EE 207 Homework2 - Haifa Altamimi#131372 Dr Sghaier Guizani...

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Haifa Altamimi #131372 Dr. Sghaier Guizani October 21 st , 2014 EE 207 Homework #2 Chapter 2 1. Problem 2.30: Find R eq for the circuit in Fig. 2.94. The 180 and 60 are in series. Therefore: Ω Ω R eq 1 = 180+60 = 240 Now the 240 and 60 are in parallel, so: Ω Ω 1 240 + 1 60 = 1 48 = ¿ R eq 2 = 48. Remaining is the 25 and 48 , which are in series=> 25+48= Ω Ω R eq = 73 Ω 2. Problem 2.36: Find i and V 0 in the circuit of Fig. 2.100 a) We combine the 60 and 20 : Ω Ω 1 60 + 1 20 = 1 15 = ¿ 15 b) 15 + 25 = 40 Ω Ω Ω c) 30 +50 = 80 Ω Ω Ω d) 1 80 + 1 20 = 1 16 = ¿ 16 e) 16 Ω + 24 Ω = 40 Ω f) 1 40 + 1 40 = 1 20 = ¿ 20 g) R eq = 20 + 80 = 100 i= V R eq = 20 100 = 0.2 A a = ¿ V ¿ 30 i 0 i 1 = 40 40 + 40 i = 0.1 A i 0 = 20 20 + 80 i 1 = 0.02 A a = ¿ V ¿ 30(0.02) = 0.6V
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Haifa Altamimi #131372 Dr. Sghaier Guizani October 21 st , 2014 3. Problem 2.48 a) Convert the circuits in Fig. 2.112 from Y to R a = R 1 R 2 + R 2 R 3 + R 3 R 1 R 1 = ( 10 ) ( 10 ) + ( 10 ) ( 10 ) +( 10 )( 10 ) 10 =30 R b = R 1 R 2 + R 2 R 3 + R 3 R 1 R 2 = ( 10 ) ( 10 ) + ( 10 ) ( 10 ) +( 10 )( 10 ) 10 =30 R c = R 1 R 2 + R 2 R 3 + R 3 R 1 R 3 = ( 10 ) ( 10 ) + ( 10 ) ( 10 ) +( 10 )( 10 ) 10 =30 4. Problem 2.49 a) Transform the circuits in Fig. 2.113 from to Y R 1 = R b R c R a + R b + R c = ( 12 ) ( 12 ) 12 + 12 + 12 = 4 R 2 = R c R a R a + R b + R c = ( 12 ) ( 12 ) 12 + 12 + 12 = 4 R 3 = R b R a R a + R b + R c = ( 12 ) (
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