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Kirchhoff’s Rules

Kirchhoff’s Rules. When series and parallel combinations aren’t enough. Some circuits have resistors which are neither in series nor parallel. They can still be analyzed, but one uses Kirchhoff’s rules. Not in series.

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Kirchhoff’s Rules

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

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

  3. 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.

  4. 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.

  5. 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!

  6. 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

  7. 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

  8. 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

  9. Branch version

  10. 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.

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

  12. 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

  13. 5 = 1 • I1 + 2.2 • (I1 – I3) I1.5  I1  I3  I1-I3  I1.5+I3 

  14. 1 • I1 + 3 • I3 = 1.5 • I1.5 I1.5  I1  I3  I1-I3  I1.5+I3 

  15. 2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3) I1.5  I1  I3  I1-I3  I1.5+I3 

  16. 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.

  17. 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

  18. Comparison with Simulation

  19. 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

  20. Loop version

  21. 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.

  22. 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

  23. 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

  24. 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

  25. 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

  26. 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.

  27. 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.

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

  29. Matrix equation

  30. 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

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

  32. 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

  33. Result of matrix inversion

  34. 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

  35. Results of Matrix Multiplication

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