ECE2040D_HW4_Soln - ECE 2040D Solutions to HW#4 Spring 2009...

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(a) P.5.2-1 ECE 2040D Solutions to HW#4 - Spring 2009 Page 1 / 16
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= 2 = 0.5 V t t R v g63 g58 g16 (b) 9 4 2 ( 0.5) 0 9 ( 0.5) 1.58 A 4 2 i i i g16 g16 g16 g14 g16 g32 g16 g14 g16 g32 g32g16 g14 (c) 9 4 9 4( 1.58) 2.67 V 1.58 a v i i i A g32 g14 g32 g14 g16 g32 g32 g32g16 ECE 2040D Solutions to HW#4 - Spring 2009 Page 2 / 16
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12 6 24 3 3 0 1 A a a a i i i g16 g16 g14 g16 g16 g32 g159 g32 P5.2-5 P5.3–1 g172 g188 g14 g171 g187 g170 g186 g32 g159 v g32 v 20 10 35 9 40 V 10 2 2 Use current division: Consider 9 A source only (open 6 A source) g172 g188 g171 g187 g170 g186 20 15 g14 30 g32 g159 6 v g32 40 V v 15 1 1 Use current division: Consider 6 A source only (open 9 A source) ECE 2040D Solutions to HW#4 - Spring 2009 Page 3 / 16
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g32 g32g16 12 2 i A g16 6 1 2 6 3 i i g14 g14 g14 2 3 i g32 0 2 2 2 g11 g12 g11 g12 Apply KVL to the supermesh: current source. be the part of i due to the 2 A 2 x Consider 2 A source only (short the 8 V source) Let i 12 3 i g32 g32 A 8 2 1 6 3 2 i i g14 g14 i g16 8 g32 0 1 1 1 Apply KVL to the supermesh: voltage source. 1 x Consider 8 V source only (open the 2 A source) Let i be the part of i due to the 8 V P5.3-5 g11 g12 g11 g12 g11 g12 i i g32 g14 i g32 g16 g32 3 2 6 A 2 1 1 x 1 2 Finally, ECE 2040D Solutions to HW#4 - Spring 2009 Page 4 / 16
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ECE 2040D Solutions to HW#4 - Spring 2009 Page 5 / 16
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P5.4-1 ECE 2040D Solutions to HW#4 - Spring 2009 Page 6 / 16
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Find v : oc Notice that v is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: oc v a = - v Apply KCL at node a: 6 3 0 8 4 4 oc oc oc v v v g16 g167 g183 g167 g183 g16 g14 g14 g16 g168 g184 g168 g184 g169 g185 g169 g185 g32 V 6 2 6 0 2 oc oc oc oc v v v v g16 g14 g14 g16 g32 g159 g32g16 Find R t : We’ll find i sc and use it to calculate R t .
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