nodal analysis

Nodal analysis with complex

              Nodal analysis defines as knowing the current and voltage using nodes, where nodes are set to be V1, V2, V3 …....and Vn, and find the current of each elements that are connected to the nodes. In nodal analysis we have to consider the cases that are given: Case 1: If the voltage source (dependent or independent) is connected between two non-reference nodes, the two non-reference nodes form a generalized node or super node. We apply both KCL and KVL to determine the node voltages. Case 2: if a voltage source is connected between the reference node and a non-reference node, we simply set the voltage at the non-reference node equal to the voltage of the voltage source.

The total current entering a node equals the total current leaving a node.

Example:

             On this circuit we are required to get the ix that passes through the capacitor 0.1F.

                                V= 20∠0   ω= 4

            In this circuit we can produced two nodes (Node 1 and Node 2). Before we going to redraw our circuit we have to express our Inductor and Capacitor Impedances into complex number.

         Let Zc be the impedance of the capacitor

         Let Zh be the impedance of 1 H inductor 

         Let Zn be the impedance of 0.5 H inductor

                Zc =   1  /  jωC    =   1  /  j(4)(0.1)    

                                          =    -j2.5

                Zh =     jωL    =    j(4)(1)    

                                          =    -j4

                Zn =     jωL    =    j(4)(0.5)    

                                          =    j2

Let's redraw our circuit;

            We have already our circuit that is expressed in complex number. We can now apply nodal analysis.

                  @ node 1

           (V2 - V1)  /  j4   =     (V1 - 20∠0) / 10   +     V1  /  (- j2.5)

          10 (  (V2 - V1)  /  j4   =     (V1 - 20∠0) / 10   +     V1  /  (- j2.5))

Yields,

          (j1.5 + 1) V1 + j2.5 V2  = 20 ----->    Equation 1

                    @ node 2

          (V2 - V1) / j4    +    (V2/j2)   =  2ix

          ix = V1/ -j2.5

          (V2 - V1) / j4    +    (V2/j2)   =  2 (V1/ -j2.5)

          20((V2 - V1) / j4    +    (V2/j2)   =  2 (V1/ -j2.5))

          

Yields,

          j11V1 + j15V2 + 0   ------------->  Equation 2

Apply matrix;

                   (j1.5 + 1)         j2.5            =      20

                        j11             j15             =      0

Δ =         (j1.5 + 1)(j15 )   -   (  j11)( j2.5)

Δ =         5 + j15

next is to solve for Δ1

Δ1 =  20( j15 )  -   ( j2.5 )(0)

Δ1 =   j300

So,

            V1=  (Δ1 /  Δ)  =    j300  /  ( 5 + j15 )

Yields,

           18.97∠18.43  V

                        I've learned that in nodal analysis we commonly used the kirchhoff's Current Law. As this Law is implemented to the circuit we first know the value of voltage since the voltages play as the variables of the equation of the given circuit. In super node, I learned that  the process in solving the circuit is just the same in solving DC circuit, but this time our voltages sources is in AC and it is expressed in time domain or phasor domain.