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 4.2-2 P 4.2-2. 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 4.2-2 KCL t node 1: KCL t node 2: KCL t node 3: v v v 1 2 1 + + 1 = 0 5 v v = 20 20 5 1 2 v v v v 1 2 2 3 + 2 = v + 3 v 2 v = 40 20 10 1 2 3 v v v 2 3 3 + 1 = 3 v + 5 v = 30 10 15 2 3 Solving gives v 1 = 2 V, v 2 = 30 V nd v 3 = 24 V. P 4.2-7 The node voltges in the circuit shown in Figure P 4.2-7 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
Solution Apply KCL t node to get Apply KCL t node b to get v v v vb 7 7 7 10 7 1 2 = + + = + + = + R = 4 Ω R 4 2 R 4 2 R 4 v vb vb vb 7 10 10 10 is + = + = is + = + is = 4 A 2 8 8 2 8 8 P 4.3-2 The voltges v, v b, v c, nd v d in Figure P 4.3-2 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 4.3-2 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 = 0.002 + i = 0.002 + i vb + 12 = 8 + 4000 i 4000 4000 KCL t the supernode corresponding to the 8 V source:
so ( ) vd 0.001 = + i 4 = vd + 4000 i 4000 v + 4= 4 v v + 8 + 4= 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 P4.3-3. Determine the vlues of the power supplied by ech of the sources in the circuit shown in Figure P4.3-3. 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
Apply KCL to the supernode to get The 12 V source supplies The 24 V source supplies ( ) 12 v 24 v 24 v + 0.6 = + 196 = 6v v = 32 V 10 40 40 ( v ) ( ) 12 24 12 32 24 12 = 12 = 4.8 W 10 10 v 32 24 0.6 + = 24 0.6 + = 4.8 W 40 40 The current source supplies ( ) 0.6v = 0.6 32 = 19.2 W P 4.3-6 The voltmeter in the circuit of Figure P 4.3-6 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
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 8 + + 0.002 + = 0 6000 R 3000 Tht is 6000 6000 3 + 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 4.3-9 Determine the vlues of the node voltges of the circuit shown in Figure P 4.3-9. Figure P 4.3-9 Express the voltge source voltges s functions of the node voltges to get v v = 5 nd v = 15 2 1 4 Apply KCL to the supernode corresponding to the 5 V source to get
v1 v3 v2 15 1.25 = + = 0 8 20 80 = 5v + 2v 5v Apply KCL t node 3 to get 1 2 3 v1 v3 v3 v3 15 = + 15v1+ 28v3 = 150 8 40 12 Solving, e.g. using MATLAB, gives So the node voltges re: 1 1 0 v 1 5 v 1 22.4 5 2 5 v = 80 v = 27.4 2 2 15 0 28 v 3 150 v 3 17.4 v = 22.4 V, v = 27.4 V, v = 17.4 V, nd v = 15 1 2 3 4 P 4.4-2 Find i b for the circuit shown in Figure P 4.4-2. Answer: i b = 12 ma Figure P 4.4-2 Write nd solve node eqution: v 6 v v 4v + + = 0 v = 12 V 1000 2000 3000 i b v 4v = = 12 ma 3000 (checked using LNAP 8/13/02)
P 4.4-5 Determine the vlue of the current i x in the circuit of Figure P 4.4-5. Answer: i x = 2.4 A Figure P 4.4-5 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 2 24 12 v2 = 3ix = 3 v2 = V 2 5 so i x = 12/5 A = 2.4 A. P 4.4-8 Determine the vlue of the power supplied by the dependent source in Figure P 4.4-8. Figure P 4.4-8
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 = = 16 16 Now the controlled voltge of the dependent source cn be expressed s v3 10 3 v1 v3 = 8 i = 8 v1 = v3 5 16 2 Apply KCL to the supernode corresponding to the dependent source to get v1 v2 v1 v3 v2 v3 + + + = 0 4 12 16 8 Multiplying by 48 nd using v 2 = 10 V gives 16v + 9v = 150 1 3 Substituting the erlier expression for v 1 3 16 3 5 9 3 150 3 6.970 V 2 v + v = v = Then v 1 = 5.455 V nd i = -0.1894 A. Applying KCL t node 2 gives v1 10 v1 = ib + 12 ib = 3+ 4 v1 = 30 + 4 5.455 12 4 So i b = 0.6817 A. Finlly, the power supplied by the dependent source is ( ) b ( ) ( ) p= 8 i i = 8 0.1894 0.6817 = 1.033 W ( )
P4.4-18 The voltges v2, v3 nd v4for the circuit shown in Figure P4.4-18 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 2 3 4 Figure P4.4-18 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 v4 15 6 R5 = v4 R5 = = 45 Ω, 15 8 6 i b 40 24 = = 2 A nd ic 12 40 16 16 = = 0.6667 A 12 12 2 2 2 v p 4 = = = 0.2667 W 15 15
P 4.5-2 The vlues of the mesh currents in the circuit shown in Figure P 4.5-2 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 4.5-2 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)
P 4.5-4 Determine the mesh currents, i nd i b, in the circuit shown in Figure P 4.5-4. Figure P 4.5-4 KVL loop 1: 25 i 2 + 250 i + 75 i + 4 + 100 ( i i ) = 0 b 450 i 100 i = 2 b KVL loop 2: 100( i i ) 4 + 100 i + 100 i + 8 + 200 i = 0 b b b b 100 i + 500 i = 4 Solving these equtions: i = 6.5 ma, i = 9.3 ma b b (checked using LNAP 8/14/02)
P 4.5-5 Find the current i for the circuit of Figure P 4.5-5. Hint: A short circuit cn be treted s 0-V voltge source. Mesh Equtions: Figure P 4.5-5 mesh 1 : 2 i + 2 ( i i ) + 10 = 0 1 1 2 mesh 2 : 2( i i ) + 4 ( i i ) = 0 2 1 2 3 mesh 3 : 10 + 4 ( i3 i2) + 6 i3 = 0 Solving: 5 i = i2 i = = 0.294 A 17 (checked using LNAP 8/14/02) P 4.6-2 Find v c for the circuit shown in Figure P 4.6-2. 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
P 4.6-4 Figure P 4.6-4. Find v c for the circuit shown in Figure P 4.6-4 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)
P 4.6-8 Determine vlues of the mesh currents, i 1, i 2, nd i 3, in the circuit shown in Figure P 4.6-8. Figure P 4.6-8 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 = 3 3 3 3 1 2 1 2 3 ( ) ( )( ) ( ) 2i + 4 i i + 1 i i = 0 1 i + 7i 4i = 0 2 2 3 2 1 1 2 3 1 0 1 i 1 2 i 1 3 1 5 5 i = 3 i = 1 2 2 1 7 4 i 3 0 i 3 1 (checked: LNAP 6/21/04)
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 ( ) 1 4000i + 2000 6i 2 = 0 i = = 0.125 ma 8 2 = 2 0.125 10 = 0.25 mw = = = 3 3 3 5i 2000 6i 60 10 0.125 10 0.9375 10 0.9375 mw The CCCS supplies ( ) ( )( ) ( )( ) 2 P 4.7-4 Determine the mesh current i for the circuit shown in Figure P 4.7-4. Answer: i = 24 ma Figure P 4.7-4 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 ) + 3 100(.006 i ) + 250 i = 0 i = 24 ma (checked using LNAP 8/14/02)
P 4.7-9 Determine the vlue of the resistnce R in the circuit shown in Figure P 4.7-9. Figure P 4.7-9 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 + 0.5 10 = 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 + 10000 + 0.5 10 25 = 0 1 5000 10 + 10000 + 6 0.5 10 = 25 125 R = 6 3 0.5 10 = 41.67 kω (checked: LNAP 6/21/04) P4.7-15 Determine the vlues of the mesh currents i 1 nd i 2 for the circuit shown in Figure P4.7-15. Figure P4.7-15 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 1 2 1 2 ( ) ( ) 2i + 2 i + 1 2 i i = 0 2= 2i + 6i 2 2 1 2 1 2
Solving these equtions using MATLAB we get i 1 = 8 A nd i 2 = 3 A P 4.8-2 The circuit shown in Figure P 4.8-2 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 4.8-2 So = 24 nd b = -.02. (b) ( ) ( ) 32i v + 96 i i = 0 1 s 1 s v + 30i + 120 i i = 0 s 2 2 s 150i2 =+ 120is vs 4 vs i2 = is 5 150 1 v = 30i = 24i v 5 o 2 s s Apply KCL to the supernode corresponding to the voltge source to get
So Then ( ) v vs + vo v vo vs + vo vo + = + 96 32 120 30 i s vs + vo vo vs vo = + = + 120 30 120 24 1 v = 24i v 5 o s s So = 24 nd b = -0.2. (checked: LNAP 5/24/04)