Chap_4_Solut_Dorf

Chap_4_Solut_Dorf - Chapter 4 Methods of Analysis of...

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Chapter 4 Methods of Analysis of Resistive Circuits Exercises Ex. 4.2-1 KCL at a: 30 5 3 1 8 32 vv v aa b ab + += ⇒ = KCL at b: 31 0 8 2 ba −−= = Solving these equations gives: v a = 3 V and v b = 11 V Ex. 4.2-2 KCL at a: aab 3 2 1 2 42 v + = KCL at a: 40 352 vvv b 4 −− = ⇒− + = Solving: v a = 4/3 V and v b = 4 V Ex. 4.3-1 Apply KCL to the supernode to get 10 25 20 30 bb + ++ = Solving: 30 V and 10 40 V v b == + =
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Ex. 4.3-2 () 81 2 3 8 V and 16 V 10 40 v v b b vv ba +− +=⇒ = = Ex. 4.4-1 Apply KCL at node a to express i a as a function of the node voltages. Substitute the result into and solve for v 4 b v = a i b . 9 6 4 4 4.5 V 2 1 2 bb iv i v ab a b + ⎛⎞ += = = = ⎜⎟ ⎝⎠ Ex. 4.4-2 The controlling voltage of the dependent source is a node voltage so it is already expressed as a function of the node voltages. Apply KCL at node a. 64 02 20 15 v aa a v a −− = V Ex. 4.5-1 Mesh equations: 12 6 3 8 0 9 3 20 11 2 1 2 ii i i i −+ + − −= ⇒ = 83 6 0 3 9 8 12 2 1 2 i i i + = + = Solving these equations gives: 13 1 A and A 66 == The voltage measured by the meter is 6 i 2 = 1 V.
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Ex. 4.6-1 Mesh equation: () 31 93 2 4 0 324 93 A 49 ii i i i ⎛⎞ +++ + = + + = = ⎜⎟ ⎝⎠ 2 V The voltmeter measures 34 i =− Ex. 4.6-2 Mesh equation: ( ) ( ) 33 2 15 3 6 3 0 3 6 15 6 3 = 3 A i i ++ += + = −− =
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PSpice Problems SP 4-1 SP 4-2 From the PSpice output file: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 -3.000E+00 V_V2 -2.250E+00
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V_V3 -7.500E-01 The voltage source labeled V3 is a short circuit used to measure the mesh current. The mesh currents are i 1 = 3 A (the current in the voltage source labeled V1) and i 2 = 0.75 A (the current in the voltage source labeled V3). SP 4-3 The PSpice schematic after running the simulation: The PSpice output file: **** INCLUDING sp4_2-SCHEMATIC1.net **** * source SP4_2 V_V4 0 N01588 12Vdc R_R4 N01588 N01565 4k V_V5 N01542 N01565 0Vdc R_R5 0 N01516 4k V_V6 N01542 N01516 8Vdc I_I1 0 N01565 DC 2mAdc I_I2 0 N01542 DC 1mAdc VOLTAGE SOURCE CURRENTS NAME CURRENT V_V4 -4.000E-03 V_V5 2.000E-03 V_V6 -1.000E-03 From the PSpice schematic: v a = 12 V, v b = v c = 4 V, v d = 4 V. From the output file: i = 2 mA. SP 4-4 The PSpice schematic after running the simulation:
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The PSpice output file: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V7 -5.613E-01 V_V8 -6.008E-01 The current of the voltage source labeled V7 is also the current of the 2 Ω resistor at the top of the circuit. However this current is directed from right to left in the 2 Ω resistor while the current i is directed from left to right. Consequently, i = +5.613 A.
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Problems Section 4-2 Node Voltage Analysis of Circuits with Current Sources P4.2-1 KCL at node 1: 44 2 112 01 . 5 86 8 6 vv v ii i i 1 . 5 A −− =+ + + = −+ ⇒= (checked using LNAP 8/13/02) P4.2-2 KCL at node 1: 12 1 10 5 2 0 20 5 v + += − =− KCL at node 2: 23 123 20 10 v v v 2 4 0 + =⇒ + = KCL at node 3: 3 13 5 10 15 v 3 0 + + = Solving gives v 1 = 2 V, v 2 = 30 V and v 3 = 24 V.
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Chap_4_Solut_Dorf - Chapter 4 Methods of Analysis of...

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