chapter 04

# chapter 04 - Chapter 4 Methods of Analysis of Resistive...

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Chapter 4 – Methods of Analysis of Resistive Circuits Exercises Ex. 4.3-1 KCL at a: 30 5 3 1 8 32 vvv aab vv ab + += = KCL at b: 31 0 8 2 ba −−= = Solving these equations gives: v a = 3 V and v b = 11 V Ex. 4.3-2 KCL at a: 3 2 1 2 42 + = KCL at a: 40 35 bab 2 4 −− = ⇒− + = Solving: v a = 4/3 V and v b = 4 V Ex. 4.4-1 Apply KCL to the supernode to get 10 25 20 30 bb + ++ = Solving: 30 V and 10 40 V v b == + = 4-1

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Ex. 4.4-2 () 81 2 3 8 V and 16 V 10 40 v v b b vv ba +− +=⇒ = = Ex. 4.5-1 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 44 4 . 5 2 1 2 bb iv i v ab a b +  += ⇒ = = ⇒ =   V Ex. 4.5-2 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. 4.6-1 Mesh equations: 12 6 3 8 0 9 3 20 11 2 12 ii i i i −+ + − −= ⇒ = 83 6 0 3 9 8 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. 4-2
Ex. 4.7-1 Mesh equation: () 31 2 4 0 324 93 A 49 ii i i i  +++ + = ⇒ ++ = − ⇒ =   2 V The voltmeter measures 34 i =− Ex. 4.7-2 Mesh equation: () ( ) ( ) 33 2 3 6 3 0 3 6 15 6 3 = 3 A ii i i ++ += ⇒ + = −− ⇒ = 15 Ex. 4.7-3 Express the current source current in terms of the mesh currents: 12 1 33 44 i i 2 = −⇒= + . Apply KVL to the supermesh: 12 2 2 2 2 3 94 3 2 0 4 5 9 9 6 4 i i i −+ + + = + + = = so 2 2 A 3 i = and the voltmeter reading is 2 4 2 V 3 i = 4-3

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Ex. 4.7-4 Express the current source current in terms of the mesh currents: 12 1 33 ii i i 2 = −⇒= + . Apply KVL to the supermesh: ( ) 2 2 2 15 6 3 0 6 3 3 15 9 3 i i i −+ + = ⇒ + + = = Finally, 2 1 A 3 =− i is the current measured by the ammeter. Problems Section 4-3 Node Voltage Analysis of Circuits with Current Sources P4.3-1 KCL at node 1: 44 2 112 01 . 5 86 vvv i i 1 . 5 A −− =+ + + = −+⇒= (checked using LNAP 8/13/02) 4-4
P4.3-2 KCL at node 1: 121 10 5 2 0 12 20 5 vv v vv + += =− KCL at node 2: 23 123 20 10 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. (checked using LNAP 8/13/02) P4.3-3 KCL at node 1: 41 5 4 2 A 11 52 0 0 ii += ⇒ = KCL at node 2: 2 51 5 5 1 8 2 5 i i −−  ⇒= + =   (checked using LNAP 8/13/02) 4-5

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P4.3-4 Node equations: 112 1 12 2 2 .003 0 500 .005 0 500 vvv R vv v R −++ = −+ −= When v 1 = 1 V, v 2 = 2 V 1 1 2 2 11 1 .003 0 200 1 500 .003 500 12 2 .005 0 667 1 500 .005 500 R R R R −++ =⇒= = + −+−= ⇒= =Ω (checked using LNAP 8/13/02) P4.3-5 Node equations: 13 2 23 13 3 0 500 125 250 .001 0 0 vv v vvvv v + += = −− = Solving gives: 12 3 0.261 V, 0.337 V, 0.239 V = == Finally, 0.022 V = (checked using LNAP 8/13/02) 4-6
Section 4-4 Node Voltage Analysis of Circuits with Current and Voltage Sources P4.4-1 Express the branch voltage of the voltage source in terms of its node voltages: 06 6 aa vv V =⇒ = KCL at node b : 6 22 1 2 3 0 61 0 61 0 6 1 0 ab bc b b v v −−− −− += ⇒−− += = 8 3 KCL at node c : 9 445 10 8 4 bc c b vv v vvv v =⇒ −= ⇒= c v Finally: 9 30 8 3 2 V 4 cc c v 

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## This note was uploaded on 06/10/2011 for the course PHYSICS physics taught by Professor Physics during the Spring '11 term at HKU.

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chapter 04 - Chapter 4 Methods of Analysis of Resistive...

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