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Problems With Assistance Module 8 – Problem 1

Problems With Assistance Module 8 – Problem 1. Filename: PWA_Mod08_Prob01.ppt. Go straight to the First Step. Go straight to the Problem Statement. Next slide. Overview of this Problem. In this problem, we will use the following concepts: Phasor Analysis Phasors with Sines and Cosines

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Problems With Assistance Module 8 – Problem 1

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  1. Problems With AssistanceModule 8 – Problem 1 Filename: PWA_Mod08_Prob01.ppt Go straight to the First Step Go straight to the Problem Statement Next slide

  2. Overview of this Problem In this problem, we will use the following concepts: • Phasor Analysis • Phasors with Sines and Cosines • Source Transformations Go straight to the First Step Go straight to the Problem Statement Next slide

  3. Textbook Coverage The material for this problem is covered in your textbook in the following chapters: • Circuits by Carlson: Chapter 6 • Electric Circuits 6th Ed. by Nilsson and Riedel: Chapter 9 • Basic Engineering Circuit Analysis 6th Ed. by Irwin and Wu: Chapter 8 • Fundamentals of Electric Circuits by Alexander and Sadiku: Chapter 9 • Introduction to Electric Circuits 2nd Ed. by Dorf: Chapter 11 Next slide

  4. Coverage in this Module The material for this problem is covered in this module in the following presentations: • DPKC_Mod08_Part01, DPKC_Mod08_Part02, andDPKC_Mod08_Part03. Next slide

  5. Problem Statement • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. Next slide

  6. Solution – First Step – Where to Start? • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. How should we start this problem? What is the first step? Next slide

  7. Problem Solution – First Step • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. • How should we start this problem? What is the first step? • Write KCL for each node • Identify the essential nodes • Write KVL for each loop • Pick the reference node • Apply superposition • Convert the circuit to the phasor domain

  8. Your choice for First Step –Write KCL for each node • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. This is not a good choice for the first step. If we were to write the KCL for each node, we would end up with a set of simultaneous intregral-differential equations. There are easier ways to solve this problem. Go back and try again.

  9. Your choice for First Step –Write KVL for each loop • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. This is not a good choice. If we were to write the KVL for each loop, we would end up with a set of simultaneous intregral-differential equations. There are easier ways to solve this problem. Go back and try again.

  10. Your choice for First Step was –Pick the reference node • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. This is not the best choice for the first step. If we choose to use the node-voltage method, we will indeed pick, and label, the reference node. However, it is not even clear at this point whether we will need to use the node-voltage method. We recommend that you go back and try again.

  11. Your choice for First Step was –Apply Superposition • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. This is not the best choice for the first step. It is true that superposition can be used to solve this problem. However, in most of these problems, using superposition does not help us solve the problem faster, and in fact makes the solution take significantly longer. This situation would be quite different if the sources had different frequencies; then we would want to use superposition. Here, we do not. Go back and try again.

  12. Your choice for First Step was –Identify the essential nodes • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. This is not the best choice for the first step. Finding the number of essential nodes will be important when we consider the best way to solve the circuit. However, before choosing solution approaches, we really should recognize that this is a steady-state problem with sinusoidal sources, which is best attacked with the phasor analysis method. Try again.

  13. Your choice for First Step was –Convert the circuit to the phasor domain • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. This is the best choice for the first step. We should recognize that this is a steady-state problem with sinusoidal sources, which is best attacked with the phasor analysis method. The first step in using this method is to convert the circuit to the phasor domain. This means converting the voltages and currents to phasors, and replacing passive elements with their impedances. Let’s convert.

  14. Convert the circuit to the phasor domain – Sine Functions? • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. • We have two sine functions and a cosine function here, and all three have the same frequency. How do we handle this kind of situation? • The difference is not significant. Ignore the type of function, and transform. • Convert sine functions to cosine functions. • Convert cosine function to a sine function.

  15. You said that the difference is not significant • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. It is true that the difference between the sine function and the cosine function is just 90 degrees of phase. However, this much phase shift is very significant in some applications. We cannot assume, without knowing what the circuit will be used for, that this phase shift is not significant. This is not a correct step. Go back and try again.

  16. You said that the cosine function should be converted to a sine function • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. It would be possible to define the phasor transform using sine functions, and not cosine functions. Some even do this when there are only sine functions present. However, this is not recommended. We recommend that you use cosine functions for the phasor transform every time. There is less confusion this way, and the process will be understood better by others. So, here we will convert sine functions to cosine functions.

  17. Convert the circuit to the phasor domain • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. We have converted this circuit to the phasor domain. Please note that we have chosen the cosine function as the basis for our phasor transform. Thus, when we have sine functions, as here, we need to convert them to cosines before transforming. Next slide

  18. Next Step in the Phasor Domain • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. • What is the next step? • Use the node-voltage method • Use the mesh-current method • Use Thevenin’s Theorem • Use Source Transformations

  19. Next Step in the Phasor Domain – Use the node voltage method • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. Your choice for the next step was to use the node-voltage method. This is a good choice. There are only two essential nodes, and so only one node-voltage equation is needed. However, for this problem, we will choose another approach. While, your choice is as good as ours, we would like you to go back and try to find another good approach.

  20. Next Step in the Phasor Domain – Use the mesh-current method • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. Your choice for the next step was to use the mesh-current method. This is not a good choice, since there are three meshes in this circuit. There are better ways to solve. Go back and try to find another approach.

  21. Next Step in the Phasor Domain – Use Thevenin’s Theorem or Source Transformations • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. You may have already noticed that Source Transformations are simply a special case of Thevenin’s and Norton’s Theorems. Perhaps your textbook does not even mention Source Transformations. Do not allow this to bother you. The two methods are about the same thing. In particular, we are going to replace the voltage sources in series with impedances with current sources in parallel with those same impedances. Let’s take this step here. The parts of the circuits to be replaced are circled in red dashed lines.

  22. Using Thevenin’s Theorem or Source Transformations In particular, we have replaced the voltage sources in series with impedances with current sources in parallel with those same impedances. Note the polarity of Is3 is a result of the reference polarity of the voltage source. The values of these new current sources can be found using the ratio of the voltage source to the impedance. For the left-hand current source we have For the right-hand current source we have Next slide

  23. Using Equivalent Sources and Impedances We note that the three current sources were in parallel, and can be replaced with a single current source with a value equal to the three current sources in parallel. This equivalent current source has the value, Note that the current source Is3 has a negative sign since the source has a reference polarity pointing in the opposite direction from the equivalent source. Similarly, we can combine the impedances in parallel and get Next slide

  24. Current Divider Rule • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. We now note that we have two impedances in parallel being fed by a current source. We can use the current divider rule to write, This leads to the answer for part a). See next slide.

  25. Inverse Transform • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. To get the expression for i(t), we need to take the inverse transform. To get this, we need the magnitude and the phase. We can write, This leads to the answer for part a), Next slide

  26. Plugging in Values -- 1 • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. To get the answer for part b), it is easier to go back to the phasor domain, and plug in each of the values for R, and then inverse transform. We get: Next slide

  27. Plugging in Values -- 2 • Find an expression for the steady-state value of the current i(t), as a function of R. • Find the steady-state value of the current for R equal to 500[W], 1[kW], 2[kW], and 4[kW]. To get the answer for part b), it is easier to go back to the phasor domain, and plug in each of the values for R, and then inverse transform. We get: Next slide

  28. What if I chose another method? Is that a big deal? • If you picked another method, such as the node-voltage method, it does not make that much difference. One advantage of the approach taken here is that the equations are all pretty simple. However, if we have access to a good calculator or a computer, the solution can be done easily taking many other approaches. • One key issue here is recognizing that the solution is done in the phasor domain, but is not complete until we inverse transform the solution back to the time domain. • Another key issue is remembering that the phasor transform is defined for cosine functions.A cosine is 90° out of phase with the sine, so we can convert sines to cosines bychanging the phase, subtracting 90°. Go to next comments slide. Go back to Overviewslide.

  29. What if I choose to use the sine function for the Phasor transform? Is that a big deal? • This is a tough call. The key is that, in any given problem, you must define your transform and inverse transform the same way. You could use cosine, or you could use sine, but not both. So, to use sine instead of cosine is really not wrong. • While in some problems with only sine functions, it would be faster not to convert them to cosine functions, we recommend againstit. Do all problems the same way. Use the cosine. It won’t cost much time, and it may avoid some confusion. Go back to Overviewslide.

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