DORFChapter 5 - Circuit Theorems

Prévia do material em texto

Problems 
Section 5-2: Source Transformations 
 
P5.2-1 
(a) 
 
 
 
 
 = 2 
 = 0.5 V
t
t
R
v
∴ Ω
− 
(b) 9 4 2 ( 0.5) 0
9 ( 0.5) 1.58 A
4 2
i i
i
− − − + − =
− + −= = −+
 
9 4 9 4( 1.58) 2.67 Vv i= + = + − = 
(c) 1.58 Aai i= = − 
(checked using LNAP 8/15/02) 
 
 
P5.2-2 
 
 
 
 
 
 
 
 
Finally, apply KVL: 1610 3 4 0 2.19 A
3a a a
i i i− + + − = ∴ = 
 
(checked using LNAP 8/15/02) 
 
 
P5.2-3 
 
 
Source transformation at left; equivalent resistor for parallel 6 and 3 Ω 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 
o6 (9 19) 36 0i v− + + − − = 
o5 / 2 42 28 (5 / 2) 28 Vi v= ⇒ = − + = 
 
(checked using LNAP 8/15/02) 
 
 
 
P5.2-4 
 
 
 4 2000 4000 10 2000 3 0
 375 A
a a a
a
i i i
i μ
− − − + − − =
∴ = 
 
 
(checked using LNAP 8/15/02) 
 
P5.2-5 
 
 
 
12 6 24 3 3 0 1 Aa a ai i i− − + − − = ⇒ = 
 
(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, ( ) ( ) ( )50 100 1000.21 0.21 7 V
50 100 3a
v = =+ = 
 
(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, 
 
1 3 8v v− = 
 
Apply KCL to the supernode corresponding to the 8-V source to get 
 
1 2 3
1 2 3
24 10
0 0.125 0.3 0.05 0.02 0.2 0
8 20 50
v v v
v v v
− −+ + = ⇒ − + + − = 
 
Apply KCL at node 2 to get 
 
1 2 2 2 3
1 2 3 =0.04 0.19 0.1 025 20 10
v v v v v
v v v
− −= + ⇒ + − = 
 
Solving, for example using MATLAB 
 
1 1
2 2
3 3
1 0 1 8 4.7873
0.125 0.05 0.02 0.5 0.6831
0.04 0.19 0.1 0 3.2127
v v
v v
v v
⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢= ⇒ = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢− − −⎣ ⎦ ⎣ ⎦ ⎣⎣ ⎦ ⎣ ⎦
⎤⎥⎥⎥⎦
 
 
The power supplied by the 8-V source is 
 
( )4.7873 0.6831 4.7873 248 4.316 W
25 8
− −⎛ ⎞−+ =⎜ ⎟⎝ ⎠
 
 
Apply KCL at node 4 of the original circuit to get 
 
( )3 4 4 3
4
2 30 2 3.2127 30
0.5 4.71 V
30 20 5 5
v v v v
v
− + − ++ = ⇒ = = = 
 
The power supplied by the 0.5 A source is 
 
( )0.5 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 + =& Ω 
 
 
 
 
Do a source transformation, then replace 
series voltage sources by an equivalent 
voltage source. 
 
 
Do two more source transformations 
 
Now current division gives 
8 23
8 8
i
R R
⎛ ⎞= =⎜ ⎟+ +⎝ ⎠
4
 
Then Ohm’s Law gives 
24
8
Rv Ri
R
= = + 
 
( )
( )
24a 2 A
8 4
24 8
(b) 12 V
8 8
24(c) 1 16 
8
24(d) 16 16 
8
i
v
R
R
R R
R
= =+
= =+
= ⇒ =+
= ⇒ =+
Ω
Ω
 
(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 
 
( )23.1 2 10.3125 2 87.45 Wp = + =⎡ ⎤⎣ ⎦
 
 
 
Section 5-3 Superposition 
 
P5.3–1 
Consider 6 A source only (open 9 A source) 
 
 
Use current division: 
 
1
1
15 6 40 V
20 15 30
v v⎡ ⎤= ⇒⎢ ⎥+⎣ ⎦ =
 
Consider 9 A source only (open 6 A source) 
 
 
Use current division: 
 
2
2
10 9 40 V
20 10 35
v v⎡ ⎤= ⇒⎢ ⎥+⎣ ⎦ =
 
1 2 40 40 80 Vv v v∴ = + = + = 
(checked using LNAP 8/15/02) 
 
 
P5.3-2 
Consider 12 V source only (open both current sources) 
 
 
 
KVL: 
 
1 1 1
1
20 12 4 12 0
 1/ 3 mA
i i i
i
+ + + =
⇒ = − 
Consider 3 mA source only (short 12 V and open 9 
mA sources) 
 
 
Current Division: 
 
2
16 43 mA
16 20 3
i ⎡ ⎤= =⎢ ⎥+⎣ ⎦ 
 
Consider 9 mA source only (short 12 V and open 3 
mA sources) 
 
 
 
 
 
 
Current Division: 
 
3
129 3 
24 12
i ⎡ ⎤= − = −⎢ ⎥+⎣ ⎦ mA 
 
1 2 3 1/ 3 4 / 3 3 2 mAi i i i∴ = + + = − + − = − 
 
(checked using LNAP 8/15/02) 
 
P5.3–3 
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. 
 
 
1
2 6 30 6 mA 2 mA
2 8 6 12a a
i i i⎛ ⎞ ⎛ ⎞= = ⇒ = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ 
 
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. 
 
 
 
2
4 6 15 6 mA 2 mA
4 6 6 12b b
i i i⎛ ⎞ ⎛ ⎞= = ⇒ = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ 
 
Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V 
voltage source. 
 
 
 
 
( )3
6 || 6 3 2.5 10 0.5 mA
6 || 6 12 3 12
i
⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
 
 
Finally, 1 2 3 2 2 0.5 3.5 mAi i i i= + + = + − = 
 
(checked using LNAP 8/15/02) 
 
 
P5.3–4 
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. 
 
( ) ( )
( )
1
100 ||100 10
100 ||100 100
50 1010 V
150 3
v = +
= =
 
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. 
 
( ) ( )
( )
1
100 ||100 8
100 ||100 100
50 88 V
150 3
v = +
= =
 
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. 
 
 
 
3 (100 ||100 ||100)(0.03)
100 (0.03) 1 V
3
v =
= = 
 
Finally, 1 2 3
10 8 1 7 V
3 3a
v v v v= + + = + + = 
 
(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: 
 ( ) ( ) ( )1 1 16 3 3 8i i i 0+ + − = 
 
1
8 2 A
12 3
i = = 
Consider 2 A source only (short the 8 V source) 
 
 
Let i2 be the part of ix due to the 2 A 
current source. 
 
Apply KVL to the supermesh: 
 ( ) ( )2 2 26 3 2 3i i i 0+ + + = 
 
2
6 1 A
12 2
i −= = − 
 
Finally, 1 2
2 1 1 A
3 2 6x
i i i= + = − = 
 
P5.3-6 
Using superposition s 2x a
1 2 1 2
v R
i i
R R R R
⎛ ⎞= + ⎜⎜+ +⎝ ⎠
⎟⎟ . Then 
2
o s
1 2 1 2
A RAv v
R R R R
= ++ + ai 
 
The equation of the straight line is o s7.5 30v v= + so we require
1 2
7.5A
R R
=+ . For example, 
we can choose Then 1 2 10 , and 150 V/A.R R A= = Ω = o s7.5 75v v ai= + so we require 
a
30 0.4 A
75
i = = . 
(Checked: LNAP 6/22/04) 
 
 
P5.3-7 
s a
x
1
v v
i
R
−= 
s a
a o x
1
v v
v v A i A
R
−− = = 
1 o s
a
1
R v A v
v
R A
+= + 
 
Apply KCL to the supernode corresponding to the CCVS to get 
 
a s a o
a
1 2 3
0
v v v v
i
R R R
− + + + = 
 
1 2 s o
a a
1 2 1 3
0
R R v v
v i
R R R R
+ − + + = 
 
1 2 1 o s s o
a
1 2 1 1 3
0
R R R v A v v v
i
R R R A R R
⎛ ⎞+ + − + + =⎜ ⎟⎜ ⎟+⎝ ⎠( )
( )
( )1 21 2 o s3 12 1 1 2 1
1 1 0
R R AR R
v v
R RR R A R R R A
⎛ ⎞ ⎛ ⎞++⎜ ⎟ ⎜ ⎟+ + − +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
ai = 
 ( ) ( )
( ) ( )3 1 2 2 1 2o s2 3 1 2 1 0
R R R R R A A R
v v
R R R A R R A
+ + + −
ai+ + =+ + 
 ( )
( ) ( )
( )
( ) ( )3 2 2 3 1o s3 1 2 2 1 3 1 2 2 1
R R A R R R A
v v
R R R R R A R R R R R A
− += −+ + + + + + ai 
 
 
When 1 2 36 , 12 and 6 R R R= Ω = Ω = Ω
( )
o s
12 612
24 24
AAv v
A A
+−= −+ + ai
9
 
 
Comparing this equation to , we requires o s2v v= +
 
12 V2 1 
24 A
A A 2
A
− = ⇔ = −+ 
 
Then so we require s o s2 9 2 6v v v+ = = + ai
 
a a9 6 1.5 Ai i= ⇒ = 
 
(checked: LNAP 6/22/04) 
 
 
 
P5.3-8 
 
 
o1 1 1
40 ||10 1 1
8 40 ||10 2 2
v v v= = ⇒+ a = 
 
 
 
 
o2 1 2
10 3 3
8 || 40 10 5 5
v v v= − = − ⇒ = −+ b 
 
 
 
 
( )o3 3 38 ||10 || 40 4 4v i i= = ⇒ c = 
 
 
(checked: LNAP 6/22/04) 
 
P5.3-9 
Using superposition: 
 
x x10v i= 
and 
x x x
x
12 cos 2
4
40 10 10
v t v v
i
− + + = 
so 
 
x
x x
10 12cos 2 122 cos 2t
40 70
i t
i i
− = ⇒ = −
 
Finally, ( )o1 x5 4 3.429 cos 2 Vv i= − = t 
 
x x10v i= 
and 
x x x
x
2
4
40 10 10
v v v
i
−+ + = 
so 
 
x x0.2 1.75 0.11429 Ai i− = ⇒ = − 
 
Finally, ( )o1 x5 4 2.286 Vv i= − = 
 
 
 
o o1 o2 3.429 cos 2 2.286 Vv v v t= + = + 
 
(checked: LNAP 6/22/04) 
 
 
P5.3-10 
Using superposition: 
 
 
 
 
 
( )1 24 0.3 7.2 Vov = = 
 
2
30 20 4 V
120 30o
v = − = −+ 
 
1 2 3.2 Vo o ov v v= + = 
 
 
(checked: LNAP 5/24/04) 
 
 
P5.3-11 
(a) 
3 1 2 2 and 1R R R R nR= =& 
 
( )
2
1
3 1
11 1
nR nR R
n R n
⎛ ⎞= = ⎜ ⎟+ +⎝ ⎠ 
 
1 1
1 3
1 1
1
2 1
1
nR R
nn
1R R Rn nR R
n
⎛ ⎞⎜ ⎟+ ⎛ ⎞⎝ ⎠= = ⎜ ⎟+⎛ ⎞ ⎝ ⎠+ ⎜ ⎟+⎝ ⎠
& 
1 1
2 3 1
1 1
1 1
1 21
11
n nnR R
nn n
1R R Rn nnR R
nn
⎛ ⎞⎜ ⎟+ ⎛ ⎞⎝ ⎠ += = = ⎜ ⎟+⎛ ⎞ ⎝ ⎠++ ⎜ ⎟ ++⎝ ⎠
& R 
 
1
2 3
1 2 3
1 1
2
2 2
2
n RR R nna
nR R R nR R
n
⎛ ⎞⎜ ⎟+⎝ ⎠= = =+ +⎛ ⎞+ ⎜ ⎟+⎝ ⎠
&
& 
 
1
1 3
2 1 3
1 1
1
12 1 2 1
1 2 21
2 12 1
n RR R n nb
nR R R nnR R
nn
⎛ ⎞⎜ ⎟+⎝ ⎠ += = = =+ +⎛ ⎞ ++ ⎜ ⎟ ++⎝ ⎠
&
& 
 
a n
b
∴ = 
 
(b) From (a), we require n =4, i.e. R2 = 4R1 and 3 1 2
4
5 1
R R R R= =& . For example 
 
1 2 310 , 40 and 8 .R R R= Ω = Ω = Ω 
 
(checked: LNAP 6/22/04) 
 
 
P5.3-12 
Using superposition 
( ) ( )o 1
|| 4 42 2
6 || 4 2 || 4 4
Rv i
R R
⎛ ⎞ ⎛= − +⎜ ⎟ ⎜⎜ ⎟ ⎜+ +⎝ ⎠ ⎝ 2
i
⎞⎟⎟+ ⎠
 
 
Comparing to , we require o 10.5 4v i= − +
 
( ) ( ) ( )
|| 42 0.5 4 || 4 6 || 4 || 4 2 4 
6 || 4
R R R R R
R
⎛ ⎞− = − ⇒ = + ⇒ = ⇒⎜ ⎟⎜ ⎟+⎝ ⎠
= Ω 
and 
 
( ) ( )2 2
4 42 4 2 4
2 || 4 4 2 4 || 4 4
i i
R
⎛ ⎞ ⎛ ⎞= ⇒ = ⇒ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠ 2
4 Ai 
 
(checked LNAP 6/12/04) 
 
 
 
P5.3-13 
Use units of mA, kΩ and V. 
 
4 + (5||20) = 8 kΩ 
(a) Using superposition 
 
( )8 82 7 2 8 48 16 k
8 8
R R
R R
⎛ ⎞= − ⇒ + = ⇒ =⎜ ⎟+ +⎝ ⎠ Ω 
 
(b) Using superposition again 
 
a
5 16 8 4 2 17 7
5 20 8 16 8 16 5 3 3
i ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = × +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥+ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ 4 mA= 
 
 
P5.3-14 
( ) ( )
( ) ( )
1
o 2
3
10 10 20
10 40 20 12 40 10 10 40 20 12 40 10
20 12
40 20 12 10 40 20 12
v
i i
v
⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞= − + − ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ + + + + +⎡ ⎤⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎣ ⎦⎝ ⎠
⎛ ⎞⎛ ⎞++ − ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟+ + + +⎡ ⎤⎝ ⎠ ⎣ ⎦⎝ ⎠
& &
&
 
 
o 1 2
1 1 1
200 10 62.5
i v i⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 3v 
So 
0.05, 0.1 and 0.016a b c= − = − = − 
(checked: LNAP 6/19/04) 
 
 
 
 
P5.3-15 
 ( )25 3 5 5 3 2 A
3 2 2 3m
i = − = − =+ + 
 
 
P5.3-16 
 ( ) ( )3 33 5 18 5 6
3 (3 3) 3 (3 3)m
v ⎡ ⎤= − = −⎢ ⎥+ + + +⎣ ⎦ 1 A= −
 
 
 
Section 5-4: Thèvenin’s 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 Ω 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 Ω and the open 
circuit voltage, voc = 2 V. 
 
(checked using LNAP 8/15/02) 
 
 
 
 
P5.4-4 
Find Rt: 
 
 
 
( )
( )
12 10 2
6 
12 10 2t
R
+= = Ω+ + 
 
Write mesh equations to find voc: 
 
 
Mesh equations: 
 ( )
( )
1 1 2 1
2 1 2
12 10 6 0
6 3 18
i i i i
i i i 0
+ − − =
− + − = 
 
1 2
2 1
28 6
9 6 1
i i
i i 8
=
− = 
 
1 1
2
136 18 A
2
14 1 7= A
3 2 3
i i
i
= ⇒ =
⎛ ⎞= ⎜ ⎟⎝ ⎠
 
 
Finally, 2 1
7 13 10 3 10 12 V
3 2oc
v i i ⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 
 
(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 3 0
8 4 4
oc oc
oc
v v v−⎛ ⎞ ⎛ ⎞− + + −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ =
 V
 
 
6 2 6 0 2oc oc oc ocv v v v− + + − = ⇒ = − 
 
Find Rt: 
 
We’ll find isc and use it to calculate Rt. Notice that 
the short circuit forces 
 
va = 0 
 
Apply KCL at node a: 
 
6 0 0 3 0 0
8 4 4 sc
i−⎛ ⎞ ⎛ ⎞− + + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ = 
 
6 3 A
8 4sc
i = = 
 
2 8
3 4 3
oc
t
sc
vR
i
−= = = − Ω 
 
 
 
 
 
(checked using LNAP 8/15/02) 
 
 
 
 
 
 
 
 
 
 
P5.4-6 
Find voc: 
 
 
Apply KCL at the top, middle node: 2 3 0 18 V
3 6
a a a
a
v v v v− = + + ⇒ = 
The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V 
 
Find isc: 
 
 
Apply KCL at the top, middle node: 2 3 1
3 6 3
a a a a
a
v v v v v 8 V− = + + ⇒ = − 
Apply Ohm’s law to the right-hand 3 Ω resistor : 18 6 V
3 3
a
sc
vi −= = = − 
Finally: 18 3
6
oc
t
sc
vR
i
= = = −− Ω 
 
 
 
(checked using LNAP 8/15/02) 
 
 
 
 
 
 
 
 
 
P5.4-7 
 
 
(a) 
( )1 21 0s a av R i d R i− + + + = 
( )1 21
s
a
vi
R d R
= + + 
( )
( )
2
1 2
1
1
s
oc
d R v
v
R d R
+= + + 
 
 
 
 
1
s
a
vi
R
= 
( ) ( )
1
1
1 ssc a
d v
i d i
R
+= + = 
 
 
 
2
0Ta a T
vi d i i
R
− − + − = 
1 a TR i v= − 
( ) ( )2 1
1 2 1 2
1
1 T TT T
R d Rv vi d v
R R R R
+ += + + = × 
( )
1 2
1 21
T
t
T
R RvR
i R d R
= = + + 
 
(b) Let R1 = R2 = 1 kΩ. Then 
1000 1000625 2 0.4 A/A
2 625t
R d
d
Ω = = ⇒ = − = −+ 
and 
( )1 0.4 25 5 13.33 V
2 0.4 1
s
oc s
d v
v v
d
+ − += = ⇒ = =+ − + 
 
(checked using LNAP 8/15/02) 
 
 
P5.4-8 
 
 
 
oc
t
Rv v
R R
= + 
From the given data: 
20006
2000 1.2 V
160040002
4000
oc
t oc
t
oc
t
v
R v
R
v
R
⎫= ⎪+ =⎧⎪ ⇒⎬ ⎨ = − Ω⎩⎪= ⎪+ ⎭
 
 
When R = 8000 Ω, 
 
( )8000 1.2 1.5 V
1600 8000
v = =− + 
 
 
 
P5.4-9 
 
 
 
oc
t
vi
R R
= + 
From the given data: 
0.004
2000 24 V
4000
0.003
4000
oc
t oc
toc
t
v
R v
Rv
R
⎫= ⎪+ =⎧⎪ ⇒⎬ ⎨ = Ω⎩⎪= ⎪+ ⎭
 
 
(a) When i = 0.002 A: 
240.002 8000 
4000
R
R
= ⇒ =+ Ω 
 
(b) Maximum i occurs when R = 0: 
24 0.006 6 mA 6 mA
4000
i= = ⇒ ≤ 
 
 
 
P5.4-10 
The currentat 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 0.002 150
3 0 tt
R
R
−− = ⇒ = − Ω− − 
 
P5.4-11 
 
 
 
12 6000 2000 1000 0
 4 3000 A
4 1000 V
3
a a a
a
oc a
i i i
i
v i
− + + +
=
= =
=
 
ia = 0 due to the short circuit 
 
12 6000 0 2 mA
4
3 667
.002
sc sc
oc
t
sc
i i
vR
i
− + = ⇒ =
= = = Ω
 
 
4
3
667b
i
R
= + 
 
ib = 0.002 A requires 
 
4
3 667 0
0.002
R = − = 
 
 
(checked using LNAP 8/15/02) 
 
 
 
P5.4-12 
 
 
 
10 0 10 A
4 2 0
2 20
oc
oc
i i
v i i
v i V
= + ⇒ =
+ − =
⇒ = − = −
 
 
10 10sc si i i i c+ = ⇒ = − 
 
4 0 2 0 0 10 Asci i i i+ − = ⇒ = ⇒ = 
 
20 2
10
oc
t
sc
vR
i
−= = = − Ω 
 
 
 
 
 
202 
2L LL
i R
R
12−− = = ⇒ = Ω− 
 
 
(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 kΩ Ω+ Ω = Ω Ω = Ω
 
 
 
 
 
a
36
12000
i
R
= + and a 3612000
Rv
R
= + 
 
2
a a
36
12000
p i v R
R
⎛ ⎞= = ⎜ ⎟+⎝ ⎠ 
 
(a) a
36 3 mA
0 12000
i = =+ when R = 0 Ω (a short circuit). 
(b) 
5
a 5
10 36 32.14 V
10 12000
v = =+ when R is as large as possible, i.e. R = 100 kΩ. 
(c) Maximum power is delivered to the adjustable resistor when t 12 kR R= = Ω . Then 
 
2
a a
36 12000 0.027 27 mW
12000 12000
p i v ⎛ ⎞= = = =⎜ ⎟+⎝ ⎠ 
 
(checked: LNAP 6/22/04) 
 
 
 
P5.4-14 
Replace the source by it’s Thevenin equivalent circuit to get 
 
oc
o
t LR +R
v
i = 
Using the given formation 
( ) (
oc
t
t t
oc
t
0.375
R 4
 0.375 R 4 0.300 R 8
0.300
R 8
v
v
⎫= ⎪+ ⎪ ⇒ + =⎬⎪= ⎪+ ⎭
)+ 
So 
( ) ( ) ( )t oc0.300 8 0.375 4R 12 and 0.3 12 8 6 V0.075 v
−= = Ω = + = 
(a) When L o
6R 10 , 0.2727 A.
12 10
i= Ω = =+ 
(b) . t12 R 48 11R R 16 Ω = = ⇒ = Ω
(checked: LNAP 5/24/04) 
 
 
 
P5.4-15 
(a) 
 
 
3 2 0.25 Ai i− = 
Apply KVL to mesh 1 to get ( ) ( )1 2 1 320 20 40 0i i i i− + − − = 
 
Apply KVL to the supermesh corresponding to the unspecified resistance to get 
 
 
( ) ( )2 3 1 3 1 240 10 20 20 0i i i i i i+ − − − − = 
Solving, for example using MATLAB, gives 
 
1 1
2 2
3 3
0 1 1 0.25 1.875
40 20 20 40 0.750
40 60 30 0 1.000
i i
i i
i i
⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢− − = ⇒ =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢−⎣ ⎦ ⎣ ⎦ ⎣⎣ ⎦ ⎣ ⎦
⎤⎥⎥⎥⎦
 
 
Apply KVL to mesh 2 to get 
( ) ( ) ( )1 2 22 2 3 1 2
2 3
20 40
40 20 0 30 
i i i
i R i i i i R
i i
− −+ − − − = ⇒ = =− Ω 
(b) 
 
 
oc
20 4040 40 12 V
20 20 10 40
v ⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ − 
 
 
 
 
 
 
t 18 R = Ω 
 
 
 
120.25 30 
18
R
R
= ⇒ = Ω+ 
 
(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 
 
 
oc
32 3020 20 5 4 1 V
32 96 120 30
v = − = − =+ + 
 
 
 
 
 
 
 
( ) ( )t 96 || 32 120 || 30 24 24 48 R = + = + = Ω 
Replacing the part of the circuit to the left of terminals a-b by its Thevenin equivalent circuit 
gives 
 
 
 
o
1 0.0125 A
48 32
i = =+ 12.5 mA= 
 
(checked: LNAP 5/24/04) 
 
 
 
P5.4-18 
Replace the circuit by its Thevenin equivalent circuit: 
 
 
 
m
m
m
5
50
R
v
R
⎛ ⎞= ⎜ ⎟⎜ ⎟+⎝ ⎠
 
 
(a) 
 mi m
m
lim 5 V
R
v v→∞= = 
 
(b) When so m m1000 , 4.763 VR v= Ω =
% error = 5 4.762 100 4.76%
5
− × = 
 
(c) 
m
m m
m
m
5 5
50
0.02 0.98 2450 
5 50
R
R R
R
R
⎛ ⎞− ⎜ ⎟⎜ ⎟+⎝ ⎠≥ ⇒ ≥ ⇒+ ≥ Ω 
(checked: LNAP 6/16/04) 
 
 
P5.4-19 
 
 
 
( )
( )
s oc
a
1 s 2
oc
oc 1 2
a a
2
1
1
v v
i
R v R b
v
v R R b
i bi
R
− ⎫= ⎪ +⎪ ⇒ =⎬ + +⎪+ = ⎪⎭
 
 
 
( ) ( )ssc a
1
1 1
v
i i b b
R
= + = + 
 
( )
( )
( ) ( )
s 2
oc 1 2 1 2
t
ssc 1 2
1
1
1
11
v R b
v R R b R R
R vi Rb
R
+
+ += = = + ++ R b
 
 
(checked: LNAP 7/22/04) 
 
 
 
Section 5-5: Norton’s 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, 
2 sci i= . The controlling current of the CCVS is expressed in terms of the mesh currents as 
1 2 1a si i i i i c= − = − 
Apply KVL to mesh 1 to get 
 
 ( ) ( )1 1 2 1 2 1 23 2 6 10 0 7 4 1i i i i i i i− − + − − = ⇒ − = 0 (1) 
 
Apply KVL to mesh 2 to get 
 
( )2 1 2 1 2 1 115 6 0 6 11 0 6i i i i i i− − = ⇒ − + = ⇒ = 2i 
 
Substituting into equation 1 gives 
 
 2 2 2
117 4 10 1.13 A 1.13 A
6 sc
i i i i⎛ ⎞ − = ⇒ = ⇒ =⎜ ⎟⎝ ⎠ 
 
 
 
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 
T
t
T
vR
i
= 
 
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, . The controlling current of the CCVS is expressed in terms of the mesh 
currents as 
2 Ti i=
1 2 1a Ti i i i i= − = + 
Apply KVL to mesh 1 to get 
 
 ( ) ( )1 1 2 1 2 1 2 1 43 2 6 0 7 4 0 7i i i i i i i i− − + − = ⇒ − = ⇒ = 2i (2) 
 
Apply KVL to mesh 2 to get 
 ( )2 1 2 1 25 6 0 6 11T Ti v i i i i v+ − − = ⇒ − + = − 
 
Substituting for i1 using equation 2 gives 
 
2 2 2
46 11 7.57
7 T T
i i v i⎛ ⎞− + = − ⇒ =⎜ ⎟⎝ ⎠ v− 
Finally, 
2
7.57T T Tt
T T
v v vR
i i i
− −= = = =− Ω 
 
 
 
Figure (b) Calculating the Thevenin resistance, Tt
T
vR
i
= , using mesh equations. 
 
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 afteradding 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, 
. The controlling current of the CCVS is expressed in terms of the mesh currents as 2 0 Ai =
 
 1 2 1 0ai i i i i1= − = − = 
Apply KVL to mesh 1 to get 
 
 
( ) ( ) ( ) ( )1 1 2 1 2 1 1 1
1
3 2 6 10 0 3 2 0 6 0 10 0
10 1.43 A
7
i i i i i i i i
i
− − + − − = ⇒ − − + − − =
⇒ = =
 
Apply KVL to mesh 2 to get 
 
 ( ) ( ) ( )2 1 2 15 6 0 6 6 1.43 8.58 Voc oci v i i v i+ − − = ⇒ = = = 
 
 
Figure (c) Calculating the open circuit voltage, voc, using mesh equations. 
 
As a check, notice that ( )( )7.57 1.13 8.55t sc ocR i v= = ≈ 
 
(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, . The voltage at node 3 is equal to the voltage across a short, 1 24 Vv = − 3 0v = . 
The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. . The 
voltage at node 3 is equal to the voltage across a short, i.e. 
2av v=
3 0v = . 
 Apply KCL at node 2 to get 
 
1 2 2 3
1 3 22 3 48 3 16 V3 6 a a
v v v v
v v v v v
− −= ⇒ + = ⇒ − = ⇒ = − 
 
Apply KCL at node 3 to get 
 
( )2 3 24 9 9 16 24 A6 3 6 6sc a sc sc
v v
v i v i i
− + = ⇒ = ⇒ = − = − 
 
 
 
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 
T
th
T
vR
i
= 
 
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. . The 
controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. . The 
voltage at node 3 is equal to the voltage across the current source, i.e. 
1 0v =
2av v=
3 Tv v= . 
 Apply KCL at node 2 to get 
 
1 2 2 3
1 3 22 33 6 T a
v v v v
v v v v v
− −= ⇒ + = ⇒ = 3 
 
Apply KCL at node 3 to get 
 
2 3
2 2 3
4 0 9 6 0
6 3
9 6 0
3 6 0 2
T T
a T T
T T T T
v v
v i v v i
v v i
v v i v i
− + + = ⇒ − + =
⇒ − + =
⇒ − + = ⇒ = −6 T
 
Finally, 
3Tt
T
vR
i
= = − Ω 
 
 
 
Figure (b) Calculating the Thevenin resistance, Tth
T
vR
i
= , using mesh equations. 
 
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, . The controlling voltage of the VCCS, va, is equal to the node voltage 
at node 2, i.e. . The voltage at node 3 is equal to the open circuit voltage, i.e. . 
1 24 Vv = −
2av v= 3 ocv v=
 Apply KCL at node 2 to get 
 
1 2 2 3
1 3 22 3 48 33 6 oc a
v v v v
v v v v v
− −= ⇒ + = ⇒ − + = 
 
 
Apply KCL at node 3 to get 
 
2 3
2 2 3
4 0 9 0 9
6 3 a o
v v
v v v v
− + = ⇒ − = ⇒ = cv 
 
Combining these equations gives 
 
( )3 48 9 72 Voc a oc ocv v v v− + = = ⇒ = 
 
 
 
 
Figure (c) Calculating the open circuit voltage, voc, using node equations. 
 
As a check, notice that 
 ( )( )3 24 72th sc ocR I V= − − = = 
 
(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 b 0 Ai = . Then 
 ( )oc b25 5000 25 Vv i= + = 
 
Apply KVL to the supermesh corresponding to 
the dependent source to get 
 ( )b b b5000 10000 3 25 0 1 mAi i i− + − = ⇒ = 
 
Apply KCL to get 
sc b3 3 mi i A= = 
 
Then 
oc
t
sc
8.33 k
v
R
i
= = Ω 
Current division gives 
 
83330.5 3 41.67 k
8333
R
R
= ⇒ =+ Ω 
 
(b) 
 
Notice that bi and 0.5 mA are the mesh currents. 
Apply KCL at the top node of the dependent 
source to get 
3
b b b
10.5 10 4 mA
6
i i i−+ × = ⇒ = 
Apply KVL to the supermesh corresponding to 
the dependent source to get 
 
( )( )3b5000 10000 0.5 10 25 0i R −− + + × − = 
( )( )3 315000 10 10000 0.5 10 256 R− −⎛ ⎞− × + + × =⎜ ⎟⎝ ⎠
3
125
6 41.67 k
0.5 10
R −= = Ω× 
 
 
 
 
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 
 
2
20.4 0.8 0
10 2 5
v v vv= + ⇒ + − = 
so 
.2 1.8 0.8, -1.0 V
2
v − ±= = 
 
Choosing the positive value of v, 
20.8 0.32 A
2
i = = . Choosing the negative value of v, 
21 0.5
2
i −= = . There are two solutions to this problem. Linear circuits are so much simpler than 
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: 
 
( ) x x10 2 3 15 0 1 Ai i+ + − = ⇒ = 
 
Then 
oc x3 3v i V= = 
 
 
 
Express the controlling current of the 
CCVS in terms of the mesh currents: 
 
x 1 si i i c= − 
The mesh equations are 
 ( ) ( )1 1 sc 1 sc 1 sc10 2 3 15 0 15 5 15i i i i i i i+ − + − − = ⇒ − = 
and 
( )sc 1 sc 1 sc43 0 3i i i i− − = ⇒ = i 
so 
 
sc sc sc
415 5 15 1 A
3
i i i⎛ ⎞ − = ⇒ =⎜ ⎟⎝ ⎠ 
The Thevenin resistance is 
t
3 3 
1
R = = Ω 
Finally, the Norton equivalent circuit is 
 
 
(checked: LNAP 6/21/04) 
 
 
P5.5-9 
Identify the open circuit voltage and short circuit current. 
 
 
1
1 3 1 V
3
v ⎛ ⎞= =⎜ ⎟⎝ ⎠ 
 
Then 
 ( )oc 1 14 2.5 9 Vv v v= − = − 
 
1 sc
13 1
3
v i⎛ ⎞= − = −⎜ ⎟⎝ ⎠ sc3 i 
 ( )1 sc sc 1
1 sc
4 2.5 5 0
9 9
v i i v
v i
+ + − =
⇒ + = 0 
( )sc sc sc 19 1 3 9 0 A2i i i− + = ⇒ = 
The Thevenin resistance is 
t
9 18 
0.5
R −= = − Ω 
Finally, the Norton equivalent circuit is 
 
 
 
(checked: LNAP 6/21/04) 
 
 
P5.5-10 
Replace the circuit by its Norton equivalent circuit: 
 
 
( )3m
m
1600 1.5 10
1600
i
R
−⎛ ⎞= ×⎜ ⎟⎜ ⎟+⎝⎠
 
(a) 
m
mi m0
lim 1.5 mA
R
i i→= = 
 
(b) When Rm = 20 Ω then so m 1.48 mAi =
 
1.5 1.48% error 100 1.23%
1.5
−= × = 
(c) 
( )
m
m
m
16000.015 0.015
1600 16000.02 0.98 32.65
0.015 1600
R
R
R
⎛ ⎞− ⎜ ⎟⎜ ⎟+⎝ ⎠≥ ⇒ ≥ ⇒+ ≤ Ω 
 
(checked: LNAP 6/18/04) 
 
P5.5-11 
 
 
2 12
3 A
6
2 6 V
a
a a
oc a
i
i i
v i
−= ⇒ =
= = −
−
 
 
 
 
( )
12 6 2 3 A
23 2 3 2
3
a a a
sc a sc
i i i
i i i
+ = ⇒ = −
= ⇒ = − = − A 
 
6 3
2t
R −= = Ω− 
 
P5.5-12 
 
 
( )
12 24 12 24 8 
12 24 36
24 30 20 V
12 24
t
oc
R
v 
× ×= = =+
= =+
Ω
 
 
 
 
 
20
8
i
R
= + 
 
 
 
Section 5-6: Maximum Power Transfer 
 
P5.6-1 
 
a) For maximum power transfer, set RL equal 
to the Thevenin resistance: 
 
100 1 101L tR R= = + = Ω 
 
b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin 
equivalent circuit: 
 
 
 
The voltage across RL is ( )101 100 50 V
101 101L
v = =+ 
Then 
2 2
max
50 24.75 W
101
L
L
vp
R
= = = 
 
 
 
P5.6-2 
Reduce the circuit using source transformations: 
 
 
 
 
 
Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value 
of that maximum power is 
2 2( ) (0.03) (60) 54 mWmax RP i R= = = 
 
 
P5.6-3 
 
 
 
L
L S
S L
2 2
L S L
L 2
L S L
 
 
( )
R
v v
R R
v v R
p
R R R
⎡ ⎤= ⎢ ⎥+⎢ ⎥⎣ ⎦
∴ = = +
 
 
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 Ω resistor is zero because of the short circuit. Consequently, isc = 10 ix. 
Apply KCL at the top-left node to get 
 
0.90.9 10 0.1 A
9x x x
i i i+ = ⇒ = = 
so 
isc = 10 ix = 1A 
Next 
 
 
 
Apply KCL at the top-left node to get 
0.90.9 10 0.1 A
9x x x
i i i+ = ⇒ = = 
 
Apply Ohm’s law to the 3 Ω resistor to get 
 
( ) ( )3 10 30 0.1 3 Voc xv i= = = 
 
For maximum power transfer to RL: 
3 3
1
oc
L t
sc
vR R
i
= = = = Ω 
 
The maximum power delivered to RL is given by 
 
( )
2 2
max
3 3 W
4 4 3 4
oc
t
vp
R
= = = 
 
 
 
 
P5.6-5 
 
 
 
 
 
 
The required value of R is 
 
( ) ( )
( ) ( )
20 120 10 50
8 50
( ) ( )
20 120 10 50t
R R
+ += = + = Ω+ + + 
 
170 3020 10 20 50
170 30 170 30
170(20)(10) 30(20)(50) 4000 20 V
200 200
ocv
⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦
−= = =
 
 
The maximum power is given by 
( )
2 2
max
20 2 W
4 4 50
oc
t
vp
R
= = = 
 
 
 
P5.6-6 
 
 
L s
o L
A
R R
i v= + 
 
( )
2 2
s L2
L L L 2
o L
A R
P R
R R
v
i= =
+
 
 
(a) so maximizes the power delivered to the load. The corresponding 
load power is 
t oR =R L oR =R 10 = Ω
( )
2
2
L 2
120 10
2P 2
10 10
⎛ ⎞⎜ ⎟⎝ ⎠= =+ .5 W . 
 
(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 is 
 
2
2
2 2 2 2
s L s
L 2
L L
120A R A 2P 12.5 W.
R R 8
v v
⎛ ⎞⎜ ⎟⎝ ⎠= = = = 
 
(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 
 
 
( )
2
2
2
1A 8
15215 A 36 49.3 V.
818
⎛ ⎞⎜ ⎟⎝ ⎠= ⇒ = = 
 
(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: 
 
 
 
oc1
150 10 4.545 V
150 180
v ⎛ ⎞= =⎜ ⎟+⎝ ⎠ 
 
t1 180 150 81.8 R = = Ω& 
 
Next, replace the right part of the circuit by its Thevenin equivalent: 
 
 
 
oc2
470 20 15.932 V
470 120
v ⎛ ⎞= =⎜ ⎟+⎝ ⎠ 
 
t2 120 470 95.6 R = = Ω& 
 
Now, combine the two partial Thevenin equivalents: 
 
oc oc1 oc2 t t1 t210.387 V and 177.4 v v v R R R= − = − = + = Ω 
 
So 
 
The power received by the adjustable resistor 
will be maximum when R = Rt = 177.4 Ω. The 
maximum power received by the adjustable 
resistor will be ( )( )
211.387
0.183 W
4 177.4 
p
−= =Ω . 
 
 
(checked LNAPDC 7/24/04) 
 
P5.6-8 
( ) ( )2
10010 10L L
t L t L t L
R R
p i v
R R R R R R
⎛ ⎞ ⎡ ⎤= = =⎜ ⎟ ⎢ ⎥⎜ ⎟+ +⎢ ⎥ +⎝ ⎠ ⎣ ⎦
 
 
The power increases as Rt decreases so choose Rt = 1 Ω. Then 
 
( )
( )max 2
100 5
13.9 W
1 5
p i v= = =+ 
 
 
P5.6-9 
From the plot, the maximum power is 5 W when R = 20 Ω. Therefore: 
 
Rt = 20 Ω 
and 
( )2max max 4 5 4 20 204oc oc tt
vp v p R
R
= ⇒ = = = 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: 
 
0.0972
0 39.9 V
410
0.0438
500
oc
t oc
toc
t
v
R v
Rv
R
⎫= ⎪+ =⎧⎪ ⇒⎬ ⎨ = Ω⎩⎪= ⎪+ ⎭
 
 
Now check the third line of the table. When R= 5000 Ω: 
39.9 7.37 mA
410 5000
oc
t
vi
R R
= = =+ + 
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: 
 12 V (line 1 of the table)
 3 mA (line 3 of the table)
so 4 k
oc
sc
oc
t
sc
v
i
vR
i
=
=
= = Ω
 
 
Next, check line 2 of the table. When R = 10 kΩ: 
( ) ( )3 3
12 0.857 mA
10 10 5 10
oc
t
vi
R R
= = =+ + 
which agrees with the data in the table. 
To cause i = 1 mA requires ( )3
120.001 8000 
10 10
oc
t
vi R
R R R
= = = ⇒ =+ + Ω 
I agree with my lab partner’s claim that R = 8000 causes i = 1 mA. 
 
 
 
P5.8-3 
 
 
1 1 1 1 11 6
2 3 6 1tt 1
RR
R R R R R
= + + = ⇒ = 
and 
2 3 3 4 6 5 18030 20 10
3 2 3 2 3 4 1 6 5 11oc
v
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= 
 
so the prelab calculation isn’t correct. 
But then 
( )
180 180
11 11 163 mA 54.5 mA6 60 40110 40
11
oc
t
vi
R R
= = = = ≠+ ++
 
 
so the measurement does not agree with the corrected prelab calculation. 
 
 
P5.8-4 
( )6000 3000 500 1500 2000 2000 1000 + = =& & & Ω 
12 12 12 mA
1000 1000
i
R
= ≤ =+ 
 
How about that?! Your lab partner is right. 
 
(checked using LNAP 6/21/05) 
 
 
P5.8-5 
(a) 
 
KVL gives ( )oc tv R R= + i 
from row 2 ( )( )oc t 10 1.333v R= + 
from row 3 ( )( )oc t 20 0.857v R= + 
So ( )( ) ( )( )t t10 1.333 20 0.857R R+ = + 
 ( ) ( )t t28 10 18 20R R+ = + 
Solving gives 
t t10 360 280 80 8 R R= − = ⇒ = Ω 
and 
( )( )oc 8 10 1.333 24 Vv = + = 
 
(b) 
oc
oc
t t
24 24 and 
8 8
v R Ri v
R R R R R R
= = = =+ + + +v 
 
 When R = 0, i = 3 A, and v = 0 V. 
 When R = 40 Ω, 1 A
2
i = . 
 When R = 80 Ω, ( )24 80 240 21.82
88 11
v = = = . 
These are the values given in the tabulated data so the data is consistent. 
(c) When R = 40 Ω, ( )24 40 20 V
48
v = = . 
 When R = 80 Ω, 24 0.2727 A
88
i = = . 
(d) First ( )t 1 18 24 18 12 24 R R R= = + ⇒ = Ω& & 
the, using superposition, 
 
( )( ) ( )( )oc 1 s s s1
2424 12 24 18 12 8 8 2 A
24 18 12
v R i i
R
= = + + = + ⇒ =+ + && i 
(checked using LNAP 6/21/05) 
 
 
 
 
	CH5sec2
	Problems 
	Section 5-2: Source Transformations 
	P5.2-1 
	P5.2-2 
	P5.2-3 
	P5.2-4 
	 P5.2-5 
	P5.2-6 
	P5.2-7 
	P5.2-8
	P5.2-9 
	CH5sec3
	Section 5-3 Superposition 
	P5.3–1
	P5.3-2
	P5.3–3 
	P5.3–4
	P5.3-5
	P5.3-6P5.3-7
	P5.3-8
	P5.3-9 
	P5.3-10 
	P5.3-11 
	P5.3-12 
	P5.3-13 
	P5.3-14 
	P5.3-15 
	P5.3-16 
	CH5sec4
	Section 5-4: Thèvenin’s Theorem 
	P5.4-1 
	P5.4-2 
	P5.4-3 
	P5.4-4 
	P5.4-5
	P5.4-6 
	P5.4-7
	P5.4-8 
	P5.4-9
	P5.4-10 
	P5.4-11
	P5.4-12
	P5.4-13 
	P5.4-14 
	P5.4-15 
	P5.4-16 
	P5.4-18 
	P5.4-19
	CH5sec5
	Section 5-5: Norton’s Theorem 
	P5.5-1 
	P5.5-2 
	P5.5-3 
	P5.5-4 
	P5.5-5 
	P5.5-6 
	P5.5-7 
	P5.5-8 
	P5.5-9 
	P5.5-10 
	P5.5-11
	P5.5-12
	CH5sec6
	Section 5-6: Maximum Power Transfer 
	P5.6-1
	P5.6-2 
	P5.6-3
	P5.6-4 
	P5.6-5
	P5.6-6 
	P5.6-7 
	P5.6-8 
	P5.6-9 
	CH5sec8
	Section 5-8 How Can We Check…? 
	P5.8-1
	P5.8-2
	P5.8-3
	P5.8-4 
	P5.8-5