# Nodal Analysis In Electrical Circuit

Nodal Analysis or Nodal Voltage Analysis is used for determining the voltage or potential difference between the nodes in the circuit.

The nodal method of circuit analysis is based on Kirchhoff’s Current Law. Nodal Voltage Analysis uses the Nodal equations of Kirchhoff’s current law to find the voltage potentials around the circuit.

Nodal analysis basically aims at choosing a reference node in the network and then finds the unknown voltages at the other nodes w.r.t. reference node.

Let’s consider the circuit as shown in Fig.1.

Fig.1

In the nodal analysis method, one of the nodes (a node is a point in a network where two or more circuit elements meet) is taken as the reference node.

The potential of all the points in the circuit are measured with respect to this reference node.

In Fig.1, A, B, C and D are four nodes and node D has been taken as the reference node.

Here, node D is is an obvious choice as it is a ground or common node.

The voltage at node A and C w.r.t reference node D are already known.

Voltage at node A w.r.t. node D = E_{1 }= 100 V

Voltage at node C w.r.t. node D = E_{2} = 50 V

The only potential of node B w.r.t. node D (V_{B})is unknown.

If we can find V_{B}, each branch current can be determined as the voltage across each resistor would be known.

In Fig.1, The voltage V_{B }can be found by applying Kirchhoff’s current law at point B.

I_{1} + I_{2 }= I_{3 }……………. Eq.(1)

In mesh ABDA, the voltage drop across R_{1 }= E_{1 }– V_{B}

Or, I_{1} R_{1 }= E_{1 }– V_{B}

Or, I_{1} = (E_{1 }– V_{B})/ R_{1}

In mesh CBDC, the voltage drop across R_{3 }= E_{2 }– V_{B}

Or, I_{2} R_{3 }= E_{2 }– V_{B}

Or, I_{2} = (E_{2 }– V_{B})/ R_{3}

Also current I_{3 }= V_{B }/ R_{2}

Now, putting the values of I_{1}, I_{2} and I_{3} in Eq.(1), we get

[(E_{1 }– V_{B})/ R_{1}] + [(E_{2 }– V_{B})/ R_{3}] = V_{B }/ R_{2}

All quantities except V_{B }are known. Hence, we can find out the value of V_{B }.

Once V_{B }is known, all branch currents can be calculated.

Example : Consider the circuit as shown in Fig.2 below.

Fig.2

In the above circuit, node D is chosen as the reference node.

Applying Kirchhoff’s current law at point B, we get ,

I_{1}+ I_{2 }= I_{3 }……………. Eq.(1)

Now, putting the values, We get,

I_{1} = (20_{ }– V_{B})/ 20

Similarly , I_{2} = (40_{ }– V_{B})/ 40

And I_{3 }= V_{B }/ 80

Putting the values of I_{1}, I_{2} and I_{3} in Eq.(1), we get

[(20_{ }– V_{B})/ 20] + [(40_{ }– V_{B})/40] = V_{B }/ 80

Or, [1_{ }– V_{B}/ 20] + [(1_{ }– V_{B}/40] = V_{B }/ 80

Or, 2 = V_{B }[(1/80)+(1/40)+(1/20)]

Or, 2 = V_{B }[7/80]

Or, V_{B }= 160/7 = 22.8 V

Hence, I_{3 }= 22.8/80

=0.285 A