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.

• DC Parallel Circuits 1
• DC Parallel Circuits 2

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!

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!

Solutions To Brain Teasers Number 2

As I promised last time, I will going to show you the solutions to teasers on our previous topic about the Factors Affecting Resistance here in Electrical Engineering. By the way, this topic is still covered on our review of basic Physics for this is very important one when we reached our major topics in this Electrical Engineering course.

Last time, I leaved you four worded problems in order for you to analyze and understand the principles fully. But if you still running out of time to solve it, I will show to you it now.

The first problem given was:

Problem 1 : The resistance of a copper wire 2, 500 cm long and a 0.090 cm in diameter is 0.67 ohm at 20 degree celcius. What is the resistivity of copper at this temperature?

The solution here is quiet simple. We just have to substitute the given values from the formula since we have uniform units:

p = RA/ l = 0.67 ohm x [ TT ( 0.090 cm) ^2/ 4 ] / 2, 500 cm
    = 1.7 x 10 ^-6 ohm.cm - answer

The unit of resistivity in the British engineering system of units differs from that just given in that different units of length and area are employed. The unit of area is the circular mil., the area of the circle 1 mil (0.001 in) in diameter, and the unit of length is the foot. Since the areas of two circles are proportional to the squares of their diameters, the area of a circle in circular mils is equal to the square of its diameter in mils. In this system of units the resistivity of a substance is numerically equal to the resistance of a sample of that substance 1 ft long and 1 circular mil in area, and is expressed in ohm-circular mils per foot.

The abbreviation CM is often used for circular mils. This should not be confused with the abbreviation used for centimeters (cm). We will use the more standard cmil.

Let's solve another problem using the principles above.

Problem 2 : Find the resistance of 100 ft of copper wire whose diameter is 0.024 in and whose resistivity is 10.3 ohm.cmils/ft.

Convert first the given diameter in mils. Since, 1 mil = 0.001 in as mentioned above.

Therefore, d= 0.024 in = 24 mils.

Then getting the area we have, A = d^2 = 24 ^2 cmils. Substituting the values:

R = pl / A = (10.3 ohm.cmils/ft)(100 ft) / 24 ^ 2 cmils.
R = 1.8 ohm - answer

Problem 3 : A silver wire has a resistance of 1.25 ohm at 0 degree celcius and the temperature coefficient of resistance of 0.00375 per degree celcius. To what temperature must the wire be raised to double the resistance?

Since we are asking for the temperature, just derive the formula of EQ . 1 in our previous post here in Electrical Engineering topics. It will be:

 

t = Rt - R0 / Ro oo = ( 2.50 -1.25 ) ohm / 1.25 ohm x 0.00375 /C
t = 266 degree celcius - answer

It should be clearly understood that R0 in the above equation ordinarily refers to the resistance at 0 degree celcius and not to the resistance of any other temperature. A value of oo based upon the resistance at room temperature, for example, is appreciably different from the value based upon 0 degree celcius. This may be made clearer by the graphic analysis of the variation of resistance with temperature. ( Click the image to enlarge)

In the above illustration, the resistance Rt of a conductor at any temperature t is plotted. For a pure metal, this curves gives a linear relation ( approximately). Note the fact that the curve does not pass through the origin; i.e. at 0 degree celcius the resistance is not zero. Hence, we cannot say that R oo t. The slope of the curve delta R/delta t is constant . Since,

oo = delta R / delta t /Ro = slope / Ro

it is clear that the value of oo depends upon the base temperature chosen for Ro. In computations involving temperature variation of resistance, the value of Ro must be obtained by using the equation below:

Rt = Ro + Ro oo t = Ro( 1 + oo t)

Understanding the principles of effect of temperature in resistance.

Problem 4 : A tungsten filament has a resistance of 133 ohm at 150 degree celcius. If oo = 0.0045/C., what is the resistance of the filament at 500 degree celcius?

From the EQ 1: ( of our previous post)

Ro = Rt / 1 + oo t = 133 ohm / 1 + (0.00450/C ) x 150 degree celcius

Ro = 79.4 ohms

Getting the resistance at 500 degree celcius,

R500 = R0( 1 + oo t500)
R500 = 79.4 ohms [ 1 + (0.00450/C) x 500 degree celcius]

R500 = 258 ohms - answer

Since it is the resistivity factor that changes with temperature. The equations of EQ 1 and EQ 2 from our previous lecture may be written with p in place of R.

pt = p0( 1 + oot) ------------------------------------- EQ. 3

I will not give the table for resistivities and temperature coefficients of resistivity of materials for it is always given in the problem. You don't have to memorize it.

I think this is now over for the basic review....

On my next post, we will now begin to discuss the real scope of Electrical Engineering.

Cheers!

Solution To Brain Teaser Number 1

As I promised last time, I will give you the detail solution to our brain teaser on my previous post here in Learn Electrical Engineering for Beginners. So for those who are not yet through, you can still catch up.

Let's proceed to solved our first problem here with this topic.

Last time, this is the worded problem I had given to you.

Problem No.1 : A process equipment contains 100 gal of water at 25 degree celcius. It is required to bring it to boiling in 10 minutes. The heat loss is estimated to be 5%. What is the KW rating of the heater?

Here's the solution:

Since, we have given the mass of water subject for boiling ( 100 degree celcius) = 100 gal. I would like to convert it to gram so that our units are compatible when we substitute it with our general formula.

Given that, 1 gal = 3.786 liters and 1 liter of water is equivalent to 1kg. We can therefore convert it:

100 gal x 3.786 liters /1 gal x 1 liter/ 1 kg = 378.6 kg of water. But since I would like to convert it to gram. It is therefore equivalent to 378,600 grams. The specific heat of water is 1cal/g-C.

In our Heat Formula, we all know that it is,

Qo = mc delta T

Qo = 378, 600 ( 1 cal/g-C) ( 100-25 ) = 28395000 cal or 28,395 kcal

Let : Wo = Qo; where Wo = energy out

but 4, 186 joules = 1 kcal

Wo = 28, 395 kcal ( 4, 186 joules/ 1 kcal) = 118,861,470 joules

In order to get the energy in of the process equipment, we will be needing the formula below to get it:

Win = Wo/ %n, which comes from the efficiency formula. This formula will be touched in various topic here in Electrical Engineering course.

Substituting the values that we have will lead to:

Win = 118,861,470 joules / (1- .05) = 125,117,336.8 joules or 125.117 x 10 ^6 joules.

We are not yet through with our answer. The problem is asking for the KW rating of the equipment. In our previous formula of Power = Energy ( joules) / Time ( seconds). Since, the water will be boiled for 10 minutes as what stated in the above problem. It is therefore,

Rating of Equipment (KW) = 125, 117, 336.8 joules / 600 seconds ( in 10 minutes)

Rating of Equipment (KW) = 208, 529 watts or 208. 529 KW - ANSWER

Notice that I did not used rounded values in my substitution. I only rounded it off in my final answer because in board exams, multiple choice answers are very near with each other. Rounding off numbers while you are still in the process of computing will take a little difference in final answer. For you to get the accurate and exact number, I advice you to round off your numbers in your final answer only.

I hope you had learned something here today. Please catch me more here in Learn Electrical Engineering for Beginners. I'll be back shortly.

Cheers!