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 = E₁ = 100 V

Voltage at node C w.r.t. node D = E₂ = 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.

I₁  + I₂ = I₃ ……………. Eq.(1)

In mesh ABDA, the voltage drop across R₁ = E₁ – V_B

Or,      I₁ R₁ = E₁ – V_B

Or,       I₁ = (E₁ – V_B)/ R₁

In mesh CBDC, the voltage drop across  R₃ = E₂ – V_B

Or,      I₂ R₃ = E₂ – V_B

Or,       I₂ = (E₂ – V_B)/ R₃

Also current  I₃ = V_B / R₂

Now, putting the values of  I₁, I₂ and I₃ in Eq.(1), we get

[(E₁ – V_B)/ R₁] + [(E₂ – V_B)/ R₃] = V_B / R₂

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 ,

I₁+ I₂ = I₃ ……………. Eq.(1)

Now, putting the values, We get,

I₁ = (20 – V_B)/ 20

Similarly , I₂ = (40 – V_B)/ 40

And I₃ = V_B / 80

Putting the values of  I₁, I₂ and I₃ 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, I₃ = 22.8/80

=0.285 A