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Mr W Presents Series/Parallel Reduction

This text explains how to analyze a complicated circuit by identifying series and parallel resistors and combining them. It also demonstrates the importance of following the current when simplifying the circuit.

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Mr W Presents Series/Parallel Reduction

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  1. Mr W Presents Series/Parallel Reduction

  2. 2kΩ 10kΩ 2kΩ 6kΩ 4kΩ 6kΩ 1kΩ 9kΩ 2kΩ Let’s Look at a Rather Nasty Circuit 6kΩ 6V Ω

  3. 2kΩ 10kΩ 2kΩ 6kΩ 4kΩ 6kΩ 1kΩ 9kΩ 2kΩ When in Doubt…Follow the Current!Current must change when it hits a branch– so let’s identify various currents and branches I2 I5 I1 I6 I7 I8 I4 I3 6kΩ 6V I9 I8 normally we’d include arrows to show the direction of current but that cluttered up my drawing too much

  4. 2kΩ 10kΩ 2kΩ 6kΩ 4kΩ 6kΩ 1kΩ 9kΩ 2kΩ 1) Identify obvious resistors in series and combine them I2 I5 I1 I6 I4 I3 6kΩ I7 I8 6V I9 I8

  5. 2kΩ 2kΩ 10kΩ 10kΩ 2kΩ 2kΩ 6kΩ 6kΩ 4kΩ 4kΩ 6kΩ 6kΩ 1kΩ 9kΩ 9kΩ 2kΩ 3kΩ I2 I5 I1 I7 I6 I4 I3 6kΩ 6V I9 I8 Current I8 moves across 1 KΩ and 2 KΩ resistors so lets redraw the circuit to reflect that I2 I5 I1 I7 I6 I8 I4 I3 6kΩ 6V I9 I8

  6. 2kΩ 10kΩ 2kΩ 6kΩ 4kΩ 6kΩ 9kΩ 3kΩ Since we started at the far right, let’s continue thereTake a look at I7 and I8… series or parallel? I2 I5 P I1 I7 I6 I4 I3 6kΩ I8 6V P2 I9 I8 Hint: Start with current flowing into point P and FOLLOW THE CURRENT through to point P2

  7. P 6kΩ 6kΩ I7 I8 3kΩ 3kΩ P2 I8 Which then reduces to: 1/3 + 1/6 = 1/Req = 2 Ω The current flowing into point P branches 1) and flows as current I7 through a 6 K ohm resistor and ends at point P2 2) and ALSO flows as current I8 through a 3 K ohm resistor to finish at poing p2 Talk to the person next to you and explain why we can rewrite the segment of the current like this:

  8. Work with your group to redraw the circuit incorporating the changes so far including I7 and I8 in parallel

  9. 2kΩ 10kΩ 2kΩ 6kΩ 4kΩ 9kΩ Does it look like this? I2 I5 P I1 I6 I7 I4 I3 6kΩ 6V 2 Ω P2 I9 Take a look at that diagram and talk with your groupies please….what shall we tackle next? I7 Ω Ω

  10. I7 2kΩ 2kΩ 10kΩ 2kΩ 2kΩ 6kΩ 6kΩ 4kΩ 4kΩ 9kΩ 9kΩ I2 I5 P Ω I1 I6 I7 I4 I3 6kΩ 6V 2 Ω P2 I9 SURE! I5 and I7 are in series I2 I5 I1 I6 I5 I4 I3 6kΩ 6V 12 Ω I9 I5 & I6 are now in parallel

  11. SURE! I5 and I6 are in parallel I7 2kΩ 2kΩ 2kΩ 4kΩ 4kΩ 9kΩ 9kΩ Ω I2 I1 I5 I4 I3 6kΩ 6V 1/6kΩ + 1/12 kΩ = 1/R = 4Ω I9 NOW WHAT <click> I2 & I5 are in SERIES <click> I1 I5 I4 I3 6kΩ 6V 4 kΩ + 2k I9

  12. I7 2kΩ 4kΩ 9kΩ I5 Ω I1 I4 I3 6kΩ 6 kΩ 6V I9 BE CAREFUL… when in doubt, FOLLOW THE CURRENT. Let’s do that from I1 after the 2 k ohm resistor: • I1 splits into I3, I4, & I5. • I3 & I5 branch and then reconnect… so they are in parallel!

  13. I7 2kΩ 2kΩ 2kΩ 4kΩ 4kΩ 4kΩ 9kΩ 9kΩ Ω I1 I4 I3 6 kΩ 6 kΩ 6V I9 Which gives us… I3 I1 I4 1/6kΩ + 1/6kΩ = 1/R =3kΩ 6V I9 I3 I1 Followed by I3 & I9 in series I4 6V 12 kΩ

  14. I7 2kΩ 2kΩ 5kΩ 4kΩ I3 Ω I1 I4 & I3 in parallel I4 6V 12 kΩ I4 6V ¼ kΩ + 1/12 kΩ = 1/R = 3 kΩ Which goes to just 6V

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