Chapter 4, Problem 32.

Chapter 4, Problem 32.Use source transformation to find ix in the circuit of Fig. 4.100.Figure 4.100Chapter 4, Solution 32.As shown in Fig. (a), we transform the dependent current source to a voltage source,15 Ω10 Ω5ix 60V 50 Ω40 Ω(a)15 Ω60V 50 Ω50 Ω0.1ix(b)ix60V 15 Ω25 Ωix(c) 2.5ix

In Fig. (b), 50 50 25 ohms. Applying KVL in Fig. (c),-60 40ix – 2.5ix 0, or ix 1.6 AChapter 4, Problem 33.Determine RTh and VTh at terminals 1-2 of each of the circuits of Fig. 4.101.Figure 4.101Chapter 4, Solution 33.(a)RTh 10 40 400/50 8 ohmsVTh (40/(40 10))20 16 V(b)RTh 30 60 1800/90 20 ohms2 (30 – v1)/60 v1/30, and v1 VTh120 30 – v1 2v1, or v1 50 VVTh 50 V

Chapter 4, Problem 39.Obtain the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.106.1A10 Ω16 Ωc a10 Ω8V5Ω c bFigure 4.106 For Prob. 4.39.Chapter 4, Solution 39.We obtain RTh using the circuit below.10 Ω5Ω10 ΩRTh 16 20 // 5 16 1620 x5 20 Ω25RTh

To find VTh, we use the circuit below.1A10 10 Ω8V16V2V1 V25VThAt node 1,[(V1–8)/10] – 1 [(V1–V2)/10] 0 or 2V1 – V2 18(1)At node 2,[(V2–V1)/10] [(V2–0)/5] 1 0 or –V1 3V2 –10Adding 3(1) to (2) gives5V1 44 or V1 8.8 VUsing (2) we get 3V2 8.8 – 10 –1.2 or V2 –400 mV.Finally,VTh V2 (–1)(16) –0.4 – 16 –16.4 V\(2)

Chapter 4, Problem 44.For the circuit in Fig. 4.111, obtain the Thevenin equivalent as seen from terminals(a) a-b(b) b-cFigure 4.111Chapter 4, Solution 44.(a)For RTh, consider the circuit in Fig. (a).RTh 1 4 (3 2 5) 3.857 ohmsFor VTh, consider the circuit in Fig. (b). Applying KVL gives,10 – 24 i(3 4 5 2), or i 1VTh 4i 4 V3Ω3Ω1Ωa2Ωb(a)a VTh4Ω24V RTh4Ω5Ω 1Ω2Ωib10V5Ω(b)

(b)For RTh, consider the circuit in Fig. (c).3Ω1Ω4Ω3Ωb24V2ΩRTh5Ω1Ω4Ωvo b 2Ω5Ω2Ac(c)VThc(d)RTh 5 (2 3 4) 3.214 ohmsTo get VTh, consider the circuit in Fig. (d). At the node, KCL gives,[(24 – vo)/9] 2 vo/5, or vo 15VTh vo 15 VChapter 4, Problem 47.Obtain the Thèvenin and Norton equivalent circuits of the circuit in Fig. 4.114with respect to terminals a and b.50 VFigure 4.114

Chapter 4, Solution 47Since VTh Vab Vx, we apply KCL at the node a and obtain50 VTh VTh 2VTh VTh 250 / 126 1.9841 V1260To find RTh, consider the circuit below.12 ΩVxa60 Ω2Vx1AAt node a, KCL givesV V1 2V x x x V x 60 / 126 0.476260 12VVI N Th 1.9841 / 0.4762 4.167 AR Th x 0.4762Ω,R Th1Thus,VTh 1.9841 V, Req RTh RN 476.2 mΩ, IN 4.167 AChapter 4, Problem 48.Determine the Norton equivalent at terminals a-b for the circuit in Fig. 4.115.44AFigure 4.115Chapter 4, Solution 48.8

To get RTh, consider the circuit in Fig. (a).10Io 42 Ω 10Io 42 Ω Io84 Ω IoV 1A 4A(a)From Fig. (a),VTh84 Ω(b)Io 1,4 – 10 8 – V 0, or V 2RN RTh 2/1 2 ohmsTo get VTh, consider the circuit in Fig. (b),Io 4, VTh -10Io 8Io –8 VIN VTh/RTh -8/2 -4AChapter 4, Problem 51.Given the circuit in Fig. 4.117, obtain the Norton equivalent as viewed from terminals(a) a-b(b) c-dFigure 4.117Chapter 4, Solution 51.(a)From the circuit in Fig. (a),RN 4 (2 6 3) 4 4 2 ohms

RTh6ΩVTh 6Ω4Ω3Ω2Ω120V4Ω 3Ω6A2Ω(b)(a)For IN or VTh, consider the circuit in Fig. (b). After some source transformations, thecircuit becomes that shown in Fig. (c). VTh2Ω40V 4Ωi2Ω12V (c)Applying KVL to the circuit in Fig. (c),-40 8i 12 0 which gives i 7/2VTh 4i 14 therefore IN VTh/RN 14/2 7 A

(b)To get RN, consider the circuit in Fig. (d).RN 2 (4 6 3) 2 6 1.5 ohms6Ω2Ω4Ωi 3Ω2Ω(d)RNVTh12V (e)To get IN, the circuit in Fig. (c) applies except that it needs slight modification as inFig. (e).i 7/2, VTh 12 2i 19, IN VTh/RN 19/1.5 12.667 AChapter 4, Problem 53.Find the Norton equivalent at terminals a-b of the circuit in Fig. 4.119.Figure 4.119Chapter 4, Solution 53.To get RTh, consider the circuit in Fig. (a).

0.25vo0.25vo2Ω2Ωa 6Ω3Ω vo2Ω1A 1/2a1/2vovab 1A bb(a) (b)From Fig. (b),vo 2x1 2V, -vab 2x(1/2) vo 0vab 3VRN vab/1 3 ohmsTo get IN, consider the circuit in Fig. (c).0.25vo6Ω2Ωa 18V 3ΩvoIsc IN b(c)[(18 – vo)/6] 0.25vo (vo/2) (vo/3) or vo 4VBut,(vo/2) 0.25vo IN, which leads to IN 1 AChapter 4, Problem 59.Determine the Thevenin and Norton equivalents at terminals a-b of the circuit in Fig.4.125.

Figure 4.125Chapter 4, Solution 59.RTh (10 20) (50 40) 30 90 22.5 ohmsTo find VTh, consider the circuit below.i1i210 Ω20 Ω VTh8A50 Ω40 Ωi1 i2 8/2 4, 10i1 VTh – 20i2 0, or VTh 20i2 –10i1 10i1 10x4VTh 40V, and IN VTh/RTh 40/22.5 1.7778 AChapter 4, Problem 66.Find the maximum power that can be delivered to the resistor R in the circuit in Fig.4.132.Figure 4.132Chapter 4, Solution 66.

We first find the Thevenin equivalent at terminals a and b. We find RTh using the circuitin Fig. (a).2Ω10V 3Ω2Ωab 3ΩaVTh5ΩbRTh 5Ω20V i30V(a)(b)RTh 2 (3 5) 2 8 1.6 ohmsBy performing source transformation on the given circuit, we obatin the circuit in (b).We now use this to find VTh.10i 30 20 10 0, or i –6VTh 10 2i 0, or VTh 2 Vp VTh2/(4RTh) (2)2/[4(1.6)] 625 m watts

Chapter 4, Problem 73.Determine the maximum power that can be delivered to the variable resistor R inthe circuit of Fig. 4.139.Figure 4.139Chapter 4, Solution 73Find the Thevenin’s equivalent circuit across the terminals of R.10 Ω25 ΩRTh20 Ω5ΩRTh 10 // 20 25 // 5 325 / 30 10.833Ω10 Ω 60 V-25 Ω Va-20 ΩVTh - 5ΩVb-

20(60) 40,30 Va VTh Vb 0Va 2p max5(60) 1030 VTh Va Vb 40 10 30 VVb V30 2 Th 20.77 W4 RTh 4 x10.833

Chapter 4, Solution 47 Since VTh Vab Vx, we apply KCL at the node a and obtain 2V V 250/126 1.9841 V 60 V 12 50 V Th Th Th Th To find RTh, consider the circuit below. 12Ω Vx a 2Vx 60Ω 1A At node a, KCL gives 60/1260.4762