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.

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!

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!

## 1 comments:

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