Chapter 5 - Circuit Theorems

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Problems Section 5-2: Source Transformations P5.2-1 (a)
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Introdução Aos Circuitos Elétricos - 7th ed - Dorf Svoboda - Resolução - Capitulo 5

Transcript of Chapter 5 - Circuit Theorems

ProblemsSection 5-2: Source Transformations P5.2-1 (a)

Rt = 2

(b)

(c)

9 4i 2i + (0.5) = 0 9 + (0.5) i = = 1.58 A 4+2 v = 9 + 4 i = 9 + 4(1.58) = 2.67 V ia = i = 1.58 A

vt = 0.5 V

(checked using LNAP 8/15/02)

P5.2-2

Finally, apply KVL:

10 + 3 ia + 4 ia

16 =0 3

ia = 2.19 A(checked using LNAP 8/15/02)

P5.2-3

Source transformation at left; equivalent resistor for parallel 6 and 3 resistors:

Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top:

Source transformation at left; series resistors at right:

Parallel resistors, then source transformation at left:

Finally, apply KVL to loop 6 + i (9 + 19) 36 vo = 0 i = 5 / 2 vo = 42 + 28 (5 / 2) = 28 V (checked using LNAP 8/15/02)

P5.2-4

4 2000 ia 4000 ia + 10 2000 ia 3 = 0 ia = 375 A

(checked using LNAP 8/15/02)

P5.2-5

12 6 ia + 24 3 ia 3 = 0 ia = 1 A (checked using LNAP 8/15/02) P5.2-6 A source transformation on the right side of the circuit, followed by replacing series resistors with an equivalent resistor:

Source transformations on both the right side and the left side of the circuit:

Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources with an equivalent current source:

Finally,

va =

50 (100 ) 100 ( 0.21) = ( 0.21) = 7 V 50 + 100 3 (checked using LNAP 8/15/02)

P5.2-7 Use source transformations to simplify the circuit:

Label the node voltages. The 8-V source is connected between nodes 1 and 3. Consequently,

v1 v 3 = 8

Apply KCL to the supernode corresponding to the 8-V source to get v1 24 8 + v2 20 + v 3 10 50 =0 0.125v1 0.3 + 0.05v 2 + 0.02v 3 0.2 = 0

Apply KCL at node 2 to get v1 v 2 25 = v2 20 + v 2 v3 10 =0.04v1 + 0.19v 2 0.1v 3 = 0

Solving, for example using MATLAB

0 1 v1 8 1 0.125 0.05 0.02 v = 0.5 2 0.04 0.19 0.1 v 3 0 The power supplied by the 8-V source is

v1 4.7873 v 2 = 0.6831 v 3 3.2127

4.7873 ( 0.6831) 4.7873 24 + 8 = 4.316 W 25 8

Apply KCL at node 4 of the original circuit to get v3 v4 30 + 0.5 = v4 20 v4 = 2v 3 + 30 5 = 2 ( 3.2127 ) + 30 = 4.71 V 5

The power supplied by the 0.5 A source is0.5 ( 4.71) = 2.355 W

(checked: LNAP 5/31/04)

P5.2-8

Replace series and parallel resistors by an equivalent resistor.18

(12 + 24 ) = 12

Do a source transformation, then replace series voltage sources by an equivalent voltage source.

Do two more source transformations Now current division gives 24 8 i = 3 = 8+ R 8+ R Then Ohms Law gives 24 R v = Ri = 8+ R

(a )

i=

24 =2A 8+ 4 24 ( 8 ) 8+8 24 8+ R 24 R 8+ R = 12 V

(b) v =

(c) 1 =

R = 16

(d) 16 =

R = 16 (checked: LNAP 6/9/04)

P5.2-9 Use source transformations and equivalent resistances to reduce the circuit as follows

The power supplied by the current source is given by

p = 23.1 + 2 (10.3125) 2 = 87.45 W

Section 5-3 Superposition P5.31 Consider 6 A source only (open 9 A source) Use current division:v1 15 = 6 v1 = 40 V 20 15 + 30

Consider 9 A source only (open 6 A source)

Use current division:v2 10 = 9 v2 = 40 V 20 10 + 35

v = v1 + v2 = 40 + 40 = 80 V(checked using LNAP 8/15/02)

P5.3-2 Consider 12 V source only (open both current sources) KVL:

20 i1 + 12 + 4 i1 + 12 i1 = 0 i1 = 1/ 3 mA

Consider 3 mA source only (short 12 V and open 9 mA sources)

Current Division:4 16 i2 = 3 = 3 mA 16 + 20

Consider 9 mA source only (short 12 V and open 3 mA sources)

Current Division: 12 i3 = 9 = 3 mA 24 + 12

i = i1 + i2 + i3 = 1/ 3 + 4 / 3 3 = 2 mA(checked using LNAP 8/15/02) P5.33 Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to the 30 mA current source.

2 ia = 30 = 6 mA 2+8

6 i1 = ia = 2 mA 6 + 12

Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i due to the 15 mA current source.

4 ib = 15 = 6 mA 4+6

6 i2 = ib = 2 mA 6 + 12

Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V voltage source.

6 || 6 3 i3 = 2.5 ( 6 || 6 ) + 12 = 10 3 + 12 = 0.5 mA

Finally,

i = i1 + i2 + i3 = 2 + 2 0.5 = 3.5 mA(checked using LNAP 8/15/02)

P5.34 Consider 10 V source only (open 30 mA source and short the 8 V source)

Let v1 be the part of va due to the 10 V voltage source.

v1 = =

100 ||100 (10 ) (100 ||100 ) + 100 50 10 (10 ) = V 150 3

Consider 8 V source only (open 30 mA source and short the 10 V source)

Let v2 be the part of va due to the 8 V voltage source.v1 = = 100 ||100 (8) (100 ||100 ) + 100 50 8 (8) = V 150 3

Consider 30 mA source only (short both the 10 V source and the 8 V source)

Let v2 be the part of va due to the 30 mA current source.

v3 = (100 ||100 ||100)(0.03) = 100 (0.03) = 1 V 3

Finally,

va = v1 + v2 + v3 =

10 8 + +1 = 7 V 3 3

(checked using LNAP 8/15/02)P5.3-5 Consider 8 V source only (open the 2 A source)

Let i1 be the part of ix due to the 8 V voltage source. Apply KVL to the supermesh:6 ( i1 ) + 3 ( i 1 ) + 3 ( i 1 ) 8 = 0 8 2 = A 12 3 Let i2 be the part of ix due to the 2 A current source. i1 =

Consider 2 A source only (short the 8 V source)

Apply KVL to the supermesh:6 ( i 2 ) + 3 ( i 2 + 2 ) + 3 i2 = 0 i2 = 6 1 = A 12 2

Finally,P5.3-6

i x = i1 + i 2 =

2 1 1 = A 3 2 6

Using superposition i x =

R2 + i a . Then R 1 + R 2 R1 + R 2 A R2 A vo = vs + ia R1 + R 2 R1 + R 2

vs

The equation of the straight line is v o = 7.5 v s + 30 so we require

A = 7.5 . For example, R1 + R 2

we can choose R1 = R 2 = 10 , and A = 150 V/A. Then v o = 7.5 v s + 75 i a so we requireia = 30 = 0.4 A . 75

(Checked: LNAP 6/22/04)

P5.3-7

ix =

vs va R1 vs va R1

va vo = A ix = A va =

R1 v o + A v s R1 + A

Apply KCL to the supernode corresponding to the CCVS to get

va vs R1 R1 + R 2 R1 R 2

+

va R2

+ ia +

vo R3

=0

va

vs R1

+ ia +

vo R3

=0

R 1 + R 2 R1 v o + A v s v s vo + ia + =0 R 1 R 2 R1 + A R 1 R3 R +R R1 + R 2 A 1 1 1 2 vo + vs + ia = 0 + R 2 R1 + A R 3 R1 R 2 R1 + A R1

(

)

(

(

)

)

R 3 R1 + R 2 + R 2 R1 + A

(

) ( R 2 R 3 ( R1 + A )R3 R 2 A

)v

o

+

A R2 R 2 R1 + A

(

)

vs + ia = 0

vo =

R 2 R 3 ( R1 + A ) ) vs ia R 3 ( R1 + R 2 ) + R 2 ( R 1 + A ) R 3 ( R1 + R 2 ) + R 2 ( R1 + A )

(

When R1 = 6 , R 2 = 12 and R 3 = 6 vo = 12 ( 6 + A ) 12 A ia vs 24 + A 24 + A

Comparing this equation to v o = 2 v s + 9 , we requires12 A =2 24 + A A = 1 2 V A

Then 2 v s + 9 = v o = 2v s + 6i a so we require

9 = 6ia

i a = 1.5 A(checked: LNAP 6/22/04)

P5.3-8

vo1 =

40 ||10 1 1 v1 = v1 a = 8 + 40 ||10 2 2

vo2 =

10 3 3 v1 = v 2 b = 8 || 40 + 10 5 5

vo3 = ( 8 ||10 || 40 ) i 3 = 4 i 3 c = 4

(checked: LNAP 6/22/04)

P5.3-9 Using superposition:

v x = 10 i xand v x 12 cos 2t 40 so 10 i x 12 cos 2t 40= 2ix ix = +

vx 10

+

vx 10

= 4ix

12 cos 2t 70

Finally, v o1 = 5 4 i x = 3.429 cos 2t V

( )

v x = 10 i xand vx 40 so+

vx 2 10

+

vx 10

= 4ix

0.2 = 1.75 i xFinally,

i x = 0.11429 A

v o1 = 5 4 i x = 2.286 V

( )

v o = v o1 + v o2 = 3.429 cos 2t + 2.286 V(checked: LNAP 6/22/04)

P5.3-10 Using superposition:

v o1 = 24 ( 0.3) = 7.2 Vv o2 = 30 20 = 4 V 120 + 30

v o = v o1 + v o 2 = 3.2 V

(checked: LNAP 5/24/04)

P5.3-11 (a)R 3 = R1 R 2 and R 2 = nR1

R3 =

n = R (1 + n ) R1 n + 1 1

nR12

R2

n R1 R1 n n +1 R1 R 3 = = R1 n 2n + 1 R1 + R1 n +1 n n nR1 R1 n +1 n R3 = = n + 1 R1 = R1 1 n n+2 nR1 + R1 1 + n + 1 n +1

a=

R 2 R3 R1 + R 2

n R1 n n+2 = = R3 2n + 2 n R1 + R1 n+2

b=

R1 R 3 R 2 + R1

n 1 R1 1 2n + 1 = = 2n + 1 = 1 R3 2n + 2 n nR1 + R1 1 + 2n + 1 2n + 1 a =n b

(b) From (a), we require n =4, i.e. R2 = 4R1 and R 3 = R1 R 2 =

4 R1 . For example 5

R1 = 10 , R 2 = 40 and R 3 = 8 .

(checked: LNAP 6/22/04)

P5.3-12 Using superposition R || 4 4 v o = 2 i1 + 2 6 + ( R || 4 ) 2 + ( R || 4 ) + 4 i 2

Comparing to v o = 0.5 i1 + 4 , we require R || 4 2 6 + ( R || 4 ) = 0.5 4 ( R || 4 ) = 6 + ( R || 4 ) R || 4 = 2 R=4

and 4 4 2 i2 = 4 2 i2 = 4 i2 = 4 A 2 + ( R || 4 ) + 4 2 + ( 4 || 4 ) + 4

(checked LNAP 6/12/04)

P5.3-13 Use units of mA, k and V.

4 + (5||20) = 8 k (a) Using superposition8 8 2= 2 ( R + 8 ) = 48 R = 16 k 7 R +8 R +8

(b) Using superposition again 8 4 2 1 5 16 ia = 7+ = 5 3 7 + 3 = 4 mA 8 + 16 5 + 20 8 + 16

P5.3-14 v1 10 10 20 io = i2 + 10 + 40 20 + 12 + ( 40 10 ) 10 + 40 20 + 12 + ( 40 10 ) v3 20 + 12 + 40 + ( 20 + 12 ) 10 + 40 ( 20 + 12 ) 1 1 1 io = v1 + i 2 + v3 200 10 62.5

So

a = 0.05, b = 0.1 and c = 0.016

(checked: LNAP 6/19/04)

P5.3-15im = 25 3 ( 5) = 5 3 = 2 A 3+ 2 2+3

P5.3-16 3 3 vm = 3 ( 5) (18) = 5 6 = 1 A 3 + (3 + 3) 3 + (3 + 3)

Section 5-4: Thvenins Theorem P5.4-1

(checked using LNAP 8/15/02)

P5.4-2 The circuit from Figure P5.4-2a can be reduced to its Thevenin equivalent circuit in four steps:

(a)

(b)

(c)

(d)

Comparing (d) to Figure P5.4-2b shows that the Thevenin resistance is Rt = 16 and the open circuit voltage, voc = 12 V.

P5.4-3 The circuit from Figure P5.4-3a can be reduced to its Thevenin equivalent circuit in five steps:

(a)

(b)

(c)

(d)

(e)

Comparing (e) to Figure P5.4-3b shows that the Thevenin resistance is Rt = 4 and the open circuit voltage, voc = 2 V. (checked using LNAP 8/15/02)

P5.4-4 Find Rt:

Rt =

12 (10 + 2 ) =6 12 + (10 + 2 )

Write mesh equations to find voc: Mesh equations:12 i1 + 10 i1 6 ( i2 i1 ) = 0 6 ( i2 i1 ) + 3 i 2 18 = 0

28 i1 = 6 i 2 9 i 2 6 i1 = 1836 i1 = 18 i1 = i2 =7 1 voc = 3 i 2 + 10 i1 = 3 + 10 = 12 V 3 2

1 A 2

14 1 7 = A 3 2 3

Finally,

(checked using LNAP 8/15/02)

P5.4-5 Find voc: Notice that voc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: va = voc Apply KCL at node a: 6 voc voc 3 + voc = 0 + 8 4 4

6 + voc + 2 voc 6 voc = 0 voc = 2 V Find Rt: Well find isc and use it to calculate Rt. Notice that the short circuit forces va = 0 Apply KCL at node a:60 0 3 + + 0 + i sc = 0 8 4 4 i sc = 6 3 = A 8 4

Rt =

voc 2 8 = = 3 i sc 3 4(checked using LNAP 8/15/02)

P5.4-6 Find voc:

2 va va va = + 3 + 0 va = 18 V 3 6 The voltage across the right-hand 3 resistor is zero so: va = voc = 18 V

Apply KCL at the top, middle node:

Find isc:

2 va va va v = + 3 + a va = 18 V 3 6 3 va 18 Apply Ohms law to the right-hand 3 resistor : = 6 V i sc = = 3 3 v 18 R t = oc = = 3 Finally: i sc 6

Apply KCL at the top, middle node:

(checked using LNAP 8/15/02)

P5.4-7 (a)

vs + R1 ia + ( d + 1) R 2 ia = 0ia = v oc = vs R1 + ( d + 1) R 2

( d + 1) R 2vs R1 + ( d + 1) R 2ia = vs R1

i sc = ( d + 1) ia =

( d + 1) vsR1

ia d ia +

vT iT = 0 R2

R1 ia = vT

iT = ( d + 1)

vT vT R 2 ( d + 1) + R1 + = vT R1 R 2 R1 R 2R1 R 2 vT = iT R1 + ( d + 1) R 2

Rt =

(b) Let R1 = R2 = 1 k. Then625 = R t = 1000 1000 d= 2 = 0.4 A/A d +2 625

and 5 = voc =

( d + 1) vsd +2

vs =

0.4 + 2 5 = 13.33 V 0.4 + 1 (checked using LNAP 8/15/02)

P5.4-8 From the given data:

6=

2000 voc R t + 2000 voc = 1.2 V 4000 R t = 1600 2= voc R t + 4000

When R = 8000 ,

v=

R voc Rt + R

v=

8000 (1.2 ) = 1.5 V 1600 + 8000

P5.4-9

From the given data: 0.004 = voc R t + 2000 voc = 24 V voc R t = 4000 0.003 = R t + 4000

i=

voc Rt + R

(a) When i = 0.002 A: 24 R = 8000 0.002 = 4000 + R (b) Maximum i occurs when R = 0: 24 = 0.006 = 6 mA i 6 mA 4000

P5.4-10 The current at the point on the plot where v = 0 is the short circuit current, so isc = 20 mA. The voltage at the point on the plot where i = 0 is the open circuit voltage, so voc = 3 V.

The slope of the plot is equal to the negative reciprocal of the Thevenin resistance, so 1 0 0.002 = R t = 150 Rt 3 0

P5.4-11

12 + 6000 ia + 2000 ia + 1000 ia = 0 ia = 4 3000 A voc = 1000 ia = 4 V 3

ia = 0 due to the short circuit

12 + 6000 isc = 0 isc = 2 mA 4 voc Rt = = 3 = 667 isc .0024 3 ib = 667 + R

ib = 0.002 A requires4 3 667 = 0 R = 0.002

(checked using LNAP 8/15/02)

P5.4-1210 = i + 0 i = 10 A voc + 4 i 2 i = 0 voc = 2 i = 20 V

i + i sc = 10 i = 10 i sc 4 i + 0 2 i = 0 i = 0 i sc = 10 A

Rt =

voc 20 = = 2 isc 10

2 = iL =

20 RL = 12 RL 2

(checked using LNAP 8/15/02)

P5.4-13 Replace the part of the circuit that is connected to the variable resistor by its Thevenin equivalent circuit:

18 k || (12 k + 24 k ) = 18 k || 36 k = 12 k

ia =

36 R and v a = 36 R + 12000 R + 12000

36 p = ia va = R R + 12000 36 = 3 mA when R = 0 (a short circuit). 0 + 12000 105 36 = 32.14 V when R is as large as possible, i.e. R = 100 k. (b) v a = 5 10 + 12000 (c) Maximum power is delivered to the adjustable resistor when R = R t = 12 k . Then

2

(a) i a =

36 p = ia va = 12000 = 0.027 = 27 mW 12000 + 12000 (checked: LNAP 6/22/04)

2

P5.4-14 Replace the source by its Thevenin equivalent circuit to get

io =

v oc R t +R L

Using the given formation 0.375 = v oc R t + 4 v oc 0.300 = R t +8 0.375 ( R t + 4 ) = 0.300 ( R t + 8 )

So Rt =

0.075 6 (a) When R L = 10 , i o = = 0.2727 A. 12 + 10 (b) 12 = R t = 48 11R R = 16 .

( 0.300 ) 8 ( 0.375) 4 = 12 and v

oc

= 0.3 (12 + 8 ) = 6 V

(checked: LNAP 5/24/04)

P5.4-15 (a)

i 3 i 2 = 0.25 A

Apply KVL to mesh 1 to get

20 ( i1 i 2 ) + 20 ( i1 i 3 ) 40 = 0

Apply KVL to the supermesh corresponding to the unspecified resistance to get

40i 2 + 10i 3 20 ( i1 i 3 ) 20 ( i1 i 2 ) = 0

Solving, for example using MATLAB, gives1 i1 0.25 1 0 40 20 20 i = 40 2 40 60 30 i 3 0 i1 1.875 i 2 = 0.750 i 3 1.000

Apply KVL to mesh 2 to get

40i 2 + R ( i 2 i 3 ) 20 ( i1 i 2 ) = 0(b)

R=

20 ( i1 i 2 ) 40i 2 i2 i3

= 30

20 40 v oc = 40 40 = 12 V 20 + 20 10 + 40

R t = 18

0.25 =

12 18 + R

R = 30

(checked: LNAP 5/25/04)

P5.4-16 Find the Thevenin equivalent circuit for the part of the circuit to the left of the terminals a-b.

Using voltage division twice

v oc =

32 30 20 20 = 5 4 = 1 V 32 + 96 120 + 30

R t = ( 96 || 32 ) + (120 || 30 ) = 24 + 24 = 48

Replacing the part of the circuit to the left of terminals a-b by its Thevenin equivalent circuit gives

io =

1 = 0.0125 A = 12.5 mA 48 + 32

(checked: LNAP 5/24/04)

P5.4-18 Replace the circuit by its Thevenin equivalent circuit:

Rm vm = 5 R m + 50 (a)

v mi = lim v m = 5 VR m

(b) When R m = 1000 , v m = 4.763 V so % error = (c) Rm 5 5 R m + 50 0.02 5 Rm R m + 505 4.762 100 = 4.76% 5

0.98

R m 2450 (checked: LNAP 6/16/04)

P5.4-19

v s v oc R1 v s R 2 (1 + b ) v oc = v oc R1 + R 2 (1 + b ) ia + bia = R2 ia =

i sc = i a (1 + b ) =v s R 2 (1 + b )

vs R1

(1 + b )

Rt =

v oc i sc

=

R1 + R 2 (1 + b ) R1 R 2 = vs R1 + R 2 (1 + b ) (1 + b ) R1 (checked: LNAP 7/22/04)

Section 5-5: Nortons Theorem P5.5-1 When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will cool off.

P5.5-2

(checked using LNAP 8/16/02)

P5.5-3

P5.5-4 To determine the value of the short circuit current, isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit current. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently, i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 i 2 = i1 isc Apply KVL to mesh 1 to get3 i1 2 ( i1 i 2 ) + 6 ( i1 i 2 ) 10 = 0 7 i1 4 i 2 = 10

(1)

Apply KVL to mesh 2 to get5 i 2 6 ( i1 i 2 ) = 0 6 i1 + 11 i 2 = 0 i1 = 11 i2 6

Substituting into equation 1 gives 11 7 i 2 4 i 2 = 10 i 2 = 1.13 A i sc = 1.13 A 6

Figure (a) Calculating the short circuit current, isc, using mesh equations. To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as v Rt = T iT In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (b), mesh current i2 is equal to the negative of the current source current. Consequently, i2 = i T . The controlling current of the CCVS is expressed in terms of the mesh currents asi a = i1 i 2 = i1 + i T

Apply KVL to mesh 1 to get3 i1 2 ( i1 i 2 ) + 6 ( i1 i 2 ) = 0 7 i1 4 i 2 = 0 i1 = 4 i2 7

(2)

Apply KVL to mesh 2 to get5 i 2 + vT 6 ( i1 i 2 ) = 0 6 i1 + 11 i 2 = vT

Substituting for i1 using equation 2 gives4 6 i 2 + 11 i 2 = vT 7 7.57 i 2 = vT

Finally,Rt =

vT vT vT = = = 7.57 iT i2 iT

Figure (b) Calculating the Thevenin resistance, R t =

vT , using mesh equations. iT

To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently, i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents asi a = i1 i 2 = i1 0 = i1

Apply KVL to mesh 1 to get 3 i1 2 ( i1 i 2 ) + 6 ( i1 i 2 ) 10 = 0 3 i1 2 ( i1 0 ) + 6 ( i1 0 ) 10 = 0 i1 = Apply KVL to mesh 2 to get5 i 2 + voc 6 ( i1 i 2 ) = 0 voc = 6 ( i1 ) = 6 (1.43) = 8.58 V

10 = 1.43 A 7

Figure (c) Calculating the open circuit voltage, voc, using mesh equations. As a check, notice that R t isc = ( 7.57 )(1.13) = 8.55 voc (checked using LNAP 8/16/02)

P5.5-5 To determine the value of the short circuit current, Isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit current. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = 24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 . Apply KCL at node 2 to get v1 v 2 3 = v2 v3 6 2 v1 + v 3 = 3 v 2 48 = 3 v a v a = 16 V

Apply KCL at node 3 to get v2 v3 6 + 4 v 2 = isc 3 9 v a = isc 6 isc = 9 ( 16 ) = 24 A 6

Figure (a) Calculating the short circuit current, Isc, using mesh equations. To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as v R th = T iT Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.

In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. v1 = 0 . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across the current source, i.e. v3 = vT . Apply KCL at node 2 to get v1 v 2 3 Apply KCL at node 3 to getv2 v3 6 + 4 v 2 + iT = 0 9 v 2 v3 + 6 iT = 0 3 9 v a vT + 6 iT = 0 3 v T vT + 6 iT = 0 2 vT = 6 iT

=

v 2 v3 6

2 v1 + v 3 = 3 v 2

vT = 3 v a

Finally,Rt = vT = 3 iT

Figure (b) Calculating the Thevenin resistance, R th =

vT , using mesh equations. iT

To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = 24 V . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc . Apply KCL at node 2 to get v1 v 2 3 = v 2 v3 6 2 v1 + v 3 = 3 v 2 48 + v oc = 3 v a

Apply KCL at node 3 to get v2 v3 6 + 4 v 2 = 0 9 v 2 v 3 = 0 9 v a = v oc 3

Combining these equations gives

3 ( 48 + voc ) = 9 v a = voc

voc = 72 V

Figure (c) Calculating the open circuit voltage, voc, using node equations. As a check, notice that

R th I sc = ( 3)( 24 ) = 72 = Voc(checked using LNAP 8/16/02)

P5.5-6 (a) Replace the part of the circuit that is connected to the left of terminals a-b by its Norton equivalent circuit: Apply KCL at the top node of the dependent source to see that i b = 0 A . Then v oc = 25 + 5000 i b = 25 V

( )

Apply KVL to the supermesh corresponding to the dependent source to get 5000 i b + 10000 3 i b 25 = 0 i b = 1 mA Apply KCL to get i sc = 3 i b = 3 mA Then

( )

Rt =

v oc i sc

= 8.33 k

Current division gives0.5 = 8333 3 R = 41.67 k R + 8333

(b)

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 i b + 0.5 103 = 4 i b i b = mA 6 Apply KVL to the supermesh corresponding to the dependent source to get

5000 i b + (10000 + R ) 0.5 103 25 = 01 5000 103 + (10000 + R ) 0.5 103 = 25 6 125 6 = 41.67 k R= 0.5 103

(

)

(

)

P5.5-7 Use source transformations to reduce the circuit to

Replace the series voltage sources by an equivalent voltage source, the series resistors by an equivalent resistance and do a couple more source transformations to reduce the part of the circuit to the left of the terminals a-b by its Norton equivalent circuit. Apply KCL at node a to get v v2 0.4 = + 10 2 sov=

v v 2 + 0.8 = 0 5

.2 1.8 = 0.8, -1.0 V 2

Choosing the positive value of v, i =

0.82 = 0.32 A . Choosing the negative value of v, 2

12 i= = 0.5 . There are two solutions to this problem. Linear circuits are so much simpler than 2 nonlinear circuits. (checked: LNAP 5/26/04)

P5.5-8 Simplify the circuit using a source transformation:

Identify the open circuit voltage and short circuit current.

Apply KVL to the mesh to get:

(10 + 2 + 3) i x 15 = 0Then

ix = 1 A

v oc = 3 i x = 3 V

Express the controlling current of the CCVS in terms of the mesh currents:i x = i 1 i sc

The mesh equations are10 i1 + 2 ( i1 i sc ) + 3 ( i1 i sc ) 15 = 0 15 i1 5 i sc = 15

andi sc 3 ( i1 i sc ) = 0 i1 =

4 i sc 3

so

4 15 i sc 5 i sc = 15 i sc = 1 A 3

The Thevenin resistance isRt =

3 =3 1

Finally, the Norton equivalent circuit is

(checked: LNAP 6/21/04)

P5.5-9 Identify the open circuit voltage and short circuit current.1 v1 = 3 = 1 V 3

Thenv oc = v1 4 ( 2.5 v 1 ) = 9 V

1 v 1 = 3 i sc = 1 3 i sc 3

4 ( 2.5 v1 + i sc ) + 5 i sc v1 = 0 9 v1 + 9 i sc = 09 (1 3 i sc ) + 9 i sc = 0 i sc = 1 A 2

The Thevenin resistance isRt =

9 = 18 0.5

Finally, the Norton equivalent circuit is

(checked: LNAP 6/21/04)

P5.5-10 Replace the circuit by its Norton equivalent circuit:

1600 im = 1.5 103 ) 1600 + R m ( (a) i mi = lim (b) When Rm = 20 then i m = 1.48 mA so% error = 1.5 1.48 100 = 1.23% 1.5R m 0

i m = 1.5 mA

(c) 1600 0.015 0.015 ) 1600 + R m ( 0.02 0.015 1600 0.98 R m 32.65 1600 + R m (checked: LNAP 6/18/04)

P5.5-11ia = i a = 3 A 6 voc = 2 i a = 6 V 2 i a 12

12 + 6 i a = 2 i a 3 i sc = 2 i a

i a = 3 A 2 ( 3 ) = 2 A 3

i sc =

Rt =

6 =3 2

P5.5-1212 24 12 24 = = 8 12 + 24 36 24 voc = ( 30 ) = 20 V 12 + 24 Rt =

i=

20 8+ R

Section 5-6: Maximum Power Transfer P5.6-1 a) For maximum power transfer, set RL equal to the Thevenin resistance:R L = R t = 100 + 1 = 101

b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin equivalent circuit:

The voltage across RL is Then

vL =

pmax

101 (100 ) = 50 V 101 + 101 2 v 502 = L = = 24.75 W R L 101

P5.6-2 Reduce the circuit using source transformations:

Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 and (b) the value of that maximum power is P = i 2 ( R) = (0.03)2 (60) = 54 mW max R

P5.6-3

RL v L = vS RS + R L pL =2 vL

RL

=

v S2 R L ( RS + R L )2

By inspection, pL is max when you reduce RS to get the smallest denominator. set RS = 0

P5.6-4 Find Rt by finding isc and voc:

The current in the 3 resistor is zero because of the short circuit. Consequently, isc = 10 ix. Apply KCL at the top-left node to getix + 0.9 = 10 ix ix = 0.9 = 0.1 A 9

so isc = 10 ix = 1A Next

Apply KCL at the top-left node to getix + 0.9 = 10 ix ix = 0.9 = 0.1 A 9

Apply Ohms law to the 3 resistor to get

voc = 3 (10 ix ) = 30 ( 0.1) = 3 VFor maximum power transfer to RL:R L = Rt = voc 3 = =3 isc 1

The maximum power delivered to RL is given by

pmax =

voc 32 3 = = W 4 R t 4 ( 3) 4

2

P5.6-5

The required value of R is

R = Rt = 8 +

( 20 + 120 ) (10 + 50 ) = 50 ( 20 + 120 ) + (10 + 50 )

30 170 voc = ( 20 ) 10 170 + 30 ( 20 ) 50 170 + 30 170(20)(10) 30(20)(50) 4000 = = = 20 V 200 200 The maximum power is given by 2 v 202 pmax = oc = =2W 4 R t 4 ( 50 )

P5.6-6

iL =

A vs Ro +RLA 2v s 2 R Lo

PL = i L R L =2

(R

+RL)

2

(a) R t =R o so R L =R o = 10 maximizes the power delivered to the load. The corresponding load power is 2 21 20 10 2 = 2.5 W . PL = 2 (10 + 10 )

(b) Ro = 0 maximizes PL (The numerator of PL does not depend on Ro so PL can be maximized by making the denominator as small as possible.) The corresponding load power is1 20 2 = 12.5 W. = 82 2

PL =

A 2v s 2 R L R L2

=

A 2v s 2 RL

(c) PL is proportional to A2 so the load power continues to increase as A increases. The load can safely receive 15 W. This limit corresponds to

1 A 8 2 15 = 2 (18)2

2

A = 36

15 = 49.3 V. 8 (checked: LNAP 6/9/04)

P5.6-7 Replace the part of the circuit connected to the variable resistor by its Thevenin equivalent circuit. First, replace the left part of the circuit by its Thevenin equivalent: 150 v oc1 = 10 = 4.545 V 150 + 180 R t1 = 180 150 = 81.8

Next, replace the right part of the circuit by its Thevenin equivalent: 470 v oc2 = 20 = 15.932 V 470 + 120 R t2 = 120 470 = 95.6

Now, combine the two partial Thevenin equivalents:v oc = v oc1 v oc2 = 10.387 V and R t = R t1 + R t2 = 177.4

So The power received by the adjustable resistor will be maximum when R = Rt = 177.4 . The maximum power received by the adjustable 2 ( 11.387 ) = 0.183 W . resistor will be p = 4 (177.4 )

(checked LNAPDC 7/24/04)

P5.6-8 10 R L 100 R L p=iv= (10 ) = 2 Rt + R L Rt + R L ( Rt + R L )

The power increases as Rt decreases so choose Rt = 1 . Thenpmax = i v =

100 ( 5 )

(1 + 5)

2

= 13.9 W

P5.6-9 From the plot, the maximum power is 5 W when R = 20 . Therefore:

Rt = 20

andpmax v = oc 4 Rt2

voc =

pmax 4 Rt = 5 ( 4 ) 20 = 20 V

Section 5-8 How Can We Check?P5.8-1 Use the data in the first two lines of the table to determine voc and Rt:

voc Rt + 0 voc = 39.9 V voc R t = 410 0.0438 = R t + 500 0.0972 = Now check the third line of the table. When R= 5000 : v 39.9 i = oc = = 7.37 mA R t + R 410 + 5000 which disagree with the data in the table.The data is not consistent.

P5.8-2

Use the data in the table to determine voc and isc: voc = 12 V (line 1 of the table)

isc = 3 mA so Rt = voc = 4 k isc

(line 3 of the table)

Next, check line 2 of the table. When R = 10 k: v 12 i = oc = = 0.857 mA 3 R t + R 10 (10 ) + 5 (103 ) To cause i = 1 mA requires which agrees with the data in the table. v 12 R = 8000 0.001 = i = oc = R t + R 10 (103 ) + R

I agree with my lab partners claim that R = 8000 causes i = 1 mA.

P5.8-3

1 1 1 1 11 = + + = R t R 2 R 3R 6 R and

Rt =

6R 11

23 34 65 180 voc = 30 + 20 + 10 = 11 3+ 2 3 2+3 4 1+ 6 5 so the prelab calculation isnt correct. But then 180 180 voc 11 i= = = 11 = 163 mA 54.5 mA R t + R 6 110 + 40 60 + 40 ( ) 11 so the measurement does not agree with the corrected prelab calculation.

P5.8-4

6000 3000i=

( 500 + 1500 ) = 2000

2000 = 1000

12 12 = 12 mA R + 1000 1000

How about that?! Your lab partner is right. (checked using LNAP 6/21/05)

P5.8-5 (a)

KVL gives from row 2 from row 3 So

v oc = ( R t + R ) i v oc = ( R t + 10 ) (1.333) v oc = ( R t + 20 ) ( 0.857 )

(R

t

+ 10 ) (1.333) = ( R t + 20 ) ( 0.857 ) 28 ( R t + 10 ) = 18 ( R t + 20 )

Solving gives10 R t = 360 280 = 80 Rt = 8

and

v oc = ( 8 + 10 )(1.333) = 24 V

(b)

i=

v oc Rt + R

=

24 R 24 R and v = v oc = 8+ R R + Rt R+8

When R = 0, i = 3 A, and v = 0 V. 1 When R = 40 , i = A . 2 24 ( 80 ) 240 When R = 80 , v = = = 21.82 . 88 11 These are the values given in the tabulated data so the data is consistent. 24 ( 40 ) = 20 V . (c) When R = 40 , v = 48 24 When R = 80 , i = = 0.2727 A . 88 (d) First 8 = R t = 24 18 ( R1 + 12 ) R1 = 24 the, using superposition, 24 = v oc =

24 + 18

( ( R + 12))1

24

12 + 24 18 ( R1 + 12 ) i s = 8 + 8i s

(

)

is = 2 A

(checked using LNAP 6/21/05)