1. Mr W Presents Series/Parallel Reduction

2. Let’s Look at a Rather Nasty Circuit 6V6V 4kΩ6kΩ 10kΩ 9kΩ2kΩ 1kΩ 6kΩ 2kΩ 6kΩ Ω 2kΩ

3. When in Doubt…Follow the Current! Current must change when it hits a branch– so let’s identify various currents and branches 6V6V 4kΩ6kΩ 10kΩ 9kΩ2kΩ 1kΩ 6kΩ 2kΩ 6kΩ 2kΩ I3I3 I2I2 I5I5 I4I4 I1I1 I7I7 I6I6 I8I8 I8I8 I9I9 normally we’d include arrows to show the direction of current but that cluttered up my drawing too much

4. 1) Identify obvious resistors in series and combine them 6V6V 4kΩ6kΩ 10kΩ 9kΩ2kΩ 1kΩ 6kΩ 2kΩ 6kΩ 2kΩ I3I3 I2I2 I5I5 I4I4 I1I1 I7I7 I6I6 I8I8 I9I9 I8I8

5. Current I 8 moves across 1 KΩ and 2 KΩ resistors so lets redraw the circuit to reflect that 6V6V 4kΩ6kΩ 10kΩ 9kΩ2kΩ 1kΩ 6kΩ 2kΩ 6kΩ 2kΩ I3I3 I2I2 I5I5 I4I4 I1I1 I7I7 I6I6 I8I8 I8I8 I9I9 6V6V 4kΩ6kΩ 9kΩ3kΩ 6kΩ I3I3 I2I2 I5I5 I4I4 I1I1 I7I7 I6I6 I8I8 I9I9 10kΩ2kΩ

6. Since we started at the far right, let’s continue there Take a look at I 7 and I 8 … series or parallel? 6V6V 4kΩ6kΩ 9kΩ3kΩ 6kΩ I3I3 I2I2 I5I5 I4I4 I1I1 I7I7 I6I6 I9I9 10kΩ2kΩ I8I8 I8I8 Hint: Start with current flowing into point P and FOLLOW THE CURRENT through to point P 2 P P2P2

7. The current flowing into point P branches Talk to the person next to you and explain why we can rewrite the segment of the current like this: 3kΩ 6kΩ 1/3 + 1/6 = 1/R eq = 2 Ω Which then reduces to: 2) and ALSO flows as current I 8 through a 3 K ohm resistor to finish at poing p2 1) and flows as current I 7 through a 6 K ohm resistor and ends at point P 2 6kΩ 3kΩ I7I7 I8I8 I8I8 P P2P2

8. Work with your group to redraw the circuit incorporating the changes so far including I 7 and I 8 in parallel

9. Does it look like this? 6V6V 4kΩ 9kΩ 6kΩ I3I3 I2I2 I5I5 I4I4 I1I1 I7I7 I6I6 I9I9 10kΩ 2kΩ P P2P2 Ω 2 Ω Ω I7I7 Take a look at that diagram and talk with your groupies please….what shall we tackle next?

10. 6V6V 4kΩ 9kΩ 6kΩ I3I3 I2I2 I5I5 I4I4 I1I1 I7I7 I6I6 I9I9 10kΩ 2kΩ P P2P2 Ω 2 Ω I7I7 SURE! I 5 and I 7 are in series 6V6V 4kΩ 9kΩ 6kΩ I3I3 I2I2 I5I5 I4I4 I1I1 I6I6 I9I9 2kΩ 12 Ω I5I5 I 5 & I 6 are now in parallel

11. I7I7 Ω SURE! I 5 and I 6 are in parallel 6V6V 4kΩ 9kΩ 1/6kΩ + 1/12 k Ω = 1/R = 4 Ω 6kΩ I3I3 I2I2 I4I4 I1I1 I5I5 I9I9 2kΩ NOW WHAT I 2 & I 5 are in SERIES 6V6V 4kΩ 9kΩ 4 k Ω + 2k 6kΩ I3I3 I4I4 I1I1 I5I5 I9I9 2kΩ

12. I7I7 Ω 6V6V 4kΩ 9kΩ 6 k Ω I3I3 I4I4 I1I1 I5I5 I9I9 2kΩ BE CAREFUL… when in doubt, FOLLOW THE CURRENT. Let’s do that from I 1 after the 2 k ohm resistor: I 1 splits into I 3, I 4, & I 5. I 3 & I 5 branch and then reconnect… so they are in parallel!

13. I7I7 Ω 6V6V 4kΩ 9kΩ 6 k Ω I3I3 I4I4 I1I1 I9I9 2kΩ 6 k Ω Which gives us… 6V6V 4kΩ 9kΩ I3I3 I4I4 I1I1 I9I9 2kΩ 1/6kΩ + 1/6k Ω = 1/R =3kΩ 6V6V 4kΩ I3I3 I4I4 I1I1 2kΩ 12 kΩ Followed by I 3 & I 9 in series

14. I7I7 Ω 6V6V 4kΩ I3I3 I4I4 I1I1 2kΩ 12 kΩ I 4 & I 3 in parallel 6V6V ¼ kΩ + 1/12 kΩ = 1/R = 3 kΩ I4I4 2kΩ Which goes to just 6V6V 5kΩ