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EE100/42 – Spring 2011 – Prof. Poolla
HW2 Solution
P2.3.
18Ω and 9Ω resistances are in parallel because they have the
same end nodes (node a and node c). Similarly 12Ω and 6Ω
resistances are in parallel:
18 Ω  9 Ω = 6 Ω
12 Ω  6 Ω = 4 Ω
The simplified circuit will be:
Now 6 Ω and 4 Ω are in series:
R
eq
= 6 Ω + 4 Ω = 10 Ω
P2.12.
Using Figure P2.12(b) we could replace the Ladder Network (a) box with a R
eq
resistance.:
Then R
eq
and 2Ω resistance are in parallel and their combination is in series with two 1 Ω
resistances.
±
ୣ୯
ൌ
ʹ±
ୣ୯
±
ୣ୯ାଶ
ͳ π ͳπ
Therefore R
eq
2
2R
eq
4=0
And R
eq
=(1+√5) Ω
where we ignored the (1√5) Ω answer because it is negative and could not be an answer.
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View Full DocumentP2.27.
The equivalent circuit seen by the current source is:
R
eq
= (12Ω6 Ω)+8 Ω+(20 Ω30 Ω)
= 4 Ω+8 Ω+12 Ω=24 Ω
Therefore
v=3R
eq
=72V
Now using currentdivision principle:
݅
ଵ
ൌ
ʹͲ
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 Spring '07
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