NETWORK ANALYSIS (4) thevenin's theorem

Thevenin’s Theorem

This is a very useful theorem which was introduced by French Engineer ML Thevenin(1857-1926).

Thevenin’s Theorem:

Any linear active bilateral network can be replaced by an equivalent of a voltage source in series with a resistance. The voltage source is open circuited voltage across the open circuited load terminals and the resistance being the internal resistance of the source network looking from the open circuited load terminals.

Of course it is little bit difficult to understand at once. It can be simplify as follows

This theorem says that, a given network when viewed from its any two terminal points can be replaced by a single voltage “V^th” source in series with a single resistance “R^th”.

See figure 10.1

Consider all the elements in the black box are not visible to the out side and we only have the two terminals (AB) as the output. All the elements in this black box can be represented by an equivalent single voltage source with a single series resistance. This voltage is called thevenin’s voltage and resistance is thevenin’s resistance. Let’s see how to find the values of this voltage and resistance.

The voltage across AB is called the thevinings voltage. We can use the analysis methods and laws we have learned so far to find the voltage across AB.

In figure 10.1 I have marked the voltage of point A as ‘V^/’ . As we have grounded the point D 

(D = B)  V^/ = V^th

So we can use the nodal equation method to find V^/

Applying nodal equation to point A

(V^/ - 0)/ R₃ + (V^/ - 0)/ R₂ + (V^/ - V₁)/ R₁ = 0

So if V₁ , R₁ , R₂ , R₃ are given, then we can find V^/ which is equal to V^th.

Note that it is not necessary to ground the circuit. But grounding makes the calculation easier. So we have to ground the suitable point according to the problem

To find the equivalent resistance first we remove the voltage sources and short circuit it. If ant current sources are available we also remove those sources and open circuit it. Then we find the equivalent resistance as seen from the two terminals A&B.

So in this problem our  R^th  = R₁//R₂//R₃

Now we have find R^th

Finally we can simplify the network as shown in figure 10.2

Wait till my next post for solved problems of thevenin’s theorem.

 In this example we don’t have any current sources however in solved problems I will add some problems with current sources.

If you have any problems regarding this post, please leave it as a comment. I will reply you as soon as possible.

Pabindu lakshitha

B.Sc(Engineering Undergraduate)

NETWORK ANALYSIS (3)

Nodal equation and superposition theorem

Nodal equation method:

Nodal equation method makes the calculation of a network easier.  This is somewhat similar to the 1^st KCL, but little bit different. Actually the first law of KCL gives a relation of current flowing in to a node. See the previous post on KCL for more. Here we use the equation to find the potential of a point referred to another point (most usually a grounded point).

As the name implies, this method is applying to a node in the network. See figure 9.1

In this case we assume that current in each branch is leaving the point. See point B in figure 9.1. Current is leaving the point.

We mark the potential of point b is V and point D is grounded. Usually in networks we ground the circuit.

So we can write following equations,

(V-V₁)/R₁ = I₁

 (V-V₂)/R₂ = I₂

 (V-0)/R₃ = I₃

From KCL 1

I₁ + I₂ + I₃ = 0

By combining these equations we get the nodal equation

Apply nodal equation to point B

(V-V₁)/R₁  +  (V-V₂)/R₂  + (V-0)/R₃ = 0

So if we know the R values, V₁ & V₂ we can find V.

This method is frequently usin0g in network analyzing. Of course it is a very easier method to apply. We will use this method in many of our examples in future.

Super position theorem:

When there are number of voltage or current sources acting simultaneously in a network, we can use this theorem to simplify the calculation.

According to this theorem each source can be treated as if it acts independently of the other. Hence we can calculate the effect of one source at a time and then add them algebraically. Figure 9.2 will clearly clarify this theorem.

 So we can write the equation,

I = I1 + I2

This way you can calculate the current in any branch.

Using this method we can simplify our calculation though the network is too much complicated.

If you have any problems please leave it as a comment. I wii reply as soon as possible.

pabindu lakshitha

B.Sc (engineering undergraduate)

SOLVED PROBLEMS Kirchhoff law ( KCL )

SOLVED PROBLEMS ( KCL )

Q1)  Determine the value current in 40 Ohms resistance. Refer figure 8.1

Answer:

• First we have to apply KCL 1 to the network. See figure 8.2   

Note that there are three unknown variables ( I , I1 , I2 ) and therefore we need to solve three equations simultaneously  to find this unknown variables.

Apply KCL to Circuit ABDA

∑ IR   =   ∑ e.m.f 

( I - I1 )*10 + I2*40 - 20I = 0 

10*I - 30*I1 +40*I2 = 0 -------------------------(1)

Apply KCL to Circuit BCDB

(I - I1 - I2)*30 -  (I1+I2)*15  - I2*40 = 0 

30*I - 45*I1 - 85*I2 = 0 -------------------------(2)

Apply KCL to Circuit ADCEA

I1*20 + (I1+I2)*15 = 2

35*I1 + 15*I2 = 2     -----------------------------(3)

Now solve these three equations

I   = 87/785 A

I1 = 41/785 A

I2 = 9/785 A

So the current in 40 Ohms resistance is  9/785 A  From  B to D.

Q2) Determine the current flowing in each of the batteries and the voltage difference between 10 Ohms resistance. Refer the following figure.

Answer :

• First we have to apply KCL 1 to the network. See figure 8.3

Applying KCL to loop CEDAC

(I1+I2)*10 + I1*2 = 12 

12*I1 + 10*I2 = 12 ----------------------------(1)

Applying KCL to loop CEDBC

(I1+I2)*10 + I2*1 = 8 

10*I1 + 11*I2 = 8------------------------------(2)

By solving (1) &(2) 

I1 = 1.625A

I2 = -0.750A

Current through 10 Ohms resistance is = I1 + I2 = 0.875A

Voltage across 10 Ohms resistance is = 0.875A x 10 Ohms = 8.750V

Q3) Determine the current in 4 Ohms resistance. Refer Figure8.4

Note: This Network has a current source. So this question will teach you something new. To study about Current sources, see my previous post about "voltage and current sources".

Answer:

• First we have to apply KCL 1 to the network. See figure 8.5

Applying KCL to loop EFADE

I1*2 - I2*10 - (I - I1 - 6)*1 = 0 

I - 3*I1 + 10*I2 = 6 ------------------------------(1)

Applying KCL to loop ABCDA

(I1+I2+6)*2 - 3*(I - I1 - I2 - 6)  + I2*10 = -10

3*I - 5*I1 +15*I2 = 6 ---------------------------(2) 

Applying KCL to loop EDCGE

(I-I1-6)*1 + (I-I1-I2-6)*3 + I*4 = 24

8*I - 4*I1 3*I2 = 48 ----------------------------(3)

By solving these three equations

I = 4.1A

So the current in 4Ohms resistance = 4.1A 

  

NETWORK ANALYSIS 1 ( kirchhof's laws )

Kirchhoff laws (KCL)

There are two Kirchhoff laws. These two laws are somewhat comprehensive than Ohm's law. And also these laws can be used in more complicated networks.

 Kirchhoff's first law (Current law):

• The algebraic sum of current to a node in a electric circuit is equal to zero.

or

• In any network of conductors the algebraic sum of the currents meeting at a point (or junction) is zero. 

This simply says that the total current leaving the junction is equal to the total current entering the junction. 

See figure 7.1

We get the current leaving the junction as (-)ve and current entering the junction as (+)ve

Using KCL1 , we can write,

(- I1 ) + (-I2) + I3 + I4 + (-I5)  = 0

Or

I1 + I2 + I5 = I3 + I4

Mathematically,

∑ _i  = 0            ..... at a junction .

See figure 7.2 .  This is the practical way of using KCL1 . 

When you have to analyse this type of circuit, don't get confused. Although there are many voltage sources available it is very easy to analyse by using KCL . 

First of all you must understand that you don't need to mark the current that draws from each and every battery. And don't get confused with the current flowing direction. Just forget those things and choose one source. Then mark the current draw from that source. You don't need to worry about the actual current flowing direction, just mark it in any direction.

If the assumed direction of current is not the actual direction, then on solving the equation, this current will be found to have a minus sign. If the answer is positive, then the assumed direction is the same as actual direction.

 In figure 7.2 I have selected the E1 Source in AB branch and marked the current I . Then that current goes to point C. Here we apply the KCL1 . In the figure I have already applied the KCL1. you can see that I1 current goes through the CD branch and I - I1 current goes through the FE branch. So the sum of the algebraic current to point C is zero. 

Then again at point D the current I1 join I - I1 current. and current I flows to the E1 source again.

Direct current 5 (voltage & current sources)

Voltage source & current source

In this post we wil discuss about both ideal and practical voltage sources and current sources. First we will discuss about voltage source. But before that we have to study about the internal resistance to understand the different of a ideal and practical source.

Internal resistance:

The materials which are used to made the source are not pure conductors (as they are not ideal) . Therefore there exists a resistance. due to this resistance, a voltage drop occurs in the source itself. So when there is a current flow from the battery, battery itself drop some voltage and rest  is feed to the circuit. This resistance is called 'Internal Resistance' of a source and it can be shown as a series connected resistance with the source.

E.M.F : 

We saw that there exists a voltage drop in the source itself and it changes with the current draw from the source. So the voltage across the source changes with the current. When there is no current flow from source, there is no internal voltage drop. The voltage across the source in this time, is called the E.M.F value of the source.

Voltage source :

Ideal Constant - Voltage Source: 

When there is no internal resistance , the output voltage of the voltage source remains constant whatever the change in load current as there is no internal voltage drop. Such a voltage source is called an 'Ideal voltage source' . In practice , none such ideal constant voltage source can be obtained. But, smaller the internal resistance of a voltage source, closer it comes to an ideal voltage source. 

Practical Voltage source:

All the voltage sources that we are using has an internal resistance. Therefore we have to make sure that we are adding the internal resistance to the account when we are doing calculations. 

Current source: 

Ideal Constant - Current Source :

When the internal resistance is infinity in a source, the internal voltage drop goes to infinity and there is no output voltage. It gives a constant current as the output.

Practical Current Source :

 In practical current source ,internal resistance is very high compared to the external resistance of the circuit    but not infinite.

 Consider 6V battery with Mega Ohm internal resistance and a load resistance with 20K to 200K. 

when load is 20K,

from ohm's law,

I = (V/R) 

I = (6V / 1.02MOhm)

I = 5.9 micro A

when load is 200K

from Ohm's law

I = (V/R) 

I = (6V / 1.2MOhm)

I = 5 micro A

you can see though the load resistance increases 10 times, current is only varying by 0.9 micro Amperes. Such a small value is negligible and therefore we can consider this source is a constant current source. 

Pabindu lakshita

B. Sc (Engineering undergraduate)

Solved problems

Solved Problems

Q 1) Calculate the equivalent resistance across AB. Find the current draw from the battery. You may refer  figure 5.1.

  a)      Then find the voltage across 9 Ohms resistance.

  b)      Find the voltage between AD.

  c)       Find the current through 18 Ohms resistance.

Answer :

Resistance across AC = (6/2) = 3 Ohms-----------( as the resistance are equal, R = r/n )

Resistance across AD = (3+6) // 18 = {9*18/ (9+18)}  = 6 Ohms

Resistance across AB = 6+9 = 15 Ohms

Now we can find the current drag from the battery by using this equivalent resistance across AB.

Ohm’s law to R_ab,

V = IR

60V = I X 15

I = 4A

    a)      As there are no parallel paths between AB , current through 9 Ohms resistance is equal to 4A .  Ohm’s law  to 9 Ohms resistor,
V = IR

Vdb = 4A x 9 Ohms
Vdb = 36V

     

     b)      Since,
V ab = V ad + V db

V ad = Vab – V db

V ad = 60 – 36

V ad = 24 V

     c)       Current through 18 Ohms  can be calculate by Ohm’s law
Vad = I x 18
     I = (24/18) A
     I = (4/3) A

Q2)  Find the equivalent resistance across AB in every circuit.

i)                    Each resistance is 3 Ohms

Figure 5.2

Answer :

                  

Figure 5.2.1

Figure 5.2.1 shows the way to calculate the equivalent resistant

Rab  =  3 + 1.5 + 1 + 3 =  8.5 Ohms

ii)                   

Answer :

Then we can simplify this circuit as shown in figure 5.3.2

After therse steps the problem becomes a very simple one.

Rab = 0.5 Ohms

Exercise

Q3)  

i)                    Find the current draw from the circuit in figure 5.2 (Q2. i )  if  60 V voltage is given to AB .

ii)                  Find the voltage difference between every  resistance

iii)                Then find the current through every resistance

Q4)i)                    Find the current draw from the circuit in figure 5.3 (Q2. ii )  if  60 V voltage is given to AB .

ii)                  Find the voltage difference between every  resistance

iii)                Then find the current through every resistance

Q5)   i) Find the total current draw from the battery in Figure 6.1     Ans (3A)

         ii) Calculate the power supplied by battery in figure 6.2           Ans(72W)

         iii) Find the total resistance seen by the battery in figure 6.3 . find the current draw from it. 

         Ans (1.5 Ohms , 8A  )

NOTE : In figure 6.1 The resistance are parallel connected (12 Ohms & 6 Ohms ) .

              In figure 6.2 Again you may see that all three resistors are parallel connected.

               In figure 6.3 Again you may see that all four resistors are parallel connected.

to understand these things you need to solve more problems and gain experience.

               

Pabindu lakshitha

B.sc (Engineering undergraduate) 

Resistance in series & parallel

Resistance in Series and parallel

So far, we have discussed up to ohm’s law. By using ohm’s law we will discuss about the connected resistors.
There are two different methods to connect resistors

1. Series connection
2. Parallel connection

First we will discuss about the series connection of resistors,

Resistance in series:

In series connection, the current goes through each and every resistor is equal since there are no other parallel paths. As there is no parallel path, electrons have only a single path to flow. Therefore, the number of electrons flow through a one resistor during a specific time period must be equal to the number of electrons flow through in other series resistors.

But as the resistors are connected in series and voltage is given to the two ends of the combination (series connected resistor combination), voltage difference between a single resistor is not equal to the supply voltage. According to figure 4.1 we can see that the supply voltage is equal to the sum of the voltage between each resistor. The voltage between a resistor is depend on the value of the resistance. To calculate this voltage, we can use the ohm’s law. As the current is equal through every resistor, the voltage difference is directly proportional to the resistance of the resistor in a series connection.

                                                                                  figure 4.1                                                              
Applying ohm’s law to every resistor
V1 = IR1  ---------- (1)
V2 = IR2  ----------(2)
V3 = IR3  ----------(3)

And also we can write that,

V = V1 + V2 + V3

From (1),(2),(3)

V = IR1   + IR2  + IR3
V = I (R1 + R2 + R3) ---------(4)

If we consider about a one single resistor which can replace for this set of series connected resistor combination, we call it as the equivalent or total resistance. Here we consider about a single resistor with ‘R’ resistance as the equivalent resistance of this combination.

Then applying ohm’s law to the equivalent resistance,
V = IR ----------(5)

Now,
(4) = (5)
IR = I (R1 + R2 + R3)
R = R1 + R2 + R3

This can be apply to ‘n’ number of resistors in series

R = R1 + R2 + R3 + ………. + Rn

Resistance in parallel: 

In parallel connections we can see that the resistors are connected in such a way that voltage difference between each and every resistor is equal. In this case the current is divided. Current through every branch is differs according to the resistance in each branch. We can use the ohm’s law to find this values.

figure 4.2

From ohm’s law,
V = IR
I = (V/R) -------(A)

And also we can write that

I = I1  + I 2 + I3

Similar to the series connection, we define a equivalent resistance ‘R’

Now using (A)

V/R = (V/R1) + (V/R2) + (V/R3)
1/R = 1/R1 + 1/R2+ 1/R3

For n numbers of resistors in parallel

1/R = 1/R1 + 1/R2+ 1/R3 +………..+ 1/Rn

Special case:

When the resistance of every resistor is equal (say ‘r’ ) and n number of resistors are connected,

• For series connections,    R =  nr
• For parallel connections, R = r/n

Where R is the equivalent resistance.

• For parallel resistance, it’s easy to use the following derived equation for two resistors

                                               R  = ( R1R2 / (R1+ R2) )

Direct current (3) 'Ohm's law'

Ohm's law

In our previous post, we discussed about the resistance. Today we are about to study the Ohm's law.

Whenever electric current flows through a conductor , the following factors are present :

1. The potential difference V across the conductor causing the current to flow.
2. The opposition or resistance R of the conductor which must be overcome.
3. The current I which in the conductor as a result of pressure overcoming the resistance.

The relationship between these three quantities are defined in Ohm's law.

Ohm's law:

The voltage difference across a conductor is directly proportional to the current which goes through it due to the voltage difference , provided the temperature of the conductor does not change.

or

The ratio of potential difference  V between any two points of a conductor to the current I flowing between them is constant, provided the temperature of the conductor does not change.

Mathematically,

V/I = constant

or

V/I = R

where R is the resistance of the conductor between the considered points.

This can be written as, 

V = IR

This law is applicable to both DC and AC circuits. But it is complicated than DC .

Derived relations from Ohm's law

• Power is given by the product of voltage and current.
  
  
  W = VI
  Using ohm's law we can rewrite this equation as,
  
  
  
  

   W = I²R 

   W = V²/R

  
  

Direct current (2) (RESISTANCE AND RESISTIVIY)

RESISTANCE AND RESISTIVIY 

          In this post, we will learn about the resistance and the resistivity of a conductor . In our 1st post about direct current, we studied about the current flowing through a battery. To carry the current we need a media. that media is a conductor. Metals , acids, and salt solutions are good conductors of electricity. Here we study about metal conductors. 

                                 This resistance to the flow of electrons occurs due to the electric friction . Just keep in mind that this friction is very similar to the friction in mechanics. Resistance is offered by 'R' .

The resistance of a conductor not only depends on the metal type. Consider two different conductors which are made of the same metal, the resistance of those may not be same. There are three factors affects on the resistance of a conductor

1. Length of the conductor
2. Cross sectional area of the conductor ( area normal to the current)
3. Metal type (resistivity or the specific resistance )

resistivity is offered by the greek letter ᵨ

 Figure 2.1

When a road gets narrowed, traffic goes higher. similar to that, when the cross section is small,then the resistance  goes higher . So the cross sectional area is inversely proportion to the resistance. And when the road length is large, number of barriers we face increase. Similar to that , resistance increases with the length of the conductor. So length of the conductor is directly proportional to the resistance. The proportion constant is the resistivity of the metal.

In our next post, we will discuss about the Ohm's law.

pabindu lakshitha
B.sc (engineering undergraduate )

DIRECT CURRENT (1)

DIRECT CURRENT (1)

                       In this post, you will see how current goes through a battery. This post is about basics of Electrical Engineering which is very easy to understand. I add these basic lessons to give a better understanding for you. 

figure 1.1

                                      First I will explain what direct current is. When there is a flow of electrons in a conductor to a specific direction (without altering the direction with respect to time) we call that there is a current flow to the opposite direction of the electron flow. 

                                                             According to figure 1.1 electrons go through the bulb as there is a potential different between the two end of the battery. Electrons flows from negative end to the positive end of the battery. The process is described as follows. 

electrons with high energy comes out from the negative end of the battery. those electrons flow in the conductor through the bulb (bulb represent the load here) to the positive end of the battery.  When the electron goes through the bulb, it absorbs the energy from the electron (potential energy contained in electron) . Then the bulb convert that energy into another type of energy  (in a bulb: "light energy").  then the electrons flows to the positive end passing the bulb.

                  Bulb absorbs only the 'energy' contained in electron, not the electron. so the number of electrons came to the bulb is equal to the number of electrons goes to the positive end. And keep in mind that number of electrons which left the negative end must be equal to the number that reach the positive end in a certain time period . otherwise the battery do not let the electrons to leave the battery. 

figure 1.2

when the less energized electrons comes to the positive end of the battery, It charges the electrons to a high energy level. this energy is gained from the chemical energy which is stored in the battery. then it goes to the negative end (see figure 1.2) . 

NOTE

these things are for your understanding. Actual process is somewhat different from this. but for beginners this clarification will give a better understanding.

please ignore my spelling mistakes. I need your comments. please leave your problem regarding this post as a comment.

pabindu lakshitha
B.sc Eng (Hons) 

This blog is for students those who are willing to study about electrical engineering. Here onward we will post a series of posts about very basics of electrical  engineering. we will discuss abouth DC , AC first. Both single and three phase. next we will move to machines