What Jobs Awaits for Electrical Engineering Graduates Today?

Let me pause for awhile in posting lectures about Electrical Engineering. Let's have a short talk about the jobs and opportunity awaits for all graduates of Electrical Engineering. Details are fact based on the present situation nowadays and also based on my own personal experienced.

According to National Statistics Coordination Board in the Philippines, unemployment rise from 7.5% of April 2009 to 8.0% of April 2010. The most common rise up of this unemployment rate is because of the newly graduates and one of these are the Electrical Engineering graduates. There are approximately about 1,611 newly graduates in the Philippines (Registered for Electrical Engineering Board Exams)which are expected to be unemployed nowadays. Still grabbing for a new job and opportunity that may suit them.

In U.S. according to Department of Labor's Bereau of Labor Statistics employment grew for Electrical Engineers by 7.8% from the fourth quarter of 2009 to the first quarter of 2010 and now 16.1% above compared to its very low employment last first quarter of 2009. This rise of employment for Electrical Engineering field was due to positive recovery in the U.S. economy.

What's the reason of giving this short comparison to you?

Sad to say, there are only limited of opportunities and jobs for Electrical Engineering here in the Philippines even before during my time way back 1999 after passing my licensure exams. If there is an opportunity or jobs available for Electrical Engineering, the salary or the compensation is very low. Besides, majority of them will required you a minimum two years experience for the position even in the Junior/Cadet Engineering position. That's why other Electrical Engineering graduates like me were all shifted to semiconductor field with attractive compensation and benefits and got their luck related to their field and some aren't got any but dive into last resort which is in the semiconductor operations/processes. Others with no luck in this field grabbed their fate in the call centers- poor engineers...

In connection to this, many Electrical Engineering graduates and experts in the Philippines try their luck to get their job somewhere in US, Australia, Taiwan, Japan, North Korea, Saudi Arabia, Abu dhabi, Kuwait, Dubai, Dammam, Doha, Qatar and sometimes in Canada but quiet limited unlike Saudi Arabia. This means that the Philippine Government should work more on jobs availability most especially to our professionals and experts.  

Take a look on this snapshot in JobStreet: Click on the picture to enlarge.

Electrical Engineering Jobs
If you click "ALL Locations" will appear only limited opportunities in the Phillipines where most of the jobs are in Saudi Arabia in Electrical Engineering.

Take a look on the National Capital Region- Philippines. There are only 97 jobs available below considering this region is the center for opportunities and jobs in any field. Would you think newly graduates of Electrical Engineering could accomodate them all? If you read the qualifications almost none of them offers for newly graduates. If there is, only FEW.


Electrical Engineering Jobs
If you click the National Capital Region, only 97 jobs available with almost no one affers jobs for Electrical Engineering newly graduates.

In my information above, I've already given you a brief idea where most of the jobs and opportunities are coming in for Electrical Engineering field.

So, what are the recommended job positions for a newly graduate of Electrical Engineering?

If you are a newly graduate of Electrical Engineering, it is recommended that you start with a Junior position for your job. Some company are required you to train with their traning program if you are a newly graduate but sad to say, NOT ALL. That's why before you accept their offer, be sure they have some training for you before you start with your current job. Some company have higher expectations with their applicants to have a certain number of years of experience for the job position although you are only a newly graduate. Make sure you won't missed this step.

Some of the known positions for Junior position are Assistant Project Engineer, Junior Design Engineer, Estimates, Electrical Inspectors, Electrical Engineer (Junior Position) and many others related to Electrical Engineering jobs. There are some company that accepts newly Electrical Engineering graduates in the middle position considering if you have an attractive qualifications like:
  • Board Topnochers
  • With above average rating in academic performance or an Honor Student.
  • With certain special skills for their requirements: i.e autocad, excellent communication skills, programming, etc...
  • Experience in certain field they required. Majority one year experience.
Not all newly graduates have these qualifications. If you did not fall under these qualifications. Don't loose HOPE.

There are also some other ways you can explore your skills in Electrical Engineering if this is only your line of interest or the job that you want. You should not depend on what the company could offer job for you. Life must go on. You can also start on your own. For example, building your own repair shop related to electrical machineries, electrical layout job for residential, job for electrical installations etc.

Examples above are the acquired skills during your college days and you can explore these things out. If you think you can shift to other line of interest, you may do so. It really depends where you can find your luck in line with Electrical Engineering jobs. Your own decision is still a MUST.

In my personal experience, I have no luck in finding jobs for Electrical Engineering here in the Philippines because of the limited opportunities. Since, I have other skills that I can explore, I shifted in the field of Semiconductor operations and now in the Supervisory position. This is my decision and choice. But still hoping I can find my luck going abroad for Electrical Engineering jobs if there is any opportunity for me. Skills that I have in Electrical Engineering is still there and I believe I can find my luck anytime if I would pursue.

Before I close this post, here are some of the sites where you can browse for jobs in Electrical Engineering:
P.S. These are the top ten sites I have chosen for you as your reference for finding jobs in Electrical Engineering. These are based on the Google search engine. My mission is to give my readers a high quality content readings for this site especially for beginners.

The list of sites above are subject for change as long as I find other sites that offers good job for electrical engineering. So much for this...

Cheers!
Donny
readmore »»   http://img.ezinemark.com/imagemanager2/files/30004252/2011/11/E10.jpg
Labels:

Electrical Engineering Tutorial Still Alive?

It's been nine months ago when I stopped making a post here at Electrical Engineeering for Beginners. I had received more comments for some other topics I published. I just want to apologized for my very late response. As you know, I'm not a full time writer of this blog related to Electrical Engineering.

For this 2010 onwards, I will be making my own website about Electrical Engineering. But that will depend on my time since there are no other support I'm receiving to complete this course outline. As of now I'm busy learning an HTML or the Hyper Text Markup Language and CSS or the Cascading Style Sheets to make a website of my own. Besides, I'm a full time worker in one of the semiconductor company where all my time are being spend only to support my family. For all of you readers out there, good news! this Online Electrical Engineering tutorial will  resume on September 2010.

I will continue making a post about Electrical Engineering topics from where I stopped. This blog is now ranking well in Google SERP's which makes me to inspire because it's now PR4 which beats other big sites out there. This will push me helping out our young wanna be Electrical Engineering students today. I would like to thank those who are continue subscribing to this site though it's not been updated for a long time. Now, its your turn to open your laptop and see my two to three times post per week.

More things to clarify out....

I also received an email about this blog was being nominated as Top Blogs conducted by other big sites out there. Ok, well then there are you again links that would like to enter my site in order to have a one way link. You are encouraging me to put your badge to my site containing your links. I checked your site and it's only PR0. Well I'm sorry, but I refused to put your badge to my site. I would like to have a quality links to my site. I just only link to higher PR sites related to Electrical Engineering.

There are also some concerns that I would like to clarify to my readers out there. One time, I received an email requesting me to post about a topic related also to Electrical Engineering. Ok, I will just make a post for your request if there are more requesting for that particular topic.

I also received an emails on how to pass Electrical Engineering board exam. They are also asking me on how to get some manuals or review materials. Well, my advice to you is to enroll yourselves to any known Electrical Engineering Review Centers. In the Philippines, the well known Review Center is the Powerline. This is where I took my review and refresher. One thing that you should put in your mind especially to our Electrical Engineers wanna be. It's not on the review center that matters most, its on how you really understand the principles behind Electrical Engineering. Review Centers will just only give you the technique. Your knowledge is still that matters most here and your confidence on how you will pass your Electrical Engineering Board. To those who are new to this site. You may read my post here:

So much for that, I also received an email on exchanged links to other Electrical Engineering sites. Today, my decision is to stop having an exchanged links because Google will detect my site as spam. I'll better do a high quality content to sustain my SERP. Please stop sending me an email for exchanged links. I only do for one way link.

Now, how do I convinced you to come back to my site?

First, Learning Electrical Engineering keyword have a Global Monthly Search of 3,600. Not too bad to build this site. Advertising competition is only medium.

See this: keyword for "Learning Electrical Engineering"


In Google Search Engine, this site is still number one considering nine months without post or update. Take a look on the other side below:




In Yahoo, Learn Electrical Engineering is still No.3 first page for the mentioned keyword. Considering my site is only hosted free, it also rank well with those mentioned big search engines. You can still find this site easily. Ranking well in search engines means this site has a higher authority and with good quality contents over the other.

For my readers out there, please help me to continue this mission of teaching online for Electrical Engineering. I want to optimized my site with a most competitive keyword "Electrical Engineering". I hope I can still make it to the top.

Stay tune for more updates here in Learn Electrical Engineering for Beginners.

Cheers!
Donny
readmore »»   http://img.ezinemark.com/imagemanager2/files/30004252/2011/11/E10.jpg
Labels:

Applications for DC Parallel Circuits

It's been a month and a half ago when I did an update on this blog. I've been busy with some things which I cannot explained to you right now. The reason why is that this project which is initiated by me is really intended for 2010. Although I'm doing a little update on this blog about Learning Electrical Engineering, the project still alive until such time I completed the course outline of this site.



I checked my stats awhile ago and I was surprised that my traffic still in tact with an average of 45 and above visitors a day. The number of subscribers is still increasing a bit compared when I started this blog. One thing you can expect from me, though I'm doing a little update on this blog, this site will still remain in search engines. I have now a PR2 from PR0. I would like to give thanks for those who supported and still visiting this site. Target keyword " learning electrical engineering " now dominates my rank in Google and Yahoo. Now let's continue of what I have left before.

The following illustrative problems below are just an applications of what I had presented on my post about DC Parallel Circuits.

From the previous post, I had illustrated to you the theory of the DC Parallel Circuits which consists of Part 1 and 2. Today, I will show you the application through problem solving.

Problem No. 1 : Let's begin solving parallel circuits in which I will show you how the current divides in each branch of circuits and how can we obtained the unknown values using Kirchhoff's Law. Suppose I have the circuit as shown in the diagram below. We need to find the unknown currents. We need to find the currents at I1, I4 , I6 and I7 . Click the image to enlarge.


Obviously, we can find the current at junction A by simply performing the KCL or the Kirchoff's Current Law. We can see that the current divides at junction A. Therefore, I1 = I2 + I3 = 7A + 3 A= 10 amperes.

Taking a look at junction C. We can see that the current entering this junction I2 divides into I4 and I5 which becomes a total current to I2. Therefore we can say, I2 = I4 + I5, which becomes 7 A = I4 + 5 A, and since we are looking for I4 = 2 Amperes.

Now, we would like to solve for I6. Since, the flow of currents I3 and I4 are flowing toward junction B. We can say that, I6 = I3 + I4 = 3 A + 2 A = 5 Amperes.

The last requirement is I7. Since the currents I5 and I6 are flowing toward junction D. we can say that I7 = I5 + I6 = 5 A + 5A = 10 Amperes.

The example above I had shown you is how the currents divides in each branch. The next example will illustrate you the application of unequal resistors in parallel.

Problem No.2 : Three loads A, B and C are connected in parallel to a 230 volt source. Load A takes 9.2 KW, load B takes a current of 60 amperes and load C is a resistance of 4.6 ohms. Calculate (a) the resistance of loads A and B, (b) the total resistance of the three paralleled loads, (c) the total current, (d) the total power. Click the image to enlarge.

Since we have different parameters given for each load, We have to solve first some missing requirements that we need in our calculations.

(a) Calculating Ra = Vt ^2 / Pa = 230 ^ 2 / 9,200 = 5.75 ohms
for Rb = Vt / Ib = 230 / 60 = 3.83 ohms
Note : The solutions shown above are just an applications of ohms law that we previously discussed.

(b) Calculating the total equivalent resistance of the circuits will follow:
Req = 1/ 1/5.75 + 1/3.83 + 1/4.6 = 1/ 0.174+ 0.261+0.217 = 1.53 ohms
Note : the formula used was just discussed on the previous post above DC PARALLEL CIRCUITS.

(c) Since we have already calculated the missing values, we can now solve for total current.
It = 230 / 1.53 = 150 Amperes

(d) To get the total power, since we already know the values of Vt and It. It will be now,
Pt = 230 x 150 = 34,500 watts or 34.5 KW.

For the above illustrative problem ( No.2 ) , I had shown you how the Ohms Law was also applied in solving the unknown quantities. This will surely applied when one value is missing. This is the best and basic technique that you can applied anytime you encountered such problems like this. The mentioned technique will also be applied when we reached complex AC circuitry.

This is the end of our basic circuitry in parallel connections. The next post will deal about series-parallel circuits lecture.

See you again on my next post.

Cheers!
readmore »»   http://img.ezinemark.com/imagemanager2/files/30004252/2011/11/E10.jpg
Labels: , , , ,

DC Parallel Circuits Part 2

Yes, let's continue of what we had left last time here in Electrical Engineering for Beginners. I was glad that you are still there and an increasing number of subscribers makes me feel more energetic in writing more in this Electrical Engineering course. But before you rolled your eyes over me, the coverage of this lesson for today is all about the unequal resistors, kirchoff's law and applying ohm's law in parallel circuits.





Last time, I had mentioned about solving the total resistance in parallel with equal resistors. I will tell you how it was derived when I reached the topic of solving unequal resistors in parallel within today. Let's begin to have a short introduction of unequal resistors in parallel then, I will insert Kirchhoff's first law before continue discussing unequal resistors in parallel. I did it that way because Kirchhoff's first law has something to do with the flow of current.


Moving on...

If the circuit contains resistors in parallel whose values are not equal or unequal, we have some difficulty in assessing the total resistance of the circuits. One easy way to get the total resistance in parallel is by using your ohmmeter to measure the total resistance. Suppose you have an R1 and R2 connected in parallel with 40 and 80 ohms respectively, you would obviously measure a total resistance of 27 ohms for that circuit.

Wondering how it was obtained?

In our previous lesson, DC Parallel Circuits Part 1. I had mentioned there that the current flowing in each branch of the parallel circuits are not equal if the resistances were also different from each other. More current will flow on the smaller resistance compared to that with bigger resistance value. All of them were mentioned in this post without some problem illustrations. I just only show you how the current divides parallel connections with varying values of resistances.

Since, it is not often possible to get the total resistance of the circuit by using an ohmmeter especially in this case our circuit connection is getting more complex, we ought to know how to get such values by using calculations. Previously, we had learned the useful concepts of Ohm's law by solving circuit values in series circuit connection. But in this case, there is another equation which you will need this time. It is what we have been waiting for. It was known as Kirchhoff's First Law - Second Law was already discussed here.

What was it all about?

Kirchhoff's Law is true in every type of circuit. The concerns of this law is not the circuit as the whole but only individual junctions where currents combine within the circuit itself. It's law states that : The sum of the currents flowing toward a junction always equal the sum of all currents flowing away from that junction.

Or, other states like this...

The algebraic sum of the currents at any junction of an electrc circuit is zero. This statement has something to do with the algebraic signs of the currents coming and moving away of the node. In order for you to understand this principle, take a look on the illustration below:

The image above is the simple representation of a circuit junctions. Suppose you have a four junctions there and all that conductors are carrying a current in the direction shown above. If you look at the image above IA are delivering stream of electrons at its node. It is obviously, that when the currents leaves that node, the current divides into IB, IC and ID which is equivalent to IA. Thus, making it IA = IB+IC+ID.

or, in other ways of expressing it...

IA- IB - IC -ID = 0, which also states on the above Kirchhoff's first law. In this case, it is important to know the direction of current. The current coming to the node is (+) positive while the current leaving, we'll assign a (-) negative sign for it.

I will be giving a pure problem illustrations of this topic on my next post this coming first week of October 2009 for you to comprehend well this topic. I reduced the frequency of posting due to my busy schedule at work.

Uhmm....


Unequal Resistors in Parallel Circuits

Here are some of the important rules to remember when dealing with unequal resistors in parallel:


1. The same voltage is impressed across all resistors.
2. The individual-resistor currents are inversely proportional to their respective magnitudes. You will understand this fully when I give you the sample problem on my next post.
3. The total current for the circuit is : It = I1 + I2 + I3+...
4. The total equivalent resistance of the circuit is:


Req = 1/ (1/R1) + (1/R2) + (1/R3) + ....


Note : When two unequal resistors are connected in parallel their equivalent resistance is equal to their product divided by their sum.

R xy = Rx x Ry / Rx + Ry


Ohms Law in Parallel Circuits


Just like series circuits, we were also need to apply Ohms Law when dealing with the parallel circuits. We will be using this law to calculate some other unknown quantities like current, voltage, and resistance in such circuits. This law would require less time and effort if you would have to know such quantities mentioned above.

Let's say you have a number of resistors connected in parallel but you like to measure the resistance of a particular resistor using your ohmmeter. Of course, you would first disconnect the resistor to be measured from the circuit otherwise, you will measure or the ohmmeter reads the total resistance of the circuits.

Another one, if you would like to know the current across the particular resistor of a combination of parallel resistors using an ammeter. Again, this time you would have to disconnect it and insert an ammeter to read only the current flow through that particular resistor.

Knowing the voltage requires no disconnection. But of course, Ohms Law is the very pratical use in knowing such quantities for electrical engineers like us.

These are just a short concepts for our Part 2 of DC Parallel Circuits. On my next post it would be a little bit lengthy for I will illustrate to solve problems related to this topic.

I will come back on first week of October 2009.



Cheers!
readmore »»   http://img.ezinemark.com/imagemanager2/files/30004252/2011/11/E10.jpg
Labels: , , , ,

DC Parallel Circuits Part 1

Before I proceed with my new post here on Learn Electrical Engineering for Beginners, I would like to thank those who sent their email for some questions. Keep those emails coming in. If you don't received an email from me that means the answer is already here on my blog or I will answer you on my future post. Please read below portion of this site for your guidance.

I hope you are now learning with this site. We are now moving on with our topic and let's study the next part of Electric Circuits which is DC Parallel Circuits.

One objective of this lesson is for you to understand that you can solve any circuits because all circuits are made of combinations of series and /or parallel circuits.

Previously, in DC Series Circuits we defined that whether resistors, lamps or cells are connected end-to-end. Today, the scenario would be completely different. Instead of being connected end-to end as in series circuit, they are connected side by side therefore it would create more than one path in which the current can flow. If this is so, we can say that resistors/resistances are said to be parallel connected or connected in parallel. The circuit would now be called a parallel circuit. The image below is one example of the parallel circuit. I show you this illustration because the diagram already explains everything.

As what I mentioned in our earlier topic about electric circuits, we cannot say a complete circuit if we do not have a source of emf connected to them. For instance, we have two resistors connected their wire in parallel or parallel connected. When any two terminals are connected across the voltage source just as what had shown above, the whole arrangement- both resistors, the wires connecting them together and the voltage source - forms complete parallel circuit.
The circuit above shows that there are more than one path of current to flow. This means that these two resistors shares on the total current drawn from the battery. A part of the total current goes through the first resistor and at the same time a part of the total current also drawn for the second resistor. If you will intend to connect a device with polarity, for example a cell or batteries, you must connect the positive terminals together and the negative terminals together.

The Voltage in Parallel Circuit

When the resistances are connected in parallel just like the diagram above and connected across the voltage source, the voltage across each resistors are always the same. Observe the diagram that the labels used were just the same. It was self-explained in the diagram that the voltage are just the same.

Since, it is the fact that voltages across each resistances are just always the same. It has a practical consequence. What do I mean with this? This means that all components which are to be connected in parallel must have the same voltage rating if they are to work properly. Did you noticed that?

The line voltage throughout the Philippines is 220 volts. In U.S. 120 volts. I think you are aware that some of our appliances are rated 120 volts or 220 volts work properly well. If in case you have a lamp or a bulb rated 20 volts. What the hell do you think will happen? The bulb will be burn immediately because the excess current will flow through it.

Since, all appliances are connected across the same voltage source, the same voltage will also experienced across each load. Each load must be properly rated to handle this voltage.

How the Current Flow in Parallel Circuits
In order to understand well the flow of current in the circuit for parallel connections. I made a little details on the diagram above. Take a look on the diagram below:

As I mentioned earlier, the flow of current in the parallel connections divides through each of the parallel paths. In the circuit diagram above, the two branches named them AB and CD connected in parallel. Observe that as the current flowing in each branches will divide and reunite them at node B returning to voltage source. You will wonder how the amount of current divides in each branches.

The amount of current that will serve in each branches will depend on the resistance value. This will be the principle: The current flowing through the several branches of a parallel circuit divides in inverse proportions, governed by the comparative resistance of the individual branches.
And so? what does this mean?

It only means that the lower resistance value in any branch circuit in proportion to the resistance of other branches in the same parallel circuit, the higher will be the current value or proportion which that branch will take.

Simply...

In parallel circuit, branches having low resistance draw more current than other branches having high resistance.

Like voltage, the flow of current connected in parallel circuit is also of a great importance. For instance, like we know that every electrical appliances connected in parallel, the current will divide unequally to each branch since they have differing value of resistances- the highest current flowing through the lowest resistance. You will learn more about this when we reached the protection against excessive flow of current.

Ohhh... I love to illustrate example again through problem solving. Let's take a sample problem for you to show that it really happens. Let's take again the circuit above and assign values for them.

Given that, I = 9 amperes ; I1 = 3 amperes ; I2 = 6 amperes

For instance, a total current I = 9 amperes is flowing through the parallel branch of R1 and R2. If you will observe, the value of R1 is twice the value of R2. We have mentioned earlier that the current divides in inverse proportion to the values of the two resistors. Therefore, only 3 amperes will flow on R1 and 6 amperes will flow on R2. If for example, R1 triple its resistance from 40 ohms to 120 ohms. The current flowing through R1 will be reduced from 3 amperes to 3/3 ampere or 1 ampere while I2 would remain unchanged. Thus the total current would be 6 amperes + 1 ampere = 7 amperes.

What does it implies?

Since, in series circuit we said that all currents are the same throughout the circuit. In parallel circuit we just add it. This can be expressed mathematically as, It = I1 + I2 + I3 +...

What if the resistances are equal in parallel circuit? This will be the next topic.

Equal Resistors in Parallel Circuits

Let's consider a water pipes connected in parallel as shown in the diagram below.

Let's say we have a constant water pressure incoming or the head as we learned in fluid dynamics. Assuming the same cross-sectional area of water pipes connected in parallel. The amount of water here would flow is equivalent to cross sectional area of pipe 1 + pipe 2.What do you think would happened to the amount of water flowing in the system if you would convert it to single pipe only? The amount of water will be less than that connected in parallel. Why? because you reduced cross sectional area in which the water would flow. The bigger the cross sectional area, the more water would flow on the system shown above.

The same thing in resistor. The bigger the cross sectional of resistor, the more current would flow because the resistor value diminishes as the cross sectional area is getting bigger.

The conclusion here is that : resistors or loads connected in parallel present a lower combined resistance or load than does any one of them individually.
This means that if you have four 400 ohms resistors connected in parallel. The resistance of the combined load will be equally divided into 4 equal resistances thereby giving you a 100 ohms total resistance. In other words, 100 ohms is your combined resistances of four 400 ohms connected in parallel.

If you don't get my point here, let's discuss it when we illustrate more problem solving in DC Parallel Circuits.

To be continued.....

Cheers!
readmore »»   http://img.ezinemark.com/imagemanager2/files/30004252/2011/11/E10.jpg
Labels: , , ,

Applications: A Few Tips in Solving DC Series Circuit Problems

Before we proceed with to the Parallel Circuits, let's study first some other worded problems that you may be encountered during your board exam using the concept of DC Series Circuit. Learn Electrical Engineering for Beginners will provide you a technique on how you will overcome those scenarios.

The first problem that you will encounter is somewhat an application of a simple transmission lines. I just want to open this topic earlier because we will be dealing with this topic on my future post. I will just show you the snapshot on how those concepts that we studied in my previous post are being applied.

Let's begin...

Problem 1: The problem states that a load resistor of 4.1 ohms, 425 ft from 240-volt generator, is to be supplied with power through a pair of standard-size copper wires. If the voltage drop in the wires is not to exceed 5 percent of the generator emf, calculate (a) the proper AWG wire that must be used, (b) the power loss in the transmission line, (c) the transmission efficiency.

This is how you will going to solve it...

Let's have a simple visual of what is being said on the above problem. It could be shown just like the simple representation of transmission line below:

This simple transmission line can be considered as a circuit consisting of a 4.1 ohms resistor in series with 850 ft length of copper wire connected to a 240 volt generator. You would wonder why I say 850 ft length of wire while in the problem states that it is 425 ft from 240 volt generator. This is because in the given problem, it only mentioned the distance of the load resistor from the generator emf. It is not pertaining to the length of wire. Since, the load resistor has 2 ends connected to the wire across the generator emf, the actual length of wire should be 2 x 425 ft = 850 ft length. Please take note on this because there are many beginners who are getting mistakes when solving this type of problem.

Moving on...

The voltage drop would not exceed 5 % of the generator emf therefore,
Line Voltage Drop = 240 x 0.05 = 12 volts

Remember that in Ohm's Law in Series Circuit, the sum of all voltages across each resistances in series circuit is equivalent to the source emf therefore,
Load Voltage = Generator Emf - Line Voltage Drop = 240 - 12 = 228 volts

You have to consider the line voltage drop when dealing with transmission line.

The Line Current across the 4.1 ohm load resistor would be,
Line Current = 228 / 4.1 = 55.6 amperes

and the Line Resistance would also be,
Line Resistance, Rl = 12 / 55.6 = 0.216 ohms

We are about to find the resistance per 1000 ft. Using the ratio and proportion to get the resistance per 1000 ft. It would be,

Resistance (1000 ft)/ 1000 ft = Line Resistance / length of transmission wire
Resistance per (1000 ft) = (1,000 ft /850 ft) x 0.216 ohm = 0.254 ohm

a.) Consulting the AWG table below it shows that the standard wire size is No. 4; this wire has a resistance of 0.253 ohm per 1, 000 ft. This is based on the table below. Click the image to enlarge.

b.) Getting the actual line resistance would be,
Actual Line Resistance / 850 ft = 0.253 ohm / 1000 ft
Actual Line Resistance = 0.85 x 0.253 = 0.21505 ohm

We need to get the following data in order to get line power loss.
Total Series-Circuit Resistance = 4.1 + 0.21505= 4.31505 ohms
Total Current I = 240 / 4.31505 = 55.6193 amperes
Line Loss Power = (55.6193)^2 x 0.21505 = 665 watts - answer

c.) Getting the efficiency of transmission line can be derived in terms of load power and total power. This can be expressed as:

Efficiency = Load Power / Total Power = [(55.6193)^2 x 4.1 / 240 x 55.6193 ] x 100 %
Efficiency = 95% - answer

This is not not formal discussion about transmission line. We will touch it more in depth on my succeeding post. I'm just showing you how the concepts are being applied with these kind of application.

You would also encountered some tricky problems like what I will going to show you. This is just a simple one but you would used a simple ohm's law solving for unknown values. Let's have this example below:

Problem 2: A dc generator may be characterized by an ideal voltage source in series with a resistor. At the terminals of the generator, voltage and current measurements for two different operating conditions are Vt = 115 V at I = 10 A and Vt = 105 V at I = 15 A. Model the generator by a voltage source in series with a resistor.

This is already an application to dc generator. The problem required you to model the generator by a voltage source in series with a resistor. So, let's do what is being said. I have here the figure that shows the simple model mentioned in the problem in order to visualize it. I always love to draw the figure first when solving problems in electrical engineering. In the first place I'm a visual person. It's the easy way to understand what the problem is asking for.

With the circuit model of the generator shown above, I will defined the symbols as Vo for the dc generator voltage, I is the current, Rg is the series field connected to the dc generator and Vt is the terminal voltage or the generator output. Since there are two conditions mentioned above, let's expressed it in ohm's law for finding the dc generator voltage Vo which has no value given in the above problem and in terms of resistance which we also need to know here.

Therefore, we can state that: Vo = Vt + IRg

Condition 1 : Given that Vt = 115 V and I = 10 A
Vo = 115 V + (10 ) Rg, will served as equation 1.

Condition 2 : Given that Vt = 105 V and I= 15 A
Vo = 105 V + (15) Rg will served as equation 2.

Let's equate 1 and 2 since the value of Vo in equation 1 and 2 are equal. We can therefore expressed it mathematically as,

115 V+ 10Rg = 105 V + 15 Rg , we can now solve for Rg.

Rg = 2 ohms- answer

Then solving Vo will give, you may substitute the value of Rg from either equation 1 or 2 above will yield, I will choose equation 1.

Vo = 115 V + (10)(2) = 135 V - answer

These are some of the illustrative problems that I could share with you. on DC series Circuit Just always remember that when solving problems like what I illustrated above:

1. Try to make an simple illustration of the problem to picture out and understand the scenario of the problem.

2. Always try collect first the given data before proceeding in solving the problem.

3. There are some problems that you need to solved first the missing data before solving what was being required. Problem number 1 and 2 above are the best examples of this one.

4. Know what is the subject matter of the given problem is also one that you should not forget.

5. Always review and finalize your answers.

My next post will continue our study of Circuits connected in Parallel.

Hope you learned some tricks for today here in Learn Electrical Engineering for Beginners.

Cheers!

readmore »»   http://img.ezinemark.com/imagemanager2/files/30004252/2011/11/E10.jpg
Labels: , , , ,

DC Series Circuit Part 2

Hello folks, I'm glad that you're still there hunting for my new post here in our study of basic electrical engineering which is very recommended for the beginners. Well, if you find this site useful for you, then tell your friends and let them subscribe to my articles.



It's a little bit hectic on my work schedule including posting my blog here. Because I have to double my effort just for you... How sweet...I will not make my introduction again get longer because I know that you're really want to learn more here. So, let's continue of what we've left last moment which is the DC Series Circuit Part 1. For those who missed it, you can still catch up with the lecture.

Let's study the continuation...

The Voltage Division in the Series Circuit

In a series circuit, you would be able to find the voltage across at any point in the series circuit. The voltage which we called it the step-down voltage.

A circuit used for this is shown in the circuit diagram below. This is called a voltage divider. You may click the picture to enlarge.

Let's assume in the given circuit above that the applied voltage with the electrical notation of E or Ein is 100 volts and the values of R1 and R2 are 15 and 20 ohms respectively. You may want to know what is the applied voltage across R2 with the electrical notation of Vout or Eout which is written in another way. Please take note that the electrical notation of Vout or Eout could be an input voltage for another circuit. which of course will also become an Ein once more). That topic would be study on my succeeding post here in Electrical Engineering.

So, this is how you will going to do it...

We know that the total resistance in the circuit is 15 + 20 = 35 ohms. Given that the given circuit voltage is 100 volts. You may now use the Ohm's Law to find the circuit current. This is:

I = Ein / Rt = 100/35 = 2.86 amperes

Then, let's go to R2 in the given circuit. We all know that the resistance is 20 ohms and we have just calculated that the current is 2.86 amperes (from the conditions mentioned earlier in Ohm's Law Series Circuit that the current in the series circuit are just the same throughout).
Therefore, you will obtain that,

Eout = I x R2 = 2.86 x 20 = 57.2 volts - answer

You may observe that with the appropriate choice of resistor values in voltage divider chain, an input voltage of 100 volts has been stepped down to an output voltage of 57.2 volts. By using ohm's law, you would be able to calculate it.

Since we have an illustrative example above, let's expressed the above example into equation. Given that the total resistance is R1 + R2 and ohm's law tells you that the circuit current would be: Ein/ R1+R2. This is also the current across R2. Using Ohm's Law again, we can calculate the Eout = I x R2 will give an equation just by substitution:

Eout = (Ein/R1+R2) x R2
Expressing it in a correct and understandable way would give you...
Eout = (R1/R1+R2) x Ein

The equation above is the simplified formula that you can use in the given condition like this in voltage division chain. To put the equation into words. The voltage across any resistor in a voltage divider chain can be calculated by multiplying the value of that resistor by the input voltage dividedby the total resistance of the circuit.

Norton's and Thevenin's Theorem are commonly used principles when solving such voltage divider problems. This will be another basic concept that we will study on my succeeding post. For the meantime, let's absorbed first what we have now.

The Variable Resistors

You can also vary the resistance of the circuit as well. If you are not aware, you always done it by yourself by adjusting volume of your radio. This is what we called a variable resistor.

The resistor can be made variable in this way by means of sliding arm made of good conducting material to be arranged so that it can be moved along the length of the resistor. The resistor is then connected into the circuit with one of its end fastened to the sliding arm. By moving its sliding arm along the resistor, the value of the resistor can be varied at between maximum and minimum (zero).

If the variable resistor is used in this way, it is called the rheostat. It is used to control the current flow in the circuit.

The maximum value of resistance was obtained when the slider moves on the lower position as what had shown on the illustration above- left portion. (You may click the image to enlarge). Likewise, when the slider moves upward would obtain the minimum value of resistance. This is the simple function of a variable resistor.

A variable resistor may have either two or three circuit connections. The first picture that you see below is the example of the three terminal teminal connections variable resisitor which are commonly known as Potentiometer.

This is one typical sample for three terminal connections...

Typical potentiometer looks like this.

The Potentiometer Connections
The circuit diagram of a potentiometer is really no more than that of a voltage divider chain. R1-R2 is a single resistor effectively divided by the sliding arm C, whose movement alters the relative values of R1 and R2. Please refer to circuit diagram above.

The output voltage can vary from zero (when C is lowered so that R2=0) to full circuit voltage (when C is moved up so that R1=0)

Variable resistors, like fixed resistors, can be made with resistance material of carbon or can be wired-wound, depending on the amount of current to be controlled - wire-wound for large currents and carbon for small currents.

Wire-wound variable resistors are constructed by winding resistance wire on a porcelain or bakelite circular form, with a contact arm which can be adjusted to any position on the circular form by means of a rotating shaft. A lead connected to this movable contact can then be used, with one or both of the end leads, to vary the resistance used.

For controlling small currents, carbon variable resistors are constructed by depositing a carbon compound on a fiber disk. A contact on a movable arm actsto vary resistance as the arm shaft is rotated.
On my next post, let's have some practical applications for DC Series Circuit here in Learn Electrical Engineering for Beginners.
Cheers!
readmore »»   http://img.ezinemark.com/imagemanager2/files/30004252/2011/11/E10.jpg
Labels: , , , ,
Subscribe to: Posts (Atom)