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Week 2

Week 2. Source transformation Mesh current analysis Thevenin and Norton equivalent circuits Transfer of Power. Combining Voltage Sources. Voltage sources are added algebraically. Combining Voltage Sources. Voltage sources are added algebraically. Combining Voltage Sources.

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Week 2

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  1. Week 2 Source transformation Mesh current analysis Thevenin and Norton equivalent circuits Transfer of Power

  2. Combining Voltage Sources Voltage sources are added algebraically

  3. Combining Voltage Sources Voltage sources are added algebraically

  4. Combining Voltage Sources Don’t do this. Why is this illogical? Whose fundamental circuit law is violated by this?

  5. Combining Current Sources Current sources are added algebraically

  6. Combining Current Sources Current sources are added algebraically

  7. Combining Current Sources Don’t do this. Why is this illogical? Whose fundamental circuit law is violated by this?

  8. Source Transformations Simplify Circuits (1/5)

  9. Source Transformations Simplify Circuits (2/5)

  10. Source Transformations Simplify Circuits (3/5)

  11. Source Transformations Simplify Circuits (4/5)

  12. Source Transformations Simplify Circuits (5/5)

  13. The Thévenin and Norton equivalent circuits both represent the same circuit • They have the same voltage-current characteristics Source Transformations 13

  14. Any voltage source in series with a resistance can be modeled as a current source in parallel with the same resistance and vice-versa Using Source Transformations in Circuit Analysis 14

  15. EXAMPLE: SOLVE BY SOURCE TRANSFORMATION In between the terminals we connect a current source and a resistance in parallel The equivalent current source will have the value 12V/3k The 3k and the 6k resistors now are in parallel and can be combined In between the terminals we connect a voltage source in series with the resistor The equivalent source has value 4mA*2k The 2k and the 2k resistor become connected in series and can be combined After the transformation the sources can be combined The equivalent current source has value 8V/4k and the combined current source has value 4mA Options at this point 1. Do another source transformation and get a single loop circuit 2. Use current divider to compute I_0 and then compute V_0 using Ohm’s law

  16. Source Transformation Show that iS=vS/R for any RL 16

  17. Application Determine whether the 6V source is absorbing or delivering the power. 17

  18. Step-by-step simplification 18

  19. Redundant Resistor (1) Prove that the left circuit and the right circuit are equivalent for any load resistor. 19

  20. Redundant Resistor (2) Prove that the left circuit and the right circuit are equivalent for any load resistor. 20

  21. Example 21

  22. Simplified Circuit 22

  23. MESH Analysis 23

  24. Technique to find voltage drops around a loop using the currents that flow within the loop, Kirchhoff's Voltage Law, and Ohm’s Law • First result is the calculation of the mesh currents • Which can be used to calculate the current flowing through each component • Second result is a calculation of the voltages across the components • Which can be used to calculate the voltage at the nodes. Mesh Analysis 24

  25. Mesh – the smallest loop around a subset of components in a circuit • Multiple meshes are defined so that every component in the circuit belongs to one or more meshes Definition of a Mesh 25

  26. Steps in Mesh Analysis Vin 26

  27. Identify all of the meshes in the circuit Step 1 Vin 27

  28. Label the currents flowing in each mesh Step 2 i2 i1 Vin 28

  29. Label the voltage across each component in the circuit Step 3 + V4 - + V2 - + V1 _ + V5 _ + V3 _ i2 i1 Vin + V6 _ 29

  30. Use Kirchhoff's Voltage Law Step 4 + V4 - + V2 - + V1 _ + V5 _ + V3 _ i2 i1 Vin + V6 _ 30

  31. Use Ohm’s Law to relate the voltage drops across each component to the sum of the currents flowing through them. • Follow the sign convention on the resistor’s voltage. Step 5 31

  32. Step 5 + V4 - + V2 - + V1 _ + V5 _ + V3 _ i2 i1 Vin + V6 _ 32

  33. Solve for the mesh currents, i1 and i2 • These currents are related to the currents found during the nodal analysis. Step 6 33

  34. Once the mesh currents are known, calculate the voltage across all of the components. Step 7 34

  35. 12V 35

  36. From Previous Slides 36

  37. Substituting in Numbers 37

  38. Substituting the results from Ohm’s Law into the KVL equations 38

  39. Chugging through the Math One or more of the mesh currents may have a negative sign.

  40. Chugging through the Math The magnitude of any voltage across a resistor must be less than the sum of all of the voltage sources in the circuit • In this case, no voltage across a resistor can be greater than 12V. 40

  41. Chugging through More Math The currents through each component in the circuit. 41

  42. None of the mesh currents should be larger than the current that flows through the equivalent resistor in series with the 12V supply. Check 42

  43. Steps in Mesh Analysis 1. Identify all of the meshes in the circuit 2. Label the currents flowing in each mesh 3. Label the voltage across each component in the circuit 4. Write the voltage loop equations using Kirchhoff's Voltage Law. 5. Use Ohm’s Law to relate the voltage drops across each component to the sum of the currents flowing through them. 6. Solve for the mesh currents 7. Once the mesh currents are known, calculate the voltage across all of the components. Summary 43

  44. Source Transformation to Solve a Circuit Find the power associated with the 6 V source. State whether the 6 V source is absorbing or delivering power to the circuit.

  45. Source Superposition 45

  46. Source Superposition This technique is a direct application of linearity. It is normally useful when the circuit has only a few sources.

  47. Source Superposition • The Superposition Theorem states that a circuit can be analyzed with only one source of power at a time, the corresponding component voltages and currents algebraically added to find out what they'll do with all power sources in effect. • To negate all but one power source for analysis, replace any source of voltage (batteries) with a wire; replace any current source with an open (break).

  48. Look at to the following circuit and apply Superposition Theorem : Since we have two sources of power in this circuit, we will have to calculate two sets of values one for the circuit with only the 28 volt battery in effect. and one for the circuit with only the 7 volt battery in effect: .

  49. Analyzing the circuit with only the 28 volt battery, we obtain the following values for voltage and current: Analyzing the circuit with only the 7 volt battery, we obtain another set of values for voltage and current:

  50. Applying these superimposed voltage figures to the circuit, the end result looks something like this: Currents add up algebraically as well:

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