is replaced by that equivalent model The solution is contained in two theorems

Is replaced by that equivalent model the solution is

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is replaced by that equivalent model. The solution is contained in two theorems due to Thévenin and Norton. Prof. C.K. Tse: Circuit Analysis Review
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27 Thévenin and Norton theorems Let’s look at the logic behind these theorems (quite simple really). If we write down KVL, KCL, and Ohm’s law equations correctly, we will have a number of equations with the same number of unknowns . Then, we can try to solve them to get what we want. Now suppose everything is linear. We are sure that we can get the following equation after elimination/substitution (some high school algebra): Case 1: a≠0 Case 2: b≠0 Thévenin Norton Prof. C.K. Tse: Circuit Analysis Review
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28 Equivalent models Thévenin equiv. ckt Voltage source in series with a resistor i.e., V + IR T = V T which is consistent with case 1 equation Norton equiv. ckt Current source in parallel with a resistor i.e., I = I N + V/R N which is consistent with case 2 equation Prof. C.K. Tse: Circuit Analysis Review
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29 How to find V T and I N Thévenin equiv. ckt Open-circuit the terminals ( I =0), we get V T as the observed value of V . Easy! V T is just the open-circuit voltage! Norton equiv. ckt Short-circuit the terminals (V=0), we get I N as the observed current I . Easy! I N is just the short-circuit current! = V T I = I N Prof. C.K. Tse: Circuit Analysis Review
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30 How to find R T and R N (they are equal) Thévenin equiv. ckt Short-circuit the terminals ( V =0), find I which is equal to V T /R T . Thus, R T = V T / I sc Norton equiv. ckt Open-circuit the terminals ( I =0), find V which is equal to I N R N . Thus, R N = V oc / I N . For both cases, R T = R N = V oc / I sc I = I sc = V oc Prof. C.K. Tse: Circuit Analysis Review
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Prof. C.K. Tse: Basic Circuit Analysis 31 Simple example Step 1: open-circuit The o/c terminal voltage is Step 2: short-circuit The s/c current is Step 3: Thévenin or Norton resistance Hence, the equiv. ckts are:
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32 Example — the bridge again Problem: Find the current flowing in R5. One solution is by delta-star conversion (as done before). Another simpler method is to find the Thévenin equivalent circuit seen from R5. Prof. C.K. Tse: Circuit Analysis Review
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33 Example — the bridge again Step 1: open circuit The o/c voltage across A and B is Step 2: short circuit The s/c current is Step 3: R T = V T Prof. C.K. Tse: Circuit Analysis Review
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34 Example — the bridge again = Prof. C.K. Tse: Circuit Analysis Review
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35 Maximum power transfer theorem We consider the power dissipated by R L . The current in R L is Thus, the power is This power has a maximum, when plotted against R L . = 0 gives R L = R T . Prof. C.K. Tse: Circuit Analysis Review
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36 A misleading interpretation It seems counter-intuitive that the MPT theorem suggests a maximum power at R L = R T . Shouldn’t maximum power occur when we have all power go to the load? That is, when R T = 0! Is the MPT theorem wrong? Discussion: what is the condition required by the theorem? Prof. C.K. Tse: Circuit Analysis Review
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37 Systematic analysis techniques So far, we have solved circuits on an ad hoc manner. We are able to treat circuits with parallel/series reduction, star-delta conversion, with the help of some theorems.
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  • Summer '16
  • Martin Chow
  • Thévenin's theorem, Voltage source, Norton's theorem, Current Source, Series and parallel circuits, Voltage drop

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