5th Week: "Mesh Analysis - AC"

 "Mesh Analysis"


Kirchhoff’s voltage law (KVL) forms the basis of mesh analysis.

Example:

The circuit shown is operating in the steady state.
Determine the four mesh currents.

First, convert all values to phasor notation.

Next, perform the mesh analysis.

Since dependent sources are present, define the parameters upon which they depend in terms of the mesh currents.

i₅ = I_C - I_D
V_L = j1K(I_A - I_B)

Since current sources are present, express their values in terms of the mesh currents, then form supermeshes.

500m0° = I_C - I_A
500m0° = - I_A + I_C (Equation 1)
12 i₅ = - I_B
12 (I_C - I_D) = - I_B
0 = I_B + 12 I_C - 12 I_D (Equation 2)

Now write KVL around each remaining mesh and supermesh.
Supermesh A/C

6 V_L + j1K(I_A - I_B) - j200 (I_C - I_D) + j5K I_C + 10K I_C + 4K I_A = 0
6 j1K(I_A - I_B) + j1K(I_A - I_B) - j200 (I_C - I_D) + j5K I_C + 10K I_C + 4K I_A = 0
0 = (4K + j7K) I_A - j 7K I_B + (10K + j4800) I_C + j200 I_D (Equation 3)

Mesh D

0 = - j400 (I_D - I_B) + 1K I_D - j200 (I_D - I_C)
0 = j400 I_B + j200 I_C + (1K - j600) I_D (Equation 4)

Inserting Equations 1 through 4 into the calculator and letting it do the difficult part yields

I_A = 429 m169°
I_B = 337 m127°
I_C = 113 m45.8°
I_D = 120 m59.2°

Converting back to the time domain gives

I_A(t) = 429 cos (10000 t + 169°) mA = - 429 cos (10000 t - 11°) mA (Either form acceptable)
I_B(t) = 337 cos (10000 t + 127°) mA = - 337 cos (10000 t - 53°) mA (Either form acceptable)
I_C(t) = 113 cos (10000 t + 45.8°) mA
I_D(t) = 120 cos (10000 t + 59.2°) mA

Reflection:

• I learned that all the concepts and process in solving the unknowns in Ac Analysis are the same with Dc Analysis. It only involves complex numbers.

Videos:

For more information, watch the video below:

That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 321

Professor:

ENGR. JAY S. VILLAN, MEP - EE

4th Week: "Nodal Analysis - AC"

"Nodal Analysis"


The basis of nodal analysis is Kirchhoff’s current law. Since KCL is validfor phasors, we can analyze ac circuits by nodal analysis.

Example:

Solution,

1. Convert the circuit to phasor domain:

Given a frequency of 1MHz, the radian frequency is 1MHz*2*pi = 6.28Mrad/s. Thus we have:

2. Solve for the desired value in the phasor domain. Let's use node voltage to find the voltage over the capacitor. Choose the bottom node as the reference and the top node as "V", giving us the following circuit:

The single KCL equation we get is:
(V - 12<0)/1k + V/-j795.8 - 20m<0 = 0

Solve for V. First, factor out V and put the constants on the right side.
V(1/1k +j/795.8) = 20m + 12/1k

Now add the constants and rearrange:
V ( 1E-3 +j1.26E-3) = 32m
V = 32m/(1E-3 + j1.26E-3)

Next, to find the final answer, we could either (a) multiply numerator and denominator by the conjugate of the denominator or (b) convert the numerator and denominator to polar format and then divide. I choose to do the latter:

V = 32m<0 / 1.6m<51.6
V = 20 < -51.6

3. Convert answer to time-domain. Simply copy over the form (cos/sin) and the frequency (1MHz) and fill in our solution for magnitude and phase:
Vc(t) = 20 sin(wt - 51.6 degrees)



Reflection:

• I learned that all the concepts and process in solving the unknowns in Ac Analysis are the same with Dc Analysis. It only involves complex numbers.

Videos:

For more information, watch the video below:

That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 321

Professor:

ENGR. JAY S. VILLAN, MEP - EE

3rd Week: "Impedance Combinations"

"Impedance Combinations"


Consider the N series-connected impedances. The same current I flows through the impedances. Applying KVL around the loop gives,

V = V1 + V2 + · · · + VN = I(Z1 + Z2 + · · · + ZN )

The equivalent impedance at the input terminals is,

Zeq = VI = Z1 + Z2 + · · · + ZN

showing that the total or equivalent impedance of series-connected impedances is the sum of the individual impedances. This is similar to the series connection of resistances.

The equivalent admittance is,

Yeq = Y1 + Y2 + · · · + YN

"Delta-to-Wye and Wye-to-Delta Transformations"
The delta-to-wye and wye-to-delta transformations that we applied to resistive circuits are also valid for impedances. The same process like how we combine resistors.

A delta or wye circuit is said to be balanced if it has equal impedances in allthree branches.


Reflection:

• I learned that all the concepts and process in solving the combinations in Ac Analysis are the same with Dc Analysis. It only involves complex numbers.

Videos:

For more information, watch the video below:

That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 321

Professor:

ENGR. JAY S. VILLAN, MEP - EE

2nd Week: "Phasor in Circuit Elements, Impedance & Admittance"

"Phasor Relationship for Circuit Elements"

Now that we know how to represent a voltage or current in the phasor or frequency domain, one may legitimately ask how we apply this to circuits involving the passive elements R, L, and C. What we need to do is to transform the voltage-current relationship from the time domain to the frequency domain for each element. Again, we will assume the passive sign convention.


We begin with the resistor. If the current through a resistor R is i = Im cos(ωt + φ), the voltage across it is given by Ohm’s law as

v = iR = RIm cos(ωt + φ)

Table 9.2 summarizes the time-domain and phasor-domain representations of the circuit elements.

"Impedance & Admittance"

In the preceding section, we obtained the voltage-current relations for the three passive elements as,

V = RI, V = jωLI, V =I/jωC

These equations may be written in terms of the ratio of the phasor voltage to the phasor current as,

V/I = R,  V/I = jωL,  V/I = 1/jωC

From these three expressions, we obtain Ohm’s law in phasor form for any type of element as

Z = V/I    or   V = ZI

The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms.

The admittance Y is the reciprocal of impedance, measured in siemens (S).


The admittance Y of an element (or a circuit) is the ratio of the phasor current through it to the phasor voltage across it, or

Y   =   1/Z   =   I/V

Reflection:

• I learned that the impedance is equal to voltage over current just like the Ohm's Law.
• The equivalent impedance of a capacitor is a negative value.
• The admittance is the reciprocal of impedance.
• Time domain is independent to time while phasor domain is dependent.

Videos:
~

 For some information, watch the videos below:

That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 321

Professor:

ENGR. JAY S. VILLAN, MEP - EE

1st Week: "Sinusoids & Phasors"

What is a Sinusoid & Phasor?


Sinusoid
 

Thus far our analysis has been limited for the most part to dc circuits, those circuits excited by constant or time-invariant sources. We have restricted the forcing function to dc sources for the sake of simplicity, for pedagogic reasons, and also for historic reasons. Historically, dc sources were the main means of providing electric power up until the late 1800s. At the end of that century, the battle of direct current versus alternating current began. Both had their advocates among the electrical engineers of the time. Because ac is more efficient and economical to transmit over long distances, ac systems ended up the winner. Thus, it is in keeping

with the historical sequence of events that we considered dc sources first. We now begin the analysis of circuits in which the source voltage or current is time-varying. In this chapter, we are particularly interested in sinusoidally time-varying excitation, or simply, excitation by a sinusoid. 

  

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

v(t) = Vm sinωt

Vm=the amplitude of the sinusoid

ω=the angular frequency in radians/s

ωt=the argument of the sinusoid
 

A periodic function is one that satisfies f(t) = f(t+nT), for all t and for all integers n.

Remember,

Conversions, 

sin (ωt±180◦ )= −sinωt

cos (ωt±180◦) = −cosωt

sin (ωt±90◦) = ±cosωt

cos (ωt±90◦) = ∓sinωt

Phasors

Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions.

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

A complex number z can be written in rectangular form as,

z = x+jy (rectangular form)

z=r<φ (Polar form)

z=re^jφ (Exponential form) 

Reflection:

• Sinusoids are in phase if the sum of their amplitudes is not equal to zero. Sinusoids are out-of-phase if the sum of  their amplitudes is equal to zero.
• A phasor is a complex number that represents the amplitude and phase of a sinusoid.

Video:

For more information, watch the video below:

That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 311

Professor:

ENGR. JAY S. VILLAN, MEP - EE