Transcript Mr W Presents Series/Parallel Reduction

Mr W Presents Series/Parallel
Reduction
Let’s Look at a Rather Nasty Circuit
2kΩ
2kΩ
6kΩ
6V
4kΩ
9kΩ
Ω
10kΩ
6kΩ
6kΩ
2kΩ
1kΩ
When in Doubt…Follow the
Current!
Current must change when it hits a
branch– so let’s identify various currents
and branches
2kΩ
I2
I1
6V
I4
4kΩ
I3
2kΩ
6kΩ
I9 9kΩ
I5
I6
6kΩ
I8
10kΩ
I7 I8
6kΩ
1kΩ
2kΩ
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
2kΩ
I2
I1
6V
I4
4kΩ
I3
2kΩ
6kΩ
I9 9kΩ
I5
I6
6kΩ
I8
10kΩ
I7
6kΩ
2kΩ
I8
1kΩ
Current I8 moves across 1 KΩ and 2 KΩ
resistors so lets redraw the circuit to reflect that
2kΩ
I2
I1
6V
I4
4kΩ
I3
2kΩ
6kΩ
I9 9kΩ
2kΩ
I2
I1
6V
I4
4kΩ
I3
2kΩ
6kΩ
I9 9kΩ
10kΩ
I5
I6
6kΩ
I8
I5
I6
6kΩ
I8
I7
6kΩ
2kΩ
10kΩ
I7
6kΩ
3kΩ
I8
1kΩ
Since we started at the far right, let’s continue
there
Take a look at I7 and I8… series or parallel?
2kΩ
I2
I1
6V
I4
4kΩ
I3
2kΩ
I5
I6
6kΩ
6kΩ
I9 9kΩ
P2
I8
10kΩ
I7
P
I8
6kΩ
3kΩ
Hint: Start with current flowing into point P and
FOLLOW THE CURRENT through to point P2
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
I7
6kΩ
P2
I8
P
I8
Talk to the person next to you and explain why we can
rewrite the segment of the current like this:
3kΩ
3kΩ
Which then reduces to:
1/3 + 1/6 = 1/Req = 2 Ω
Work with your group to redraw the circuit
incorporating the changes so far including I7 and I8
in parallel
Does it look like this?
2kΩ
I2
I1
6V
I4
4kΩ
I3
2kΩ
I5
I6
6kΩ
6kΩ
I9 9kΩ
10kΩ
P
I7
2Ω
P2
Take a look at that diagram and talk with your
groupies please….what shall we tackle next?
I7
Ω
Ω
I7
2kΩ
I2
I1
6V
I4
4kΩ
I3
2kΩ
I5
I6
6kΩ
6kΩ
10kΩ
Ω
P
I7
2Ω
I9 9kΩ
P2
SURE! I5 and I7 are in series
2kΩ
I2
I1
6V
I4
4kΩ
I3
2kΩ
6kΩ
I5
I6
6kΩ
I9 9kΩ
I5 & I6 are now in parallel
I5
12 Ω
I7
SURE! I5 and I6 are in parallel
2kΩ
I2
I1
I4
4kΩ
6V
I3
Ω
2kΩ
I5
1/6kΩ + 1/12 kΩ = 1/R = 4Ω
6kΩ
I9 9kΩ
NOW WHAT
<click>
I2 & I5 are in SERIES
<click>
2kΩ
I1
6V
I4
4kΩ
I3
6kΩ
I9 9kΩ
I5
4 kΩ + 2k
I7
2kΩ
I5
I1
6V
I4
4kΩ
I3
6kΩ
Ω
6 kΩ
I9 9kΩ
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
2kΩ
Ω
I1
I4
4kΩ
6V
Which gives us…
2kΩ
6 kΩ
6 kΩ
I9 9kΩ
I3
I1
6V
I3
I4
4kΩ
1/6kΩ + 1/6kΩ = 1/R =3kΩ
I9 9kΩ
2kΩ
I3
I1
Followed by I3
& I9 in series
6V
I4
4kΩ
12 kΩ
I7
2kΩ
I3
I1
I4 & I3 in parallel
I4
4kΩ
6V
12 kΩ
2kΩ
6V
I4
¼ kΩ + 1/12 kΩ = 1/R = 3 kΩ
5kΩ
Which goes to just
6V
Ω