Chapter 4 Homework solution: P4.2-2, 7 P4.3-2, 3, 6, 9 P4.4-2, 5, 8, 18 P4.5-2, 4, 5 P4.6-2, 4, 8 P4.7-2, 4, 9, 15 P4.8-2

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1 Chpter 4 Homework solution: P4.2-2, 7 P4.3-2, 3, 6, 9 P4.4-2, 5, 8, 18 P4.5-2, 4, 5 P4.6-2, 4, 8 P4.7-2, 4, 9, 15 P4.8-2 P P Determine the node voltges for the circuit of Figure Answer: v 1 = 2 V, v 2 = 30 V, nd v 3 = 24 V Figure P KCL t node 1: KCL t node 2: KCL t node 3: v v v = 0 5 v v = v v v v = v + 3 v 2 v = v v v = 3 v + 5 v = Solving gives v 1 = 2 V, v 2 = 30 V nd v 3 = 24 V. P The node voltges in the circuit shown in Figure P re v = 7 V nd v b = 10 V. Determine vlues of the current source current, i s, nd the resistnce, R. Figure P 4.2-7

2 Solution Apply KCL t node to get Apply KCL t node b to get v v v vb = + + = + + = + R = 4 Ω R 4 2 R 4 2 R 4 v vb vb vb is + = + = is + = + is = 4 A P The voltges v, v b, v c, nd v d in Figure P re the node voltges corresponding to nodes, b, c, nd d. The current i is the current in short circuit connected between nodes b nd c. Determine the vlues of v, v b, v c, nd v d nd of i. Answer: v = 12 V, v b = v c = 4 V, v d = 4 V, i = 2 ma Figure P Express the brnch voltge of ech voltge source in terms of its node voltges to get: KCL t node b: v = 12 V, v = v = v + 8 b c d ( 12) vb v vb = i = i vb + 12 = i KCL t the supernode corresponding to the 8 V source:

3 so ( ) vd = + i 4 = vd i 4000 v + 4= 4 v v = 4 v v = 4 V b d d d d Consequently 4 vd vb = vc = vd + 8 = 4 V nd i = = 2 ma 4000 Figure P4.3-3 P Determine the vlues of the power supplied by ech of the sources in the circuit shown in Figure P First, lbel the node voltges. Next, express the resistor currents in terms of the node voltges. Identify the supernode corresponding to the 24 V source

4 Apply KCL to the supernode to get The 12 V source supplies The 24 V source supplies ( ) 12 v 24 v 24 v = = 6v v = 32 V ( v ) ( ) = 12 = 4.8 W v = = 4.8 W The current source supplies ( ) 0.6v = = 19.2 W P The voltmeter in the circuit of Figure P mesures node voltge. The vlue of tht node voltge depends on the vlue of the resistnce R. () (b) Determine the vlue of the resistnce R tht will cuse the voltge mesured by the voltmeter to be 4 V. Determine the voltge mesured by the voltmeter when R = 1.2 kω = 1200 Ω. Answer: () 6 kω (b) 2 V Figure P 4.3-6

5 Lbel the voltge mesured by the meter. Notice tht this is node voltge. Write node eqution t the node t which the node voltge is mesured. 12 vm vm vm = R 3000 Tht is v m = 16 R = R 16 3 vm () The voltge mesured by the meter will be 4 volts when R = 6 kω. (b) The voltge mesured by the meter will be 2 volts when R = 1.2 kω. P Determine the vlues of the node voltges of the circuit shown in Figure P Figure P Express the voltge source voltges s functions of the node voltges to get v v = 5 nd v = Apply KCL to the supernode corresponding to the 5 V source to get

6 v1 v3 v = + = = 5v + 2v 5v Apply KCL t node 3 to get v1 v3 v3 v3 15 = + 15v1+ 28v3 = Solving, e.g. using MATLAB, gives So the node voltges re: v 1 5 v v = 80 v = v v v = 22.4 V, v = 27.4 V, v = 17.4 V, nd v = P Find i b for the circuit shown in Figure P Answer: i b = 12 ma Figure P Write nd solve node eqution: v 6 v v 4v + + = 0 v = 12 V i b v 4v = = 12 ma 3000 (checked using LNAP 8/13/02)

7 P Determine the vlue of the current i x in the circuit of Figure P Answer: i x = 2.4 A Figure P First, express the controlling current of the CCVS in v 2 terms of the node voltges: i x = 2 Next, express the controlled voltge in terms of the node voltges: v v2 = 3ix = 3 v2 = V 2 5 so i x = 12/5 A = 2.4 A. P Determine the vlue of the power supplied by the dependent source in Figure P Figure P 4.4-8

8 Lbel the node voltges. First, v 2 = 10 V due to the independent voltge source. Next, express the controlling current of the dependent source in terms of the node voltges: i v3 v2 v3 10 = = Now the controlled voltge of the dependent source cn be expressed s v v1 v3 = 8 i = 8 v1 = v Apply KCL to the supernode corresponding to the dependent source to get v1 v2 v1 v3 v2 v = Multiplying by 48 nd using v 2 = 10 V gives 16v + 9v = Substituting the erlier expression for v V 2 v + v = v = Then v 1 = V nd i = A. Applying KCL t node 2 gives v1 10 v1 = ib + 12 ib = 3+ 4 v1 = So i b = A. Finlly, the power supplied by the dependent source is ( ) b ( ) ( ) p= 8 i i = = W ( )

9 P The voltges v2, v3 nd v4for the circuit shown in Figure P re: Determine the vlues of the following: () The gin, A, of the VCVS (b) The resistnce R 5 (c) The currents i b nd i c (d) The power received by resistor R 4 v = 16 V, v = 8 V nd v = 6 V Figure P Given the node voltges v2 = 16 V, v3 = 8 V nd v4 = 6 V Av 16 8 V A = = = 4 v 8 6 V ( ) v3 v R5 = v4 R5 = = 45 Ω, i b = = 2 A nd ic = = A v p 4 = = = W 15 15

10 P The vlues of the mesh currents in the circuit shown in Figure P re i 1 = 2 A, i 2 = 3 A, nd i 3 = 4 A. Determine the vlues of the resistnce R nd of the voltges v 1 nd v 2 of the voltge sources. Answers: R = 12 Ω, v 1 = 4 V, nd v 2 = 28 V The mesh equtions re: Top mesh: 4 (2 3) + R(2) + 10(2 4) = 0 so R = 12 Ω. Figure P Bottom, right mesh: 8 (4 3) + 10 (4 2) + v 2 = 0 so v 2 = 28 V. Bottom left mesh v1 + 4 (3 2) + 8 (3 4) = 0 so v 1 = 4 V. (checked using LNAP 8/14/02)

11 P Determine the mesh currents, i nd i b, in the circuit shown in Figure P Figure P KVL loop 1: 25 i i + 75 i ( i i ) = 0 b 450 i 100 i = 2 b KVL loop 2: 100( i i ) i i i = 0 b b b b 100 i i = 4 Solving these equtions: i = 6.5 ma, i = 9.3 ma b b (checked using LNAP 8/14/02)

12 P Find the current i for the circuit of Figure P Hint: A short circuit cn be treted s 0-V voltge source. Mesh Equtions: Figure P mesh 1 : 2 i + 2 ( i i ) + 10 = mesh 2 : 2( i i ) + 4 ( i i ) = mesh 3 : ( i3 i2) + 6 i3 = 0 Solving: 5 i = i2 i = = A 17 (checked using LNAP 8/14/02) P Find v c for the circuit shown in Figure P Answer: v c = 15 V Mesh currents: mesh : i = 0.25 A mesh b: ib = 0.4 A Ohm s Lw: v = 100( i i ) = 100(0.15) =15 V c b Figure P 4.6-2

13 P Figure P Find v c for the circuit shown in Figure P Express the current source current in terms of the mesh currents: i b = i 0.02 Apply KVL to the supermesh: 250 i+ 100 ( i 0.02) + 9 = 0 i =.02 A = 20 ma v = 100( i 0.02) = 4 V c (checked using LNAP 8/14/02)

14 P Determine vlues of the mesh currents, i 1, i 2, nd i 3, in the circuit shown in Figure P Figure P Use units of V, ma nd kω. Express the currents to the supermesh to get Apply KVL to the supermesh to get Apply KVL to mesh 2 to get Solving, e.g. using MATLAB, gives i 1 i3 = 2 ( ) ( ) ( )( ) 4 i i + 1 i 3+ 1 i i = 0 i 5i + 5i = ( ) ( )( ) ( ) 2i + 4 i i + 1 i i = 0 1 i + 7i 4i = i 1 2 i i = 3 i = i 3 0 i 3 1 (checked: LNAP 6/21/04)

15 P4.7-2 Determine the vlues of the power supplied by the voltge source nd by the CCCS in the circuit shown Figure P4.7-2 Figure P4.7-2 First, lbel the mesh currents, tking dvntge of the current sources. Next, express the resistor currents in terms of the mesh currents: Apply KVL to the left mesh: ( ) 3 The 2 A voltge source supplies i ( ) i i 2 = 0 i = = ma 8 2 = = 0.25 mw = = = i i mw The CCCS supplies ( ) ( )( ) ( )( ) 2 P Determine the mesh current i for the circuit shown in Figure P Answer: i = 24 ma Figure P Express the controlling voltge of the dependent source s function of the mesh current: Apply KVL to the right mesh: v b = 100 (.006 i ) [ ] 100 (.006 i ) (.006 i ) i = 0 i = 24 ma (checked using LNAP 8/14/02)

16 P Determine the vlue of the resistnce R in the circuit shown in Figure P Figure P Notice tht i b nd 0.5 ma re the mesh currents. Apply KCL t the top node of the dependent source to get 3 1 ib = 4 ib ib = ma 6 Apply KVL to the supermesh corresponding to the dependent source to get b 3 ( R)( ) 3 3 ( R)( ) 5000 i = = R = = kω (checked: LNAP 6/21/04) P Determine the vlues of the mesh currents i 1 nd i 2 for the circuit shown in Figure P Figure P Expressing the dependent source currents in terms of the mesh currents we get: Apply KVL to mesh 2 to get ( ) i = 4i = 4 i + 1 4= i 4i ( ) ( ) 2i + 2 i i i = 0 2= 2i + 6i

17 Solving these equtions using MATLAB we get i 1 = 8 A nd i 2 = 3 A P The circuit shown in Figure P hs two inputs, v s nd i s, nd one output v o. The output is relted to the inputs by the eqution v o = i s + bv s where nd b re constnts to be determined. Determine the vlues nd b by () writing nd solving mesh equtions nd (b) writing nd solving node equtions. () Apply KVL to meshes 1 nd 2: Figure P So = 24 nd b = (b) ( ) ( ) 32i v + 96 i i = 0 1 s 1 s v + 30i i i = 0 s 2 2 s 150i2 =+ 120is vs 4 vs i2 = is v = 30i = 24i v 5 o 2 s s Apply KCL to the supernode corresponding to the voltge source to get

18 So Then ( ) v vs + vo v vo vs + vo vo + = i s vs + vo vo vs vo = + = v = 24i v 5 o s s So = 24 nd b = (checked: LNAP 5/24/04)