Engr. Jay S. Villan

Electric Circuits is one of the hardest subject, we really don’t know what does it mean or what does circuit all about. We are so very nervous on what will going to happen when we encounter different methods and different topics under you. We are so thankful because we were given a chance to meet you sir, we are so lucky and blessed because you are our professor. We are so thankful. We coudldn’t deny that we learned a lot, even though we met for just a short time, We are so Happy. You are one of the teacher that is ready to “jamming ” with the students :))

An Inspirer. An Empower. An Engager. These three characteristics are just a short sample of the many you demonstrate with all of your students, including me, every single day.Thank you for your patience. Thank you for your time. Thank you for everything sir.

Godbless you on your Journey, May God Bless you and We will miss you :)) Thank you sir.

Nodal Analysis

What is NODAL ANALYSIS?

Nodal analysis provides a general procedure for analyzing circuits using node voltages as the circuit variables. Choosing node voltages instead of element voltages as circuit variables is convenient and reduces the number of equations one must solve simultaneously. To simplify matters, we shall assume in this section that circuits do not contain voltage sources. Circuits that contain voltage sources will be analyzed in the next section.

In nodal analysis, we are interested in finding the node voltages. Given a circuit with “n” nodes without voltage sources, the nodal analysis of the circuit involves taking the following three steps.

“NODAL ANALYSIS W/OUT VOLTAGE SOURCES”

Steps to Determine Node Voltages:
Select a node as the reference node. Assign voltages v1,v2,…,vn−1 to the remaining n−1 nodes. The voltages are referenced with respect to the reference node.

Apply KCL to each of the n−1 nonreference nodes. Use Ohm’s law to express the branch currents in terms of node voltages.

Solve the resulting simultaneous equations to obtain the unknown node voltages.

Take Note:

• Current flows from a higher potential to a lower potential in a resistor.

We can express this principle as:

i = (Vhigher−Vlower) / R

“NODAL ANALYSIS WITH VOLTAGE SOURCES”

We now consider how voltage sources affect nodal analysis. Consider the following two possibilities.

CASE 1:
If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source.

Thus our analysis is somewhat simplified by this knowledge of the voltage at this node.

CASE 2:
If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a generalized node or supernode; we apply both KCL and KVL to determine the node voltages.

• A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it.

Note the following properties of a supernode:

1. The voltage source inside the supernode provides a constraint equation needed to solve for the node voltages.
2. A supernode has no voltage of its own.
3. A supernode requires the application of both KCL & KVL.

Example:

First, select a reference node and label the other nodes. Since each node has the same number of connected branches (4), we’ll simply choose the bottom node as the reference. There are only two other nodes, which we will label V1 and V2.

Now write KCL at each node (except the reference):

KCL at V1:

-5A + V1/5 + (V1-V2)/10 + [V1-(V2+4)]/10 = 0

Note that there are four terms in the equation, one for each branch leaving the node. The terms list the current leaving right, down, left, and up.

KCL at V2:

(V2-V1)/10 + V2/2 – 2A + [V2-(V1-4)]/10 = 0

Note that there are four terms in the equation, one for each branch leaving the node. The terms list the current leaving right, down, left, and up.

Now gather terms (multiplying through by 10 to clear up the fractions):

4V1 – 2V2 = 54
-2V1 + 7V2 = 16

Now solve the set of 2 equations with 2 unknowns.

V1 = 17.08V
V2 = 7.17V

We can now determine the current through the 5 ohm by Ohm’s law:

I = V1/5 = 3.41A

The current through the 4V source can be found as:

I = [V1-(V2+4)]/10 = 0.59A

SUMMARY:

Studying Nodal Analysis and the knowledge in KCL and KVL are very important

1. First we discuss what is Nodal analysis and How important is the application of Nodal analysis in every circuit.
2. We also discuss what is the process in getting the Voltage using nodal analysis with Voltage and without voltage source or in the other term for With voltage source is Supernode. Supernode occurs when there is a voltage source between to corresponding nodes.

LEARNINGS :

The aim of nodal analysis is to determine the voltage at each node relative to the reference node (or ground). A node is all the points in a circuit that are directly interconnected. We assume the interconnections have zero resistance so all points within a node have the same voltage.

1. The knowledge in KCL & KVL are very important to undergo with the Nodal Analysis.
2. We can assign one reference and non-reference nodes anywhere we want in a given number of nodes.
3. Shortcut method in getting the equations can executed in the circuit with or without supernodes.

Wye-Delta Transformation

The Delta-Wye transformation is an extra technique for transforming certain resistor combinations that cannot be handled by the series and parallel equations. This is also referred to as a Pi – T transformation.

What is wye-delta Transformation?

This topic is very important. Sometimes we are not sure in electric circuits  that the resistors are neither parallel or series. In many circuit applications, we encounter components connected together in one of two ways to form a three-terminal network: the “Delta,” or Δ (also known as the “Pi,” or π) configuration, and the “Y” (also known as the “T”) configuration.

There are several equations used to convert one network to the other:

EXAMPLE:

EXAMPLE:

Each resistors in Y network is the product of two adjacent branches divided by the 3 resistors.

EXAMPLE:

After the Δ-Y conversion

If we perform our calculations correctly, the voltages between points A, B, and C will be the same in the converted circuit as in the original circuit, and we can transfer those values back to the original bridge configuration.

Learnings:

Wye Delta Transformation is very important in Circuits, sometimes we are not sure in electric circuits  that the resistors are neither parallel or series. In many circuit applications, we encounter components connected together in one of two ways to form a three-terminal network: the “Delta,” or Δ (also known as the “Pi,” or π) configuration, and the “Y” (also known as the “T”) configuration.

• “Delta” (Δ) networks are also known as “Pi” (π) networks.
• “Y” networks are also known as “T” networks.
• Δ and Y networks can be converted with the proper resistance equations. By “equivalent,” I mean that the two networks will be electrically identical as measured from the three terminals (A, B, and C).
• A bridge circuit can be simplified to a series/parallel circuit by converting half of it from a Δ to a Y network. After voltage drops between the original three connection points (A, B, and C) have been solved for, those voltages can be transferred back to the original bridge circuit, across those same equivalent points.

Basic Laws in Series and Parallel Resistor (Voltage and Current Division)

Individual resistors can be connected together in either a series connection, a parallel connection or combinations of both series and parallel, to produce more complex resistor networks whose equivalent resistance is the mathematical combination of the individual resistors connected together. Circuits consisting of just one battery and one load resistance are very simple to analyze, but they are not often found in practical applications. Usually, we find circuits where more than two components are connected together.

There are two basic ways in which to connect more than two circuit components: Series& Parallel:

SERIES RESISTORS AND VOLTAGE DIVISION

A series circuit is a circuit in which resistors are arranged in a chain, so the current has only one path to take. The current is the same through each resistor. The total resistance of the circuit is found by simply adding up the resistance values of the individual resistors: equivalent resistance of resistors in series : R = R1 + R2 + R3 + …

A series circuit is shown in the diagram above. The current flows through each resistor in turn. In this circuit the electrons flow in a counter-clockwise direction, from point 4 to point 3 to point 2 to point 1 and back around to 4.
If the values of the three resistors are:

With a 10 V battery, by V = I R the total current in the circuit is:
I = V / R = 10 / 20 = 0.5 A. The current through each resistor would be 0.5 A.

Voltage Division

• v1= iR1 & v2 = iR2

KVL:

• v-v1-v2=0
• v= i(R1+R2)
• i = v/(R1+R2 ) =v/Req
• or v= i(R1+R2 ) =iReq
• iReq = R1+R2

VOLTAGE DIVISION FORMULA

v1 = iR1   &   v2 = iR2

i = v/(R1+R2 )

Thus:

v2=vR2/(R1+R2)

To solve for Req:

Req = R1+R2 …

Parallel Resistors & Current Division

A parallel circuit has more than one resistor (anything that uses electricity to do work) and gets its name from having multiple (parallel) paths to move along . Charges can move through any of several paths. If one of the items in the circuit is broken then no charge will move through that path, but other paths will continue to have charges flow through them.

A parallel circuit is a circuit in which the resistors are arranged with their heads connected together, and their tails connected together. The current in a parallel circuit breaks up, with some flowing along each parallel branch and re-combining when the branches meet again. The voltage across each resistor in parallel is the same.
The total resistance of a set of resistors in parallel is found by adding up the reciprocals of the resistance values, and then taking the reciprocal of the total:
equivalent resistance of resistors in parallel:
1 / R = 1 / R1 + 1 / R2 + 1 / R3 +…

A parallel circuit is shown in the diagram above. In this case the current supplied by the battery splits up, and the amount going through each resistor depends on the resistance. If the values of the three resistors are:

v = i1R1 = i2R2
i   = i1+ i2

= v/R1+ v/R2

= v(1/R1+1/R2)

=v/Req

• v  =iReq
• 1/Req = 1/R1+1/R2
• Req = R1R2 / (R1+R2 )

CURRENT DIVISION FORMULA

v = i1R1 = i2R2

v=iReq = iR1R2 / (R1+R2 )

and i1 = v /R1  &  i2 =v/ R2

Thus:

i2= iR1/(R1+R2 )

To solve for Req:

Req = 1/R1+  1/R2 …  1/Rn

Example:

1. Total resistor value:

2. The total current can be calculated as

3. The current  in each branch

verify that,

4. The power dissipated by each resistor

or

or

or

CONDUCTANCE

Conductance is an expression of the ease with which electric current flows through a substance. In equations, conductance is symbolized by the uppercase letter G. The standard unit of conductance is the siemens (abbreviated S), formerly known as the mho.

Series conductance:

1/Geq = 1/G1 +1/G2+…

Geq = G1 +G2+…

LEARNINGS:

• In a series circuit, all components are connected end-to-end, forming a single path for electrons to flow.
• In a parallel circuit, all components are connected across each other, forming exactly two sets of electrically common points.
• When two resistors R1 = (1/G1) and R2 = (1/G2)  are in series, their equivalent resistance Req and equivalent conductance Geq are:
•  When two resistors R1 = (1/G1) and R2 = (1/G2)  are in parallel, their equivalent resistance Req and equivalent conductance Geq are:
•  The voltage division principle for two resistors in series is:
•  The current division principle for two resistors in parallel is:

FUNDAMENTAL LAWS THAT GOVERN ELECTRIC CIRCUITS

Introduction:

Fundamental laws that govern electric circuits:

• Ohm’s Law
• Kirchoff’s Law

These laws form the foundation upon which electric circuit analysis is built.

Common techniques in circuit analysis and design:

• Combining resistors in series and parallel.
• Voltage and current divisions.
• wye to delta and delta to wye transformations

THESE TECHNIQUES ARE RESTRICTED TO RESISTIVE CIRCUITS.

What is Ohms law?

Relationship between current and voltage with in a circuit element.

The voltage across an element is directly proportional to the current flowing through it      → v α i

Thus:: v=iR   and R= v/i

where:

• R is called resistor
• Has the ability to resist the flow of electric current
• Measured in Ohms (Ω)

NOTE THAT:

•  RESISTOR HAS NO POLARITY
• THE RESISTANCE VARIES FROM  0 TO ∞
• ONLY LINEAR RESISTORS OBEY THE LAW

Conductance

Conductance is the extrinsic property. This means that conductance is the property of an object dependent of its amount/mass or physical shape and size.

• The unit used is mho or siomens (S).
• It has the ability to conduct electric current

Power

R and G are positive quantities, thus power is always positive (+)  such that R absorbs power.

Solutions:

For voltage:

v=iR

v = (2*10^-3)(10*10^3)

v = 20 volts

For Conductance:

G=i/R

=1/10*10^3

G=1*10^-4

For Power:

P=t^2R

=v^2/R

=(2*10^-3)^2 (10*10^3)

=0.04 w

=40mW

=(20)^2/10*10^3

P=0.04 w

Solution:

(a) i = 3/100 = 30 mA

(b) i = 3/150 = 20 mA

NODES, BRANCHES AND LOOPS

Elements of circuit can be interconnected in several ways and we need to understand the basic topology of this.

A node is the point of connection between two or more branches.

Node is indicated by dot sign. When a short circuit has two nodes it actually becomes one node. If we redraw the first circuit as it has two common points shown in black color filled.

After redrawing the circuit  becomes as below circuit. It shows three nodes a, b, c.

Solutions

Re draw the given figure:

5 branches

• 1 voltage source
• 1 current source
• 3 resistors

3 nodes

• a
• b
• c

Another example:

There were 7 branches, 4 nodes, and 10 loops all in all.

Kirchoffs Circuit Law

Kirchhoff’s First Law

The law states that at any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.

Examples:

• Given the following circuit, write the equation for currents.

• Current in a close boundary.

• Use KCL to obtain currents i1, i2, and i3 in the circuits.

A= 1A+i1=2A+10A

i1=12A-1A

i1=11A

B= 3A+2A=1A+i2

5A=1A+i2

i2=4A

C= i3+10A=11A

i3=11A-10A

i3=1A

Kirchhoff’s Second Law

This animation shows a multiple loop circuit which illustrates Kirchhoff’s second law which states that the sum of the potential differences as you go around the loop is zero.

Kirchoff’s Voltage Law (KVL)

• Applied to a loop in a circuit.
• According to KVL ( the algebraic sum of voltage rises and drops in a loop is zer0.

+20+25-10-v1=0

V1=35v

+10-15+v2=0

v2=-10+15

v2=5V

+v1-5-v3=0

35-5-v3=0

v3=30V

Solution:

+10v+12v+6v-v1-v1=0

28v-2V1=0

28V/2=2V1/2

V1= 14V

+V2-12V-10V=0

+12V-22V=0

V2= 22V

Solution:

Power= iv = i²R = v²/ R

V 5Ω = 5i

45-10i + 3vo – 5i =0

vo = 10i

45-Vo + 3vo -5i = 0

i = Vo/10

45-Vo +3vo -5 (Vo/10) = 0

Summary:

There are two basic laws that were introduced to us.

The first law is the “Ohms Law” which states that the voltage across a resistor is directly proportional to the current flowing to through the resistor, represented by and equation V=IR. This law deals with the resistance (R).

An element with R=0 is called “Short Circuit” which means the resistance is almost or approaching to zero. If the resistance of the circuit element is approaching to infinity it is called “Open Circuit” this means that the current cannot flow because the path has been interrupted.

The second law is the “Kirchhoff’s Law“, which is divided into two parts, the “Kirchhoff’s Current Law“, which is the sum of the currents entering a node is equal to the sum of the current leaving the node, and the “Kirchhoff’s Voltage Law”, which states that the sum of voltage drop around the loop is equal to the sum of the voltage rises in the same loop.

There are three elements in a circuit.

The first element is the “Branch”; it represents a single element (Volt source or Resistor). Second is the “Node”, it is the point of connection between two or more branches. And lastly the “Loops“, which is any closed path in a circuit.

Ohm’s law states that the current through a conductor between two points is directly proportional to the potential difference across those two points. It means that more the resistance lesser current would flow. I=V/R This would apply to any component of a circuit. For example conductors would increase the current flow and the inductors would decrease it.

Introduction to Electric Circuits

Electric Circuit is an interconnection of Electrical elements. Electric circuit theory and     electromagnetic theory are the two fundamental theories upon which all branches of   electric engineering are built. The basic Electric circuit theory course is the most             important course for an engineering student.

Objectives:

• To define what is Electric Circuit
• To be able to know what are the Basic concepts in a Circuit
• Define Circuit Elements
• and To solve problems involving charge and current

Electric Circuits

An electrical circuit is a network consisting of a closed loop, giving a return path for the current. Linear electrical networks, a special type consisting only of sources (voltage or current), linear lumped elements (resistors, capacitors, conductors), and linear distributed elements (transmission lines), have the property that signals are linearly super imposable.

Here is an example of a Simple Circuit. Composed of three things: the Power source or the battery , the Switch and the Load. A circuit is simply a closed loop through which charges can continuously move.

Here is another example of a closed circuit and an open circuit. There is a difference between the two which the electrons flow. Electrons do not flow through an open circuit. Electrons continuously flow through a closed circuit

Basic Concepts:

CHARGE -q(t)

• Basic quantity in an Electric circuit and defined as electrical property of materials.
• Each electron carries energy with it.
• Composed of 2 charges: Negative (electron) and Positive (proton)
• Measured in coulumbs (c)
• One electron has a charge of -1.602 * 10 ^-19 c
• Charge may be constant or varrying.

In a circuit, the total Charge Q, is calculated by finding how much charge flows in the circuit: ( derive the formula given)

CURRENT – i(t)

• Charge flow rate. In electric circuits this charge is often carried by moving electrons in a wire.
• Measured in Ampere (A)
• There are two types of current: DC (Direct Current) and AC (Alternating Current)

VOLTAGE -v(t)

Is the electric potential difference between two points, or the difference in electric potential energy of a unit charge transported between two points.

• charge rate of doing work
• Energy required to move a unit charge through an Element
• Measured in Volts (v)

POWER p(t)

• time rate of doing work
• measured in watts

Power can be absorbed or supplied by circuit elements.

Positive power- element absorb power

Negative power- element supplies power

• sign determined by voltage
• An ideal circuit is ΣPsupplied + ΣPabsorbed = 0

ENERGY

Is energy newly derived from electrical potential energy. When loosely used to describe energy absorbed or delivered by an electrical circuits “electrical energy” refers to energy which has been converted from electrical potential energy. – Measured in joules (J).

CIRCUIT ELEMENTS

an element is the basic building block of a circuit.

Types of Elements:

1. Active Elements – Capable of generating energy. Examples are batteries & generators. –Negative Power
2. Passive Elements – Absorbs energy. Examples are capacitors &inductors. – Positive Power

SOURCE

• 2 types of source: Independent and dependent Source.

Independent Source- does not depend to other elements to supply voltage or current.

Dependent Source- Depend to other elements to supply Voltage or Current.

EXERCISES

1. Determine the current flowing through an element if the charge flow is given by: q(t) = (9t² +2t-2) C

Solution

2(t) = dq/dt = 18t +2

2. Find the charge q(t) flowing through a device if the current (i(t)) = (2t + 5) mA and                   a(0)=0.

Solution

i=dq/dt                                                 q(t) = t² + 5t +c                             q(0)= (0)²+ 5(0) +c+0

∫dq= ∫i dt                                             q(0) =0                                           c=0

q= ∫dt                                                    q(0)=t² +5t +c=0

3. The change entering a certain element is shown. Determine the current at:
t= 1 ms, 6 ms and 10 ms.

(a) At t = 1ms,
i= dq/dt = 80/2=  40 A
(b) At t = 6ms,
i= dq/dt= 0 A

(c) At t = 10ms,
i= dq/dt = 80/4 = 20 A

LEARNINGS :
This is our first topic in our first Electrical subject.We all know about electricity. It is the flow electrons. Hence the word electricity is derived from the word electrons. A battery itself doesn’t work, if you hold it up in air, there would be no passage of electric current. Hence you need a circuit. In Simple terms an electronic circuit is a closed pathway for electrons to flow.
The Electric Current in a circuit flows from positive to negative while electrons flow from negative to positive. So when the switch is on the path is complete and electricity passes through enabling the bulb to light up, while when the switch is not on, there is a break in the flow of electricity and the bulb does not light up.
You would find circuits in every electrical device you use – from your switch board, television, refrigerator to your laptop.