Solution to Homework 10

# Solution to Homework 10 - Homework 10 Solutions P10.11-2...

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Unformatted text preview: Homework 10 Solutions P10.11-2 Use superposition 1 3 0.5 mA 6000 = = I 3 2 1 45 ( ) 0.166 10 45 A 6000 0.2 j ϖ-- ∠ ° = = - × ∠ ° + I 2 1 ( ) ( ) ( ) 0.166cos(4 45 ) 0.5 mA 0.166cos(4 135 ) 0.5 mA i t i t i t t t = + = - + ° + =- ° + P10.11-4 Find oc V : ( 29 ( 29 oc 80 80 5 30 80 80 20 80 2 21.9 5 30 100 36.90 4 2 21.9 V j j j + = ∠- ° +- ∠- ° = ∠- ° ∠ ° = ∠ - ° V Find t Z : ( 29 ( 29 t 20 80 80 23 81.9 20 80 80 j j j j- + = = ∠ - ° Ω- + + Z The Thevenin equivalent is P10.11-6 First, determine oc V : The node equation is: oc oc oc (6 8) (6 8) 3 4 2 2 2 j j j j j - +- + +- = - V V V oc 3 4 5 53.1 V j = + = ∠ ° V s 10 53 6 8 V j = ∠ ° = + V Next, determine sc I : The node equation is: (6 8) 3 (6 8) 2 4 2 2 2 j j j j j - +- + + +- = - V V V V 3 4 1 j j + =- V sc 3 4 2 2 2 j j + = =- V I s 10 53 6 8 V j = ∠ ° = + V The Thevenin impedance is oc T sc 2 2 3 4 2 2 3 4 j j j j - = = + =- Ω + V Z I The Thevenin equivalent is P10.11-8 In general: oc L oc t L t L and + + = = V Z I V V Z Z Z Z In the three given cases, we have 1 1 1 1 25 50 0.5 A 50 = Ω ⇒ = = = V Z I Z 2 2 2 6 2 1 1 100 200 0.5 A0....
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Solution to Homework 10 - Homework 10 Solutions P10.11-2...

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