Chap_5_Solut_Dorf

Chap_5_Solut_Dorf - Chapter 5 Circuit Theorems Exercises Ex...

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Chapter 5 Circuit Theorems Exercises Ex 5.2-1 R = 10 Ω and i s = 1.2 A. Ex 5.2-2 R = 10 Ω and i s = 1.2 A. Ex 5.2-3 R = 8 Ω and v s = 24 V. Ex 5.2-4 R = 8 Ω and v s = 24 V. Ex 5.3-1 () 20 10 2 15 20 2 6 20( ) 2 V 10 20 20 10 (20 20) 5 m v ⎡⎤ =+ = + ⎢⎥ ++ + + ⎣⎦ =
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Ex 5.4-1
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Ex 5.4-2 21 2 3 A 6 26 V a aa oc a i ii vi =⇒ = == () 12 6 2 3 A 2 32 3 a sc a sc i i += = = = A 6 3 2 t R = Ex 5.5-1
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Ex 5.6-1 () 6 18 12 V 63 oc v == + ( )( ) 36 24 t R = += + Ω
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For maximum power, we require 4 Lt RR = Then () 2 2 max 12 9 W 44 4 oc t v p R == = Ex 5.6-2 1 50 3 5.6 5.6 5 A 11 1 50 1 5 3 150 30 sc i ++ = ( ) 150 30 33 2 5 150 30 t R =+ =+ = Ω + 2 8 ( ) 2 2 max 28 5 175 W ts c Ri p =
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Problems Section 5-2: Source Transformations P5.2-1 (a)
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= 2 = 0.5 V t t R v Ω (b) 9 4 2 ( 0.5) 0 9 ( 0.5) 1.58 A 42 ii i −−−+ − = −+− == + 9 4 9 4( 1.58) 2.67 V vi =+ =+ − = (c) 1.58 A a (checked using LNAP 8/15/02) P5.2-2 Finally, apply KVL: 16 10 3 4 0 2.19 A 3 aa a i −+ + − = ∴= (checked using LNAP 8/15/02)
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P5.2-3 Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors: Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top: Source transformation at left; series resistors at right: Parallel resistors, then source transformation at left:
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Finally, apply KVL to loop o 6 ( 91 9 )3 6 0 iv −+ + − − = o 5/ 2 42 28 (5/ 2) 28 V =⇒ = + = (checked using LNAP 8/15/02) P5.2-4 4 2000 4000 10 2000 3 0 375 A aa a a ii i i μ −− +− = ∴= (checked using LNAP 8/15/02)
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P5.2-5 12 6 24 3 3 0 1 A aa a ii i −− + − − = ⇒ = (checked using LNAP 8/15/02) P5.2-6 A source transformation on the right side of the circuit, followed by replacing series resistors with an equivalent resistor: Source transformations on both the right side and the left side of the circuit:
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Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources with an equivalent current source: Finally, ( ) () 50 100 100 0.21 0.21 7 V 50 100 3 a v == + = (checked using LNAP 8/15/02) P5.2-7 Use source transformations to simplify the circuit:
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Label the node voltages. The 8-V source is connected between nodes 1 and 3. Consequently, 13 8 vv = Apply KCL to the supernode corresponding to the 8-V source to get 12 3 3 24 10 0 0.125 0.3 0.05 0.02 0.2 0 82 0 5 0 v v −− ++ = −+ + −= Apply KCL at node 2 to get 2 23 3 =0.04 0.19 0.1 0 25 20 10 vv v vv v =+ + = Solving, for example using MATLAB 11 22 33 1 0 1 8 4.7873 0.125 0.05 0.02 0.5 0.6831 0.04 0.19 0.1 0 3.2127 ⎡⎤ ⎢⎥ =⇒ = ⎣⎦ The power supplied by the 8-V source is () 4.7873 0.6831 4.7873 24 8 4.316 W 25 8 ⎛⎞ += ⎜⎟ ⎝⎠ Apply KCL at node 4 of the original circuit to get ( ) 34 4 3 4 0 2 3.2127 30 0.5 4.71 V 30 20 5 5 v v v = = = The power supplied by the 0.5 A source is ( ) 0.5 4.71 2.355 W = (checked: LNAP 5/31/04)
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P5.2-8 Replace series and parallel resistors by an equivalent resistor. () 18 12 24 12 += & Ω Do a source transformation, then replace series voltage sources by an equivalent voltage source.
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This note was uploaded on 08/07/2009 for the course ECE 35 taught by Professor Staff during the Spring '08 term at UCSD.

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Chap_5_Solut_Dorf - Chapter 5 Circuit Theorems Exercises Ex...

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