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.

impedance and admittance

Impedance (Z)

                       - of a circuit is the ratio of the phasor voltage v to the phasor current I, measured in ohms                                 (Ω).

Admittance (Y)

                       - is the reciprocal of impedance, measured in Siemens (S).

Impedances and Admittances of passive elements and Voltage relationships.

Element	Impedance	Admittance	Time domain	Frequency Domain
R	Z = R	Y = 1/R	v=Ri	V= RI
L	Z = jωL	Y= 1/(jωL)	v= L (di/dt)	V=jωLI
C	Z= 1/(jωC)	Y= jωC	v= C (dv/dt)	V=I/(jωC)

This table contains the important parameters in solving our circuits for this entire topic.

                         The impedance and admittance that were introduced to us, which means that in AC circuit we no longer computing only for the resistance but  the admittance of the circuit which is comprise of the resistance at the real value and then reactance at the imaginary value. And remember that a reactance is composed of both the positive reactance and inductive reactance that means the circuit has a capacitor involved and inductance there is an inductor involved.

Phasors

A Phasor is a complex number that represents the amplitude and phase of a sinusoid.

In our advanced mathematics we know that a complex number z can be written in rectangular form as z = x + iy where i = √-1 the imaginary number. For the whole discussion we will let i = j for our convention for the next topics. The real part  is x and y is Imaginary.

z = r ∠ θ = r(cos θ + j sin θ )   is the polar form.

z = x+jy                                    is rectangular form.

 To convert rectangular to polar form we use: r = √(x² + y²) and θ = tan^-1(y/x).

To convert polar to rectangular form we use:  x = r cos θ and y = r sin  θ.

Phase difference of voltage and current

Before we start, let’s consider first what is the meaning of LEADING and LAGGING in engineering world. They will say current is leading or current is lagging. What do you mean by that? The arrow in red represents the red phase of the sinusoidal of either current or voltage and the blue arrow represents the blue phase of the sinusoidal of either current or voltage of the same frequency.

We represent our phasors are stationary but in reality this phasors are rotating.

Now, the red phasor is leading the blue phasor and the blue phasor is lagging the red phasor at some angle. In other words;

Red leads blue by θ.

Blue lags red by θ.

There are some cases that blue leads red. It is by 360-θ degrees. To understand see figure below.

IF the given is, v(t)= 12 cos (50t + 10º)

We have to consider this sinusoidal voltage v(t) = Vm sin ωt were:

Vm = the amplitude of the sinusoid

 ω = the angular frequency in radians per second.

ωt = the argument of the sinusoidal

Example:

        

   10∠-30 + (3-j4)       /       (2+j4)(3-j5)'

                     

Solution:

            In the numerator our first term is in the form of polar form. In adding and subtracting complex numbers, it is better to perform it in rectangular form. So, we have to transform the polar form to rectangular form.

So,

               10 (cos (-30) + jsin (-30)) = 8.66-j5

  numerator;                        (8.66-j5) + (3-j4), (add like term)

                              Yields,

                                        

                                         11.6-j9  ---->  14.73∠-37.66 

              

             In the denominator the two terms are expressed in rectangular form but in multiplying and dividing complex number, it is better to perform the operation using the polar form. So, we have to transform rectangular form to polar form.

             (3-j5)' has prime, it could be Asterisk or Bar. This means that we have to express this into its conjugate.  (3-j5)' ----->  (3+j5).

So,

          (2+j4)(3+j5) =   26.08∠122.47

our new equation is :  

                        (14.73∠-37.66)  /     26.08∠122.47

Yields,

                        (0.56∠-160.13)

            

               I've learned that in analyzing AC circuit, complex number can be use and satisfies the conditions every circuit that we used to solve. The voltage and current impedance can be expressed into complex number which can be in rectangular form, polar form and exponential form.

sinusoids

Sinusoid is a signal that has the form of the sine or cosine function.

Sine Function

Cosine Function

The waveform repeats in certain interval we call it the period (T).

Sinusoid can be form voltage and current which is we usually used in our field.

EXAMPLE:

                   (11.66-j9)    /    (26.06 ∠22.47°)

to get the final answer, transform (11.66-j9) to a polar form so that it is easy because our denominator is in polar form. It is easy to multiply or divide a complex number when it forms in polar form.

                   (11.66-j9)  ----->  (14.73∠-37.66°)

so,       

                   (14.73∠-37.66°)    /    (26.06 ∠22.47°)

yields,

                   

                    (0.565∠-60.133°) 

                       I learned that a sine and cosine functions are very applicable in the circuit especially the signal which we called the sinusoid. And this sinusoid is occurring naturally or this can be seen in nature.