Nodal Analysis:

From topic name it is understood that there is something about nodes. The variables in the circuit are selected to be the node voltages. The node voltages are defined with respect to a common point in the circuit. One node is selected as a reference node and all other nodes are defined with respect to that one. The selected one is grounded to make it’s potential zero usually we select a node with more number of branches. To better understand this thing let’s take an example

 

In this figure our reference node is node 5 to simplify first node

      -Vs+V1+Va=0

V1=Vs – Va

                                                                                                 V1=12 – 3            =9V

for Node 2 it is

   V3=Va – Vb

for Node 3 it is

V5=Vb – Vc

for current I1 it is

I1=V1/R       =Vs – Va / R

for current I3 it is

I3=V3/R     =Va – Vb / R

for current I5 it is

 I5=V5/R       =Vb – Vc / R

for current I2 it is

 I2=Va – 0 / R

for current I4 it is

I4=Vb – 0 / R

Circuits with Independent Current Source:

We already discussed difference between independent and dependent sources. Apply KCl for Node 1

Is1=I1 + I2  -(1)

Is1=(V1 – 0) / R + (V1 – V2) / R

for Node 2 eq. takes form

I2 = Is2 + I3    -(2)

Is2 = I2 – I3

Is2 = (V1 – V2) / R – (V2 – 0) / R

Note:We can also write these equations in form of G (conductance)

Circuits with Independent Voltage Source:

In this figure we have two independent voltage sources same as previous one but apply KVL in this figure we can clearly see that V1=4V because the source voltage is on the node if there is not any more branch same for V3 which is 2V applying KVL

(V2 – V1) / R + (V2 – 0) / R + (V2 – V3) / R  -(1)

by simplifying we got                                                        V2=3 / 2 V

 Loop Analysis:

What are loops we already know but for which purpose we use them now we learn. Complex circuit solving require a lot of techniques and loop analysis is one of them. A loop analysis require KVL to determine number of loops in the circuit.

it is clearly seen that i1 = iA , i3 = iB , i8 = – iD , i6 = – iC, i8 and i6 are negative because there directions are opposite to each other.

i2 = iA – iB

i4 = iA – iC

i5 = iB – iD

i7 = iC – iD

so it is very simple and easy way to solve a complex problem with loop analysis.

Circuits with Independent Voltage Source:

This circuit have two loops , six nodes and seven branches. There are two independent voltage sources in this circuit let’s start and simplify them. Applying KVL on first loop we got equation of form

– vs1 + v1 + v3 + v2 = 0 -(1)

Applying KVL on second loop

vs2 + v4 + v5 – v3 = 0 -(2)

so applying ohm’s law

v1 = i1 . R1 , v2 = i1 . R2 , v3 = (i1 – i2) . R3 , v4 = i2 . R4 , v5 = i2.R5

inserting these values in above equations we can easily calculate values of vs1 and vs2.

 

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