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!

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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!

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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!

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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!

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