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
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
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
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
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
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:
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
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
34 Example — the bridge again = Prof. C.K. Tse: Circuit Analysis Review
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
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
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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