) and (ii) and simplifying, we get

V_{L} ( 1 /5 + 1/5 + 1/ 10 ) – 50 / 5 – V_{M} / 5 = 0

2.5 V_{L} – V_{M} – 50 = 0

V_{M} ( 1/5 + 1/ 10 + 1/ 5) – 20 / 5 – V_{L} / 5 = 0

2.5 V_{M} – V_{L} -20 = 0

– V_{L} + 2.5 V_{M} – 20 = 0

Solving (iii) and (iv), we get

V_{L} = 27.6 V, V_{M} = 19.05 V

Current through R_{3} = V_{L} / R_{3} = 27 .6 / 10 = 2.76 A

Current through R_{4} = V_{M} / R4 = 19 .05 / 10 = 1.905 A

**Example 11.** Using Node voltage method, find the current in 6 Q resistance for the network shown in Fig. 44.

**Solution.** Refer Fig. 44. Considering node 2 as the reference node and using node voltage method, we have

V_{1} [ 1 / ( 6 +4) + 1/4 + 1/4 ] – 8 / 4 – ( 8 + 4 / 10) = 0

(The reason for adding the two battery voltages of 4 V and 8 Vis because they are connected in additive series).

V1(0.1 + 0.25 + 0.25) – 2 -12 = 0

V_{1} = 5.33 V

The current flowing through the 6 Q resistance towards node 1 is

= 12 – 5.33 / 6 + 4= 0.667 A.