HW2Solution

# HW2Solution - EE100/42 Spring 2011 Prof Poolla HW2 Solution...

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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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P2.27. The equivalent circuit seen by the current source is: R eq = (12Ω||6 Ω)+8 Ω+(20 Ω||30 Ω) = 4 Ω+8 Ω+12 Ω=24 Ω Therefore v=3R eq =72V Now using current-division principle: ݅ ʹͲ
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## This note was uploaded on 02/09/2011 for the course EE 100 taught by Professor Boser during the Spring '07 term at Berkeley.

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HW2Solution - EE100/42 Spring 2011 Prof Poolla HW2 Solution...

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