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Norton and Thévenin Theorems Part 1-Basic Electronics-Lecture Slides, Slides of Fundamentals of Electronics

This lecture was delivered by Kantimoy Baag at Bengal Engineering and Science University for Fundamentals of Electronics course. Its main points are: Norton, Thevenin, Theorems, Indpendent, Current, Voltage, Source, Series, Parallel, Resistance

Typology: Slides

2011/2012

Uploaded on 07/19/2012

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Download Norton and Thévenin Theorems Part 1-Basic Electronics-Lecture Slides and more Slides Fundamentals of Electronics in PDF only on Docsity! Lecture Seventeen  Norton and Thévenin Theorems  The Thévenin theorem states that   We  can  replace  the entire network exclusive of  load with an equivalent  circuit  containing  independent voltage source (VTh) with a series resistor (RTh)  in such a  way that the current‐voltage relationship across the load remain unchanged.   The Norton  theorem  is essentially  the  same except  that  the  independent voltage source and  series resistor is replaced with independent current source (IN) and a parallel resistor.   Problem solving strategy:  • Remove  the  load  and  calculate  the  Voltage  VTh  at  these  open  circuited  terminals  by  applying  all  circuit  analysis  techniques  (KVL, KCL,  loop  and node  analysis, Ohm’s  law,  voltage and current division).  • For circuit with  independent sources only, determine RTh by open circuiting all current  sources and short circuiting all voltage sources and calculate the equivalent resistance  RTh using series‐parallel combination of pure resistive network.  • Thévenin  equivalent  circuit  consists  of  independent  voltage  source  VTh  and  a  series  resistor RTh.  • For Norton Equivalence, convert the Thévenin equivalent to Norton equivalent through  simple source transformation.  INDEPENDENT SOURCES  Example: Calculate V0 using Thévenin theorem  Since we want to calculate V0 across 8 kΩ resistors, we remove it. Since there is no current  across 4kΩ so in this case RL = 8k + 4k Ω.   + V 0 -   docsity.com Removing  the  load  and  using  simple  source  transformations  and  adding  algebraically  the  sources  8 2m 4k 16V 4kTh V ⎛ ⎞= + × =⎜ ⎟ ⎝ ⎠   By  short  circuiting  the  12  V  supply  and  open  circuiting the 2 mA source, we get  3 63k 6k 2k 2 4k 3 6Th R ×= Ω Ω+ Ω = + = Ω +   With Thevenin Equivalent Circuit  to  load again, we  can calculate V0 through voltage division  0 8k 8V 8k 4k ThTh V V R = = + +   Example:   1. Calculate V0 using Thévenin theorem.  2. Confirm your result through Norton’s principle.  3. Calculate the power transferred to load resistance.  4. What should be the load resistance to be replaced with 6kΩ so that maximum power  is transferred?  Thévenin Theorem:   In this case 6 kΩ is designated as RL      docsity.com
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