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• Anderson Niño Velasco

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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

P 4.2-2 Determine the node voltages for the circuit of Figure P 4.2-2. Answer: v1 = 2 V, v2 = 30 V, and v3 = 24 V

Figure P 4.2-2

Solution: KCL at node 1: KCL at node 2: KCL at node 3:

v −v v 1 2 1 + + 1 =0 ⇒ 5 v − v =−20 1 2 20 5 v −v v −v 1 2 2 3 = +2 ⇒ −v +3v = −2v 40 1 2 3 20 10 v −v v 2 3 3 += 1 ⇒ − 3 v + 5 v= 30 2 3 10 15

Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V.

P 4.2-7 The node voltages in the circuit shown in Figure P 4.2-7 are va = 7 V and vb = 10 V. Determine values of the current source current, is, and the resistance, R.

Figure P 4.2-7

Solution Apply KCL at node a to get 2=

va R

+

va 4

+

va − vb 2

=

7 7 7 − 10 7 1 + + = + ⇒ R= 4Ω R 4 R 4 2

Apply KCL at node b to get is +

va − vb 2

vb vb 7 − 10 10 10 = + =i s + = + ⇒ i s =4 A 8 8 2 8 8

P 4.3-2 The voltages va, vb, vc, and vd in Figure P 4.3-2 are the node voltages corresponding to nodes a, b, c, and d. The current i is the current in a short circuit connected between nodes b and c. Determine the values of va, vb, vc, and vd and of i. Answer: va = –12 V, vb = vc = 4 V, vd = –4 V, i = 2 mA

Figure P 4.3-2

Solution:

Express the branch voltage of each voltage source in terms of its node voltages to get: va = −12 V, vb = vc = vd + 8 KCL at node b: vb − va = 0.002 + i ⇒ 4000

vb − ( −12 ) = 0.002 + i ⇒ vb + 12= 8 + 4000 i 4000

KCL at the supernode corresponding to the 8 V source:

0.001 = so

vb + 4 = 4 − vd

⇒

vd + i ⇒ 4 = vd + 4000 i 4000

( vd + 8) + 4 = 4 − vd

Consequently vb = vc = vd + 8 = 4 V and i =

⇒ vd = −4 V

4 − vd = 2 mA 4000

Figure P4.3-3 P4.3-3. Determine the values of the power supplied by each of the sources in the circuit shown in Figure P4.3-3. Solution: First, label the node voltages. Next, express the resistor currents in terms of the node voltages.

Identify the supernode corresponding to the 24 V source

Apply KCL to the supernode to get 12 − ( v a − 24 ) 10

+ 0.6=

v a − 24 40

+

va 40

⇒ 196= 6 v a

⇒ v a= 32 V

The 12 V source supplies

 12 − ( v a − 24 )   12 − ( 32 − 24 )    12 12 = =   4.8 W   10 10    

The 24 V source supplies

va   32   24  −0.6 +  = 24  −0.6 +  = 4.8 W 40  40   

The current source supplies

= 0.6 v a 0.6 = ( 32 ) 19.2 W

P 4.3-6 The voltmeter in the circuit of Figure P 4.3-6 measures a node voltage. The value of that node voltage depends on the value of the resistance R. (a)

Determine the value of the resistance R that will cause the voltage measured by the voltmeter to be 4 V.

(b)

Determine the voltage measured by the voltmeter when R = 1.2 kΩ = 1200 Ω.

Answer: (a) 6 kΩ (b) 2 V

Figure P 4.3-6

Solution: Label the voltage measured by the meter. Notice that this is a node voltage. Write a node equation at the node at which the node voltage is measured.

 12 − v m  v m v −8 − + 0.002 + m =0 + 3000  6000  R That is

6000  6000  3 +  v m = 16 ⇒ R = 16 R   −3 vm

(a) The voltage measured by the meter will be 4 volts when R = 6 kΩ. (b) The voltage measured by the meter will be 2 volts when R = 1.2 kΩ.

P 4.3-9

Determine the values of the node voltages of the circuit shown in Figure P 4.3-9.

Figure P 4.3-9

Solution: Express the voltage source voltages as functions of the node voltages to get

v= 5 and = v 4 15 2 − v1 Apply KCL to the supernode corresponding to the 5 V source to get

1.25 =

v1 − v 3 8

+

v 2 − 15 20

=0

⇒

80 = 5v1 + 2v 2 − 5v 3

Apply KCL at node 3 to get v1 − v 3 8

=

v3 40

+

v 3 − 15 12

⇒

− 15v1 + 28v 3 =150

Solving, e.g. using MATLAB, gives

 −1 1 0   v1   5 2 −5 v =   2  −15 0 28   v 3 

 5   80    150 

⇒

 v1    v 2 = v 3   

 22.4   27.4    17.4 

So the node voltages are:

= v1 22.4 = V, v 2 27.4 = V, v 3 17.4 V, = and v 4 15

P 4.4-2

Find ib for the circuit shown in Figure P 4.4-2.

Answer: ib = –12 mA

Figure P 4.4-2 Solution: Write and solve a node equation:

va − 6 v v − 4va + a + a = 0 ⇒ va = 12 V 1000 2000 3000 ib =

va − 4va = 3000

−12 mA

(checked using LNAP 8/13/02)

P 4.4-5 Determine the value of the current ix in the circuit of Figure P 4.4-5. Answer: ix = 2.4 A

Figure P 4.4-5 Solution: First, express the controlling current of the CCVS in v2 terms of the node voltages: i x = 2 Next, express the controlled voltage in terms of the node voltages: v2 24 ⇒ v2 = 12 − v 2 = 3 i x = 3 V 2 5 so ix = 12/5 A = 2.4 A.

P 4.4-8 Determine the value of the power supplied by the dependent source in Figure P 4.4-8.

Figure P 4.4-8

Solution: Label the node voltages. First, v2 = 10 V due to the independent voltage source. Next, express the controlling current of the dependent source in terms of the node voltages: = ia

v 3 − v 2 v 3 − 10 = 16 16

Now the controlled voltage of the dependent source can be expressed as  v 3 − 10  v1 − v 3 = 8 i a = 8    16 

⇒

v1 =

3 v3 − 5 2

Apply KCL to the supernode corresponding to the dependent source to get v1 − v 2

+

4

v1 12

+

v3 − v2 16

+

v3 8

= 0

Multiplying by 48 and using v2 = 10 V gives

16v1 + 9v 3 = 150 Substituting the earlier expression for v1

3  16  v 3 − 5  + 9= v 3 150 2 

⇒

= v 3 6.970 V

Then v1 = 5.455 V and ia = -0.1894 A. Applying KCL at node 2 gives v1 12 So

=i b +

10 − v1 4

⇒

12 i b =−3 + 4 v1 =−30 + 4 ( 5.455 )

i b = −0.6817 A.

Finally, the power supplied by the dependent source is

p= 8 ( −0.1894 ) ( −0.6817 ) = 1.033 W (8 i a ) i b =

P4.4-18 The voltages v 2 , v 3 and v 4 for the circuit shown in Figure P4.4-18 are:

= v 2 16 = V, v 3 8 V and = v4 6 V Determine the values of the following: (a) The gain, A, of the VCVS (b) The resistance R 5 (c) The currents i b and i c (d) The power received by resistor R 4

Figure P4.4-18 Solution: Given the node voltages

= v 2 16 = V, v 3 8 V and = v4 6 V = A

Av a 16 − 8 V = = 4 va 8−6 V

 v3 − v4  R5  v4 =  15 

= ib

15 ( 6 ) 45 Ω , ⇒ R5 = = 8−6

40 − 24 40 − 16 16 ic − = 0.6667 A = 2 A and = 12 12 12 p= 4

va2 2 2 = = 0.2667 W 15 15

P 4.5-2 The values of the mesh currents in the circuit shown in Figure P 4.5-2 are i1 = 2 A, i2 = 3 A, and i3 = 4 A. Determine the values of the resistance R and of the voltages v1 and v2 of the voltage sources. Answers: R = 12 Ω, v1 = –4 V, and v2 = –28 V

Figure P 4.5-2 Solution: The mesh equations are: Top mesh: so Bottom, right mesh: so Bottom left mesh so

4 (2 − 3) + R(2) + 10 (2 − 4) = 0 R = 12 Ω. 8 (4 − 3) + 10 (4 − 2) + v 2 = 0 v2 = −28 V.

−v1 + 4 (3 − 2) + 8 (3 − 4) = 0 v1 = −4 V. (checked using LNAP 8/14/02)

P 4.5-4 Determine the mesh currents, ia and ib, in the circuit shown in Figure P 4.5-4.

Figure P 4.5-4 Solution: KVL loop 1:

KVL loop 2: Solving these equations:

25 ia − 2 + 250 ia + 75 ia + 4 + 100 (ia − ib ) =0 450 ia −100 ib = −2 −100(ia − ib ) − 4 + 100 ib + 100 ib + 8 + 200 ib = 0 −100 ia + 500 ib = −4 ia = − 6.5 mA , ib = − 9.3 mA (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 can be treated as a 0-V voltage source.

Figure P 4.5-5 Solution: Mesh Equations: mesh 1 : 2i1 + 2 (i1 − i2 ) + 10 = 0 mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0 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 vc for the circuit shown in Figure P 4.6-2. Answer: vc = 15 V

Figure P 4.6-2 Solution: Mesh currents: mesh a: ia = − 0.25 A mesh b: ib = − 0.4 A Ohm’s Law: = vc 100(ia − ib ) = 100(0.15) =15 V

P 4.6-4

Find vc for the circuit shown in

Figure P 4.6-4.

Figure P 4.6-4 Solution: Express the current source current in terms of the mesh currents: ib=

ia − 0.02

Apply KVL to the supermesh: 250 ia + 100 (ia − 0.02) + 9 = 0 ∴ ia = − .02 A = − 20 mA −4 V vc = 100(ia − 0.02) = (checked using LNAP 8/14/02)

P 4.6-8

Determine values of the mesh currents, i1,

i2, and i3, in the circuit shown in Figure P 4.6-8.

Figure P 4.6-8

Solution: Use units of V, mA and kΩ. Express the currents to the supermesh to get

i1 − i 3 = 2 Apply KVL to the supermesh to get

4 ( i 3 − i 3 ) + (1) i 3 − 3 + (1) ( i1 − i 2= ) 0

⇒

i1 − 5 i 2 + 5 i= 3 3

Apply KVL to mesh 2 to get

2i 2 + 4 ( i 2 − i 3 ) + (1) ( i 2 − i1 )= 0

( −1) i1 + 7i 2 − 4i 3=

⇒

0

Solving, e.g. using MATLAB, gives

 1 0 −1  i1   2   1 −5 5  i  =  3    2    −1 7 −4  i 3   0 

⇒

 i 1   3     i 2  = 1 i 3  1   (checked: LNAP 6/21/04)

P4.7-2 Determine the values of the power supplied by the voltage source and by the CCCS in the circuit shown Figure P4.7-2

Figure P4.7-2 Solution: First, label the mesh currents, taking advantage of the current sources. Next, express the resistor currents in terms of the mesh currents:

Apply KVL to the left mesh:

4000 i a + 2000 ( 6 i a ) − 2 = 0 ⇒ i a =

The 2 A voltage source supplies The CCCS supplies

2 i a= 2 ( 0.125 ×10−3 = ) 0.25 mW

( 5 i a ) ( 2000 ) ( 6 i a ) =( 60 ×10 )( 0.125 ×10 ) 3

−3 2

1 = 0.125 mA 8

=0.9375 ×10−3 =0.9375 mW

P 4.7-4 Determine the mesh current ia for the circuit shown in Figure P 4.7-4. Answer: ia = –24 mA

Figure P 4.7-4 Solution: Express the controlling voltage of the dependent source as a function of the mesh current: = vb 100 (.006 − ia ) Apply KVL to the right mesh: −100 (.006 − ia ) + 3 [100(.006 − ia ) ] + 250 ia = −24 mA 0 ⇒ ia = (checked using LNAP 8/14/02)

P 4.7-9

Determine the value of the resistance R

in the circuit shown in Figure P 4.7-9.

Figure P 4.7-9 Solution: Notice that i b and 0.5 mA are the mesh currents. Apply KCL at the top node of the dependent source to get

1 mA 6 Apply KVL to the supermesh corresponding to the dependent source to get i b + 0.5 ×10−3= 4 i b

⇒ i b=

(

)

−5000 i b + (10000 + R ) 0.5 ×10−3 − 25 = 0

(

)

1  25 −5000  ×10−3  + (10000 + R ) 0.5 ×10−3 = 6  125 6 = = R 41.67 kΩ 0.5 ×10−3 (checked: LNAP 6/21/04)

P4.7-15 Determine the values of the mesh currents i1 and i2 for the circuit shown in Figure P4.7-15.

Figure P4.7-15 Solution: Expressing the dependent source currents in terms of the mesh currents we get:

i1 =4 i a =4 ( i 2 + 1) ⇒ 4 =i1 − 4 i 2 Apply KVL to mesh 2 to get

2 i 2 + 2 ( i 2 + 1) − 2 ( i1 − i 2 ) =0 ⇒ − 2 =−2 i1 + 6 i 2

Solving these equations using MATLAB we get i1 = −8 A and i2 = −3 A

P 4.8-2 The circuit shown in Figure P 4.8-2 has two inputs, vs and is, and one output vo. The output is related to the inputs by the equation vo = ais + bvs where a and b are constants to be determined. Determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations. Figure P 4.8-2 Solution: (a)

Apply KVL to meshes 1 and 2:

32i1 − v s + 96 ( i1 − i s ) = 0

v s + 30i 2 + 120 ( i 2 − i s ) = 0

150i 2 = +120i s − v s = i2

vs 4 is − 5 150

1 = v o 30 = i 2 24i s − v s 5 So a = 24 and b = -.02. (b) Apply KCL to the supernode corresponding to the voltage source to get

va − (vs + vo ) 96

+

va − vo 32

=

vs + vo 120

+

vo 30

So is =

vs + vo 120

+

vo 30

=

vs 120

+

vo 24

Then

1 = v o 24i s − v s 5

So a = 24 and b = -0.2. (checked: LNAP 5/24/04)