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Combinations of Resistors

Combinations of Resistors. Series, Parallel and Kirchhoff. Simplifying circuits using series and parallel equivalent resistances. Analyzing a combination of resistors circuit.

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Combinations of Resistors

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  1. Combinations of Resistors Series, Parallel and Kirchhoff

  2. Simplifying circuits using series and parallel equivalent resistances

  3. Analyzing a combination of resistors circuit • Look for resistors which are in series (the current passing through one must pass through the other) and replace them with the equivalent resistance (Req = R1 + R2). • Look for resistors which are in parallel (both the tops and bottoms are connected by wire and only wire) and replace them with the equivalent resistance (1/Req = 1/R1 + 1/R2). • Repeat as much as possible.

  4. Look for series combinations Req=3k Req=3.6 k

  5. Look for parallel combinations Req = 1.8947 k Req = 1.1244 k

  6. Look for series combinations Req = 6.0191 k

  7. Look for parallel combinations Req = 2.1314 k

  8. Look for series combinations Req = 5.1314 k

  9. Equivalent Resistance I = V/R = (5 V)/(5.1314 k) = 0.9744 mA

  10. Backwards 1 V= (3)(.9744) = 2.9232 V= (2.1314)(.9744) = 2.0768

  11. Backwards 2 V = 2.0768=I (3.3) I=0.629mA V = 2.0768=I (6.0191) I=0.345mA

  12. Backwards 3 V=(.345)(3)=1.035 V=(.345)(1.8947)=0.654 V=(.345)(1.1244)=0.388

  13. Kirchhoff’s Rules When series and parallel combinations aren’t enough

  14. Some circuits have resistors which are neither in series nor parallel They can still be analyzed, but one uses Kirchhoff’s rules.

  15. Not in series The 1-k resistor is not in series with the 2.2-k since the some of the current that went through the 1-k might go through the 3-k instead of the 2.2-k resistor.

  16. Not in parallel The 1-k resistor is not in parallel with the 1.5-k since their bottoms are not connected simply by wire, instead that 3-k lies in between.

  17. Kirchhoff’s Node Rule • A node is a point at which wires meet. • “What goes in, must come out.” • Recall currents have directions, some currents will point into the node, some away from it. • The sum of the current(s) coming into a node must equal the sum of the current(s) leaving that node. • I1 + I2 = I3  I2 I1   I3 The node rule is about currents!

  18. Kirchhoff’s Loop Rule 1 • “If you go around in a circle, you get back to where you started.” • If you trace through a circuit keeping track of the voltage level, it must return to its original value when you complete the circuit • Sum of voltage gains = Sum of voltage losses

  19. Batteries (Gain or Loss) • Whether a battery is a gain or a loss depends on the direction in which you are tracing through the circuit Loop direction Loop direction Loss Gain

  20. Resistors (Gain or Loss) • Whether a resistor is a gain or a loss depends on whether the trace direction and the current direction coincide or not. I I Loop direction Loop direction Current direction Current direction Loss Gain

  21. Branch version

  22. Neither Series Nor Parallel I1.5  I1  I3  I1.7  I2.2  Assign current variables to each branch. Draw loops such that each current element is included in at least one loop.

  23. Apply Current (Node) Rule I1.5  I1  * * I3  I1-I3  I1.5+I3  *Node rule applied.

  24. Three Loops • Voltage Gains = Voltage Losses • 5 = 1 • I1 + 2.2 • (I1 – I3) • 1 • I1 + 3 • I3 = 1.5 • I1.5 • 2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3) • Units: Voltages are in V, currents in mA, resistances in k

  25. 5 = 1 • I1 + 2.2 • (I1 – I3) I1.5  I1  I3  I1-I3  I1.5+I3 

  26. 1 • I1 + 3 • I3 = 1.5 • I1.5 I1.5  I1  I3  I1-I3  I1.5+I3 

  27. 2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3) I1.5  I1  I3  I1-I3  I1.5+I3 

  28. Simplified Equations • 5 = 3.2 • I1 - 2.2 • I3 • I1 = 1.5 • I1.5 - 3 • I3 • 0 = -2.2 • I1 + 1.7 • I1.5 + 6.9 • I3 • Substitute middle equation into others • 5 = 3.2 • (1.5 • I1.5 - 3 • I3) - 2.2 • I3 • 0 = -2.2 • (1.5 • I1.5 - 3 • I3) + 1.7 • I1.5 + 6.9 • I3 • Multiply out parentheses and combine like terms.

  29. Solving for I3 • 5 = 4.8 • I1.5 - 11.8 • I3 • 0 = - 1.6 I1.5 + 13.5 • I3 • Solve the second equation for I1.5 and substitute that result into the first • 5 = 4.8 • (8.4375 I3 ) - 11.8 • I3 • 5 = 28.7 • I3 • I30.174 mA

  30. Comparison with Simulation

  31. Other currents • Return to substitution results to find other currents. • I1.5 = 8.4375 I3 = 1.468 mA • I1 = 1.5 • I1.5 - 3 • I3 • I1 = 1.5 • (1.468)- 3 • (0.174) • I1 = 1.68 mA

  32. Loop version

  33. Neither Series Nor Parallel JB JA JC Draw loops such that each current element is included in at least one loop. Assign current variables to each loop. Current direction and lop direction are the same.

  34. Loop equations • 5 = 1  (JA - JB) + 2.2  (JA - JC) • 0 = 1 (JB - JA) + 1.5  JB + 3 (JB - JC) • 0 = 2.2 (JC - JA) + 3 (JC - JB) + 1.7 JC • “Distribute” the parentheses • 5 = 3.2 JA – 1 JB - 2.2 JC • 0 = -1 JA + 5.5 JB – 3 JC • 0 = -2.2JA – 3 JB + 6.9 JC

  35. Algebra • JC = (2.2/6.9)JA + (3/6.9)JB • JC = 0.3188 JA + 0.4348 JB • 5 = 3.2 JA – 1 JB - 2.2 (0.3188 JA + 0.4348 JB) • 0 = -1 JA + 5.5 JB – 3 (0.3188 JA + 0.4348 JB) • 5 = 2.4986 JA – 1.9566 JB • 0 = -1.9564 JA + 4.1956 JB

  36. More algebra • JB = (1.9564/4.1956) JA • JB = 0.4663 JA • 5 = 2.4986 JA – 1.9566 (0.4663 JA) • 5 = 1.5862 JA • JA = 3.1522 mA

  37. Other loop currents • JB = 0.4663 JA = 0.4663 (3.1522 mA) • JB = 1.4699 mA • JC = 0.3188 JA + 0.4348 JB • JC = 0.3188 (3.1522) + 0.4348 (1.4699) • JC = 1.644 mA

  38. Branch Variables I1.5  I1  I3  I1.7  I2.2  Assign current variables to each branch. Draw loops such that each current element is included in at least one loop.

  39. Loop Variables JB JA JC Draw loops such that each current element is included in at least one loop. Assign current variables to each loop. Current direction and lop direction are the same.

  40. Branch Currents from Loop currents • I1 = JA – JB = 3.1522 – 1.4699 = 1.6823 mA • I1.5 = JB = 1.4699 mA

  41. Matrix equation

  42. Loop equations as matrix equation • 5 = 3.2 JA – 1 JB - 2.2 JC • 0 = -1 JA + 5.5 JB – 3 JC • 0 = -2.2JA – 3 JB + 6.9 JC

  43. Enter matrix in Excel, highlight a region the same size as the matrix.

  44. In the formula bar, enter =MINVERSE(range) where range is the set of cells corresponding to the matrix (e.g. B1:D3). Then hit Crtl+Shift+Enter

  45. Result of matrix inversion

  46. Prepare the “voltage vector”, then highlight a range the same size as the vector and enter =MMULT(range1,range2) where range1 is the inverse matrix and range2 is the voltage vector. Then Ctrl-Shift-Enter. Voltage vector

  47. Results of Matrix Multiplication

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