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EE3301 Electrical Network Analysis

EE3301 Electrical Network Analysis. Dr. Jeanne Pitz Part 2. Chapter 4. Nodal analysis Mesh analysis Automated methods Pspice Capacitors Inductors Other analysis techniques Source transformations, superposition Thevenin and Norton equivalents Max power transfer. Nodal analysis.

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EE3301 Electrical Network Analysis

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  1. EE3301 Electrical Network Analysis Dr. Jeanne Pitz Part 2

  2. Chapter 4 • Nodal analysis • Mesh analysis • Automated methods • Pspice • Capacitors • Inductors • Other analysis techniques • Source transformations, superposition • Thevenin and Norton equivalents • Max power transfer.

  3. Nodal analysis • Identify the essential nodes of the circuit and assign voltage variables • Write independent set of nodal equations at each essential nodes using the variables assigned • Solve the set of independent equations • This is the technique used by spice • See table 4.1 for more definitions

  4. Example Assessment 4.1 • For the circuit below use Node Voltage method to find v1,v2, and i1 • How much power is delivered to the circuit by the15A source • How much power is delivered to the circuit by the 5A source V2 V1

  5. Assessment 4.1 a. • Nodes are v1 and v2 (v3 is ground node at bottom) so only 2 equations are needed

  6. Simplify equations Multiply both side by 60 to get rid of fractions Multiply both side by 5 to get rid of fractions Solve using determinants

  7. Assessment 4.1 b,c • How much power is delivered to the circuit by the15A source P15A=V1*15=60*15=900W • How much power is delivered to the circuit by the 5A source P5A=-V2*5=-10*5=-50W

  8. Mesh analysis • Mesh current analysis • Define mesh currents one around each mesh. • Some may be combinations of branch currents. • write mesh equations around each independent mesh and solve for mesh currents. • Solve for the currents by writing independent equations and multiple equation methods

  9. Example: assessment 4-7 • Define 3 mesh currents: i1,i2, i3. • Write 3 mesh equations: • A. Find power delivered by 80V source. • B. find the power dissipated by the 8  resistor. i3 i1 i2

  10. Example: assessment 4-7 • Solve with determinants: • Power delivered by 80V source : • Power in the 8 resistor:

  11. Some special cases: example • Fixed node voltage source • Dependent sources • Find va=v2 in the circuit • Write node voltage equations for node 1 and 2 • Substitute in fixed voltages and solve for va V2 V3 V1

  12. Solution • Write node equations at v1 and v2 • Substitute in : • Combine like terms and multiply equation (1) by 30 and(2) by 40 to remove fractions.

  13. Solution • After combining terms and eliminating fractions we have Solve these equations simultaneously: • V2=24V=Va • i=-3.2A

  14. Mesh current Analysis • Applies to planer circuits • Depending on the circuit it may lead to fewer equations to solve simultaneously.

  15. Example:4.10 assessment • Find power in 2 resistor: • Write mesh equations in 2 loops: • Which simplifies to: • Solve for i1 and i2:

  16. Assessment 4.10 • Now the power in 2 resistor is • V2 *I2 : • I2=i1-i2= 2 - (-4)=6A • V2=I2*R=12V • P2= 12*6=72W

  17. Do this with pspice • Follow the directions to set up a new “project” as an “Analog or Mixed A/D”, then enter the schematic and run pspice.

  18. Pspice results • After simulation completes print the voltages and currents on the schematic • Press the v and I buttons to print all voltages and currents on the schematic

  19. Pspice results: voltages and currents

  20. Pspice results: power

  21. LTSpice • Same spice engine as p-spice • Free on analog technology web page • Linear Technology was bought by AT • Full spice version ( no restrictions) • Favors LT parts ( no issue for EE3301) • Need a bit more spice knowledge • Analysis : • .op – operating point • .tran - transient analysis

  22. LT Spice continued Enter the schematic Set the analysis type

  23. LTspice continued • Most components are on the banner. • Use the “and” gate symbol to bring up the sources and other devices

  24. Running the operating point • Brings up the window with all the voltages and currents for the circuit.

  25. LT Spice continued • The voltages can be annotated on the schematic by right clicking on the mouse to bring up a set of choices including: • “place .op data label”

  26. Thevenin Equivalent • Some times we want to analyze a circuit and we’re not really interested in the whole circuit but just a part in the circuit. • So we want to reduce the complexity by finding the equivalent circuit that will mimic the complex circuit so we can focus on a single circuit element or part. • First we should define the terms open circuit voltage (Voc) and short circuit current (Isc).

  27. Open circuit voltage /Short circuit current Short across the load resistor and compute the current shown as the short circuit current Load Resistor Remove the load resistor to compute open circuit voltage Vab isc Vab

  28. Reduction of circuit • We reduce a circuit at the nodes on either side of the component and consider “looking” into the pins at the circuit. There are several ways of approaching a solution based on the circuit • Circuit contains resistors and independent sources. • Deactivate the independent sources ,i.e., open circuit all current sources and short circuit all voltage sources now there is only a network of resistors and find the Thevenin equivalent resistance. • Then solve the original circuit to get the voltage across the terminals (if unloaded.) • Circuit contains resistors and both independent and dependent sources. • Solve for Voc and Isc to compute Rth • Circuit contains resistors and dependent sources. • Insert a 1A current or 1V voltage at the nodes of interest.

  29. Circuit contains resistors and independent sources Open circuit 3A source Short circuit the 25V source Solve for Rab = (5||20) + 4 = = 5*20/25 + 4 = 8 Ω Procedure • Find Vth and Rth for the circuit shown:

  30. Example 1 continued: Solve for Vab Solve the nodal equation for V1 Note: since no current flows in 4 ohm Thevenin Equivalent circuit: V1 + - 3A = 0 V1= 32V

  31. Thevenin Eq. in terms of open circuit voltage and short circuit current Open circuit voltage = vabShort Circuit current = isc Vth= Rth*IscRth= Vth/Isc

  32. Example 2: Containsresistors, independent and dependent sources. Find Isc and Voc Since Ix must be zero: solve each node in terms of v and i And Voc = Vth = v = Vth

  33. Example 2 cont’ : Solve Isc • Shorting these terminals means control voltage v is 0. Thus the LHS dependent voltage source is also 0. isc =-20i

  34. Example 2 cont’ : • Solve the following 2 equations for both halves of the circuit : • isc =-20i • 5 = 2000i • Thus isc = -20*5/2000 = -.05A=-50mA • And so Rth = Vth/Isc = 5/.05= 100 Ω

  35. Circuit contains resistors and dependent sources. • Now the tricky one , If the circuit only has resistors and dependent as shown, we put IT and VT • Now write node equations on each side: and solve for Rth = VT/IT Note since there is there is no independent source the Vth = 0.

  36. Solve for Rth • A nodal equation on RHS gives: • 20i + VT/25 =iT • On the LHS yields • i = -3VT /2K • now we just want to solve for VT/IT • Rth = 100 Ω

  37. Another example • Find the Thevenin equivalent circuit by solving for Voc =Vth and Rth at a-b . Write a node equation at node a. • Short 24 V and open 4A let iT enter a, write a node equation at a = 8V

  38. Assessment 4.9 • Solve for Vo: use mesh currents • Using linear eq solver: i1=4; i2= 2 and i3 =2.5 • So v =8*(4-2)=16V Note:

  39. Source Transformations • Another useful technique is source transformation; • Source transformations are an application of OHMs law. • Recall v= i*r

  40. Source transformation cont • We can use this to transform a Thevenin equivalent circuit into a Norton equivalent circuit. • Vs=is*R=> is= Vs/R or similarly Vth=Rth*IN • The Thevenin equivalent resistance can be found the same way we described earlier, then the Norton source can be found. • Or you can find the Thevenin equivalent circuit first then transform it to the Norton equivalent circuit.

  41. Max Power Transfer • In a resistive circuit, the Maximum power is transfer is achieved when the load resistance is equal to the circuit equivalent resistance. • So first step is to replace the network by its Thevenin equivalent at the nodes of the load resistance • Then when the load, RL is equal to the Rth power transfer is maximized.

  42. Max power

  43. Max power • To find a maximum with respect to RL , take the derivative with resect to RL • Since either term can be 0 to make the equation true. Use the second one…

  44. Max power con’t. • Solve for RL and after much simplification: • So to maximize power transfer we must make the load match the effective resistance of the network.

  45. Examples • If the circuit is replaced by the Thevenin equivalent, Vth = 50, Rth=100 see what the power is for other resistances:

  46. Superposition • When a linear circuit is driven by more than one independent source, the response can be found by superposition i.e., by finding the response to each source while zeroing the effect of the other source. The combine the effects together. • So the total response is the sum of the individual responses. • voltage source it is short circuited. • current source it is open circuited.

  47. Example Problem 4.91 • First eliminate the 110V supply by shorting it; • Now the 5 and 10  resistors are in parallel • Write 2 node equations v1 and v2 where v=v1

  48. Example Problem 4.91 • Now v1 an v2 due to the 4A current sourcev1 = -9.23V and v2 = -14.78

  49. Example Problem 4.91 • Now compute the contribution from the 110V supply, eliminate the current source by open circuiting it. Make Vs =110V and write 2 nodal equations as before

  50. Example Problem 4.91 • After simplifying the two equations become: • Which yields the contribution from the 110V source:

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