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Solve steady-state circuit for current expression using phasor analysis and superposition. Apply correct step for efficient solution. |
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Problems With AssistanceModule 8 – Problem 3 Filename: PWA_Mod08_Prob03.ppt This problem is adapted from question #4 on the Final Examination from the fall of 1995 in ECE 2300 Circuit Analysis, in the Department of Electrical and Computer Engineering at the University of Houston 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 • Multiple Frequencies and Superposition • Equivalent Circuits in the Phasor Domain Go straight to the First Step Go straight to the Problem Statement Next slide
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
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
Problem Statement The circuit given is in steady-state. Find a numerical expression for the current, i(t). Next slide
Solution – First Step – Where to Start? The circuit given is in steady-state. Find a numerical expression for the current, i(t). How should we start this problem? What is the first step? Next slide
Problem Solution – First Step The circuit given is in steady-state. Find a numerical expression for the current, i(t). • How should we start this problem? What is the first step? • Write KCL for each node • Apply superposition • Convert the circuit to the phasor domain • Write KVL for each loop
Your choice for First Step –Write KCL for each node The circuit given is in steady-state. Find a numerical expression for the current, i(t). 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.
Your choice for First Step –Write KVL for each loop The circuit given is in steady-state. Find a numerical expression for the current, i(t). 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.
Your choice for First Step was –Convert the circuit to the phasor domain The circuit given is in steady-state. Find a numerical expression for the current, i(t). This is a bad choice for the first step. The problem here is with the frequencies of the three sources. Each source has a different frequency. If we were to transform the circuit, which frequency should we use? There can only be a single frequency when we transform. Go back and try again.
Your choice for First Step was –Apply superposition The circuit given is in steady-state. Find a numerical expression for the current, i(t). This is the best choice for the first step. The problem here is with the frequencies of the three sources. Each source has a different frequency. By applying superposition, we can take each frequency, one at a time, and use the phasor transform for each frequency. Let’s begin.
Apply Superposition – Step 1 using the First Source The circuit given is in steady-state. Find a numerical expression for the current, i(t). When we apply superposition, we need to take one source at a time, and set all other sources equal to zero. We will then add the contributions of each source together to get the total response. Let’s start with the iS1 source, and set the others equal to zero. This has been done in this figure. Notice that voltage sources when set equal to zero become short circuits, and current sources when set equal to zero become open circuits. Let’s solve.
Apply Superposition – Step 2 using the First Source The circuit given is in steady-state. Find a numerical expression for the current, i(t). This is a dc source, so the capacitors become open circuits, and inductors become short circuits. Phase has no meaning at dc. The solution is straightforward, and we do not actually need the phasor transform.. Using the current divider rule, we can write Next slide
Apply Superposition – Step 1 using the Second Source The circuit given is in steady-state. Find a numerical expression for the current, i(t). Next, we take the iS2 source, and set the others equal to zero. This has been done in this figure. With this done, we will want to transform to the phasor domain.
Apply Superposition – Step 2 using the Second Source The circuit given is in steady-state. Find a numerical expression for the current, i(t). Here, we have transformed to the phasor domain. Note that we have a current divider here, and we can write Next slide
Apply Superposition – Step 3 using the Second Source The circuit given is in steady-state. Find a numerical expression for the current, i(t). To complete this calculation, we need to take the inverse phasor transform, and we get Next slide
Apply Superposition – Step 1 using the Third Source The circuit given is in steady-state. Find a numerical expression for the current, i(t). Next, we take the vS3 source, and set the others equal to zero. This has been done in this figure. With this done, we will want to transform to the phasor domain.
Apply Superposition – Step 2 using the Third Source The circuit given is in steady-state. Find a numerical expression for the current, i(t). Here, we have transformed to the phasor domain. Note that we have six series impedances connected to a voltage source, and we can write Next slide
Apply Superposition – Step 3 using the Third Source The circuit given is in steady-state. Find a numerical expression for the current, i(t). To complete this calculation, we need to take the inverse phasor transform, and we get Next slide
The Solution The circuit given is in steady-state. Find a numerical expression for the current, i(t). The solution is the sum of the solutions for each source. Therefore, the answer is, Go to notes slide
Why does this solution take so long? • This solution seems to take a long time, because we have to transform to the phasor domain, and inverse transform, for each source. However, when compared with techniques that do not use superposition, this approach is pretty efficient. Go to next comments slide. Go back to Overviewslide.
Why don’t we just add the solutions in the phasor domain? Wouldn’t that save time? • If we were to add together the solutions in the phasor domain, that is, if we added I1(w)+I2(w)+I3(w)=I(w), we would only succeed in getting a wrong answer quickly. • Even if we were to do it, we would still need to inverse transform to get the solution. What frequency wouldwe use? The solutions must be added in the time domain. Go back to Overviewslide.