Mr W Presents Series/Parallel Reduction

Let’s Look at a Rather Nasty Circuit

6V 4kΩ 6kΩ

10kΩ

9kΩ 2kΩ

1kΩ 6kΩ

2kΩ

6kΩ

Ω

2kΩ

When in Doubt…Follow the Current!

Current must change when it hits a branch– so let’s identify various currents

and branches

6V 4kΩ 6kΩ

10kΩ

9kΩ 2kΩ

1kΩ 6kΩ

2kΩ

6kΩ

2kΩ

I3

I2 I5

I4

I1

I7I6 I8

I8I9

normally we’d include arrows to show the direction of current but that cluttered up my drawing too much

1) Identify obvious resistors in series and combine them

6V 4kΩ 6kΩ

10kΩ

9kΩ 2kΩ

1kΩ 6kΩ

2kΩ

6kΩ

2kΩ

I3

I2 I5

I4

I1

I7I6

I8I9

I8

Current I8 moves across 1 KΩ and 2 KΩ resistors so lets redraw the circuit to reflect that

6V 4kΩ 6kΩ

10kΩ

9kΩ 2kΩ

1kΩ 6kΩ

2kΩ

6kΩ

2kΩ

I3

I2 I5

I4

I1 I7I6 I8

I8I9

6V 4kΩ 6kΩ

9kΩ 3kΩ

6kΩ 6kΩ I3

I2 I5

I4

I1 I7I6

I8I9

10kΩ 2kΩ 2kΩ

Since we started at the far right, let’s continue thereTake a look at I7 and I8… series or parallel?

6V 4kΩ 6kΩ

9kΩ 3kΩ

6kΩ 6kΩ I3

I2 I5

I4

I1 I7I6

I9

10kΩ 2kΩ 2kΩ

I8

I8

Hint: Start with current flowing into point P and FOLLOW THE CURRENT through to point P2

P

P2

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/Req = 2 ΩWhich then reduces to:

2) and ALSO flows as current I8 through a 3 K ohm resistor to finish at poing p2

1) and flows as current I7 through a 6 K ohm resistor and ends at point P2

6kΩ

3kΩ

I7

I8

I8

P

P2

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

in parallel

Does it look like this?

6V 4kΩ

9kΩ

6kΩ 6kΩ I3

I2 I5

I4

I1

I7

I6

I9

10kΩ 2kΩ 2kΩ P

P2

Ω

2 Ω

Ω

I7

Take a look at that diagram and talk with your groupies please….what shall we tackle next?

6V 4kΩ

9kΩ

6kΩ 6kΩ I3

I2 I5

I4

I1

I7

I6

I9

10kΩ 2kΩ 2kΩ P

P2

Ω

2 Ω

I7

SURE! I5 and I7 are in series

6V 4kΩ

9kΩ

6kΩ 6kΩ I3

I2 I5

I4

I1

I6

I9

2kΩ 2kΩ

12 ΩI5

I5 & I6 are now in parallel

I7

Ω

SURE! I5 and I6 are in parallel

6V 4kΩ

9kΩ

1/6kΩ + 1/12 kΩ = 1/R = 4Ω 6kΩ I3

I2

I4

I1

I5

I9

2kΩ 2kΩ

NOW WHAT <click>

I2 & I5 are in SERIES <click>

6V 4kΩ

9kΩ

4 kΩ + 2k6kΩ I3I4

I1

I5

I9

2kΩ

I7

Ω

6V 4kΩ

9kΩ

6 kΩ6kΩ I3I4

I1

I5

I9

2kΩ

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!

I7

Ω

6V 4kΩ

9kΩ

6 kΩI3I4

I1

I9

2kΩ

6 kΩ

Which gives us…

6V 4kΩ

9kΩ

I3

I4

I1

I9

2kΩ

1/6kΩ + 1/6kΩ = 1/R =3kΩ

6V 4kΩ

I3

I4

I1

2kΩ

12 kΩFollowed by I3 & I9 in series

I7

Ω

6V 4kΩ

I3

I4

I1

2kΩ

12 kΩI4 & I3 in parallel

6V ¼ kΩ + 1/12 kΩ = 1/R = 3 kΩI4

2kΩ

Which goes to just6V

5kΩ