Solucionario circuitos eléctricos - dorf, svoboda - 6ed

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LIBROS UNIVERISTARIOS
Y SOLUCIONARIOS DE
MUCHOS DE ESTOS LIBROS

LOS SOLUCIONARIOS
CONTIENEN TODOS LOS
EJERCICIOS DEL LIBRO
RESUELTOS Y EXPLICADOS
DE FORMA CLARA

VISITANOS PARA
DESARGALOS GRATIS.

Solution Manual

to accompany

Introduction to Electric Circuits, 6e

By R. C. Dorf and J. A. Svoboda

1

Table of Contents

Chapter 1 Electric Circuit Variables

Chapter 2 Circuit Elements

Chapter 3 Resistive Circuits

Chapter 4 Methods of Analysis of Resistive Circuits

Chapter 5 Circuit Theorems

Chapter 6 The Operational Amplifier

Chapter 7 Energy Storage Elements

Chapter 8 The Complete Response of RL and RC Circuits

Chapter 9 The Complete Response of Circuits with Two Energy Storage Elements

Chapter 10 Sinusoidal Steady-State Analysis

Chapter 11 AC Steady-State Power

Chapter 12 Three-Phase Circuits

Chapter 13 Frequency Response

Chapter 14 The Laplace Transform

Chapter 15 Fourier Series and Fourier Transform

Chapter 16 Filter Circuits

Chapter 17 Two-Port and Three-Port Networks

2

Errata for Introduction to Electric Circuits, 6th Edition

Page 18, voltage reference direction should be + on the right in part B:

Page 28, caption for Figure 2.3-1: "current" instead of "cuurent"

Page 41, line 2: "voltage or current" instead of "voltage or circuit"

Page 41, Figure 2.8-1 b: the short circuit is drawn as an open circuit.

Page 42, line 11: "Each dependent source ..." instead of "Each dependent sources..."

Page 164, Table 5.5-1: method 2, part c, one should insert the phrase "Zero all independent sources,
then" between the "(c)" and "Connect a 1-A source. . ." The edited phrase will read:

"Zero all independent sources, then connect a 1-A source from terminal b to terminal a. Determine Vab.
Then Rt = Vab/1."

Page 340, Problem P8.3-5: The answer should be .

Page 340, Problem P8.3-6: The answer should be .

Page 341, Problem P.8.4-1: The answer should be

Page 546, line 4: The angle is instead of .

Page 554, Problem 12.4.1 Missing parenthesis:

Page 687, Equation 15.5-2: Partial t in exponent:

http://www.clarkson.edu/~svoboda/errata/6th.html (1 of 2)5/10/2004 7:41:43 PM


Page 757, Problem 16.5-7: Hb(s) = V2(s) / V1(s) and Hc(s) = V2(s) / Vs(s) instead of Hb(s) = V1(s) / V2
(s) and Hc(s) = V1(s) / Vs(s).

http://www.clarkson.edu/~svoboda/errata/6th.html (2 of 2)5/10/2004 7:41:43 PM

Chapter 1 – Electric Circuit Variables
Exercises
Ex. 1.3-1
i (t ) = 8 t 2 − 4 t A
t t 8 t 8
q(t ) = ∫ 0
i dτ + q(0) = ∫ 0
(8τ 2 − 4τ ) dτ + 0 = τ 3 −2τ 2 = t 3 − 2 t 2 C
3 0 3
Ex. 1.3-3
t t 4 4 4
q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4sin 3τ dτ + 0 = −
t
cos 3τ 0 = − cos 3 t + C
0 0 3 3 3

Ex. 1.3-4

0 t <0
dq ( t ) 
i (t ) = i (t ) = 2 0< t < 2
dt  −2( t − 2 )
−2e t >2

Ex. 1.4-1
i1 = 45 µA = 45 × 10-6 A < i2 = 0.03 mA = .03 × 10-3 A = 3 × 10-5 A < i3 = 25 × 10-4 A

Ex. 1.4-2
∆ q = i∆ t = ( 4000 A )( 0.001 s ) = 4 C

Ex. 1.4-3
∆ q 45 × 10−9
i= = −3
= 9 × 10−6 = 9 µA
∆t 5 × 10

Ex. 1.4-4
 electron   −19 C   9 electron   −19 C 
i = 10 billion
 s  1.602 ×10 electron  =
  10×10
 s  1.602 × 10 electron 
 
electron C
= 1010 × 1.602 ×10−19
s electron
C
= 1.602 × 10−9 = 1.602 nA
s

1-1

Ex. 1.6-1
(a) The element voltage and current do not adhere to the passive convention in
Figures 1.6-1B and 1.6-1C so the product of the element voltage and current
is the power supplied by these elements.
(b) The element voltage and current adhere to the passive convention in Figures
1.6-1A and 1.6-1D so the product of the element voltage and current is the
power delivered to, or absorbed by these elements.
(c) The element voltage and current do not adhere to the passive convention in
Figure 1.6-1B, so the product of the element voltage and current is the power
delivered by this element: (2 V)(6 A) = 12 W. The power received by the
element is the negative of the power delivered by the element, -12 W.
(d) The element voltage and current do not adhere to the passive convention in
Figure 1.6-1B, so the product of the element voltage and current is the power
supplied by this element: (2 V)(6 A) = 12 W.
(e) The element voltage and current adhere to the passive convention in Figure
1.6-1D, so the product of the element voltage and current is the power
delivered to this element: (2 V)(6 A) = 12 W. The power supplied by the
element is the negative of the power delivered to the element, -12 W.

Problems
Section 1-3 Electric Circuits and Current Flow

P1.3-1
d
i (t ) =
dt
( )
4 1 − e −5t = 20 e −5t A

P1.3-2
4 4
( )
t t t t
q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4 1 − e −5τ dτ + 0 = ∫ 4 dτ − ∫ 4 e−5τ dτ = 4 t + e−5t − C
0 0 0 0 5 5

P1.3-3
t t
q ( t ) = ∫ i (τ ) dτ = ∫ 0 dτ = 0 C for t ≤ 2 so q(2) = 0.
−∞ −∞
t t
q ( t ) = ∫ i (τ ) dτ + q ( 2 ) = ∫ 2 dτ = 2 τ 2 = 2 t − 4 C for 2 ≤ t ≤ 4. In particular, q(4) = 4 C.
t
2 2
t t
q ( t ) = ∫ i (τ ) dτ + q ( 4 ) = ∫ −1 dτ + 4 = − τ 4 + 4 = 8 − t C for 4 ≤ t ≤ 8. In particular, q(8) = 0 C.
t
4 4
t t
q ( t ) = ∫ i (τ ) dτ + q ( 8 ) = ∫ 0 dτ + 0 = 0 C for 8 ≤ t .
8 8

1-2

P1.3-4
C
i = 600 A = 600
s
C s mg
Silver deposited = 600 ×20 min×60 ×1.118 = 8.05×105 mg=805 g
s min C

Section 1-6 Power and Energy

P1.6-1
a.) q = ∫ i dt = i∆t = (10 A )( 2 hrs )( 3600s/hr ) = 7.2×10
4
C
b.) P = v i = (110 V )(10 A ) = 1100 W
0.06$
c.) Cost = × 1.1kW × 2 hrs = 0.132 $
kWhr

P1.6-2
P = ( 6 V )(10 mA ) = 0.06 W
∆w 200 W⋅s
∆t = = = 3.33×103 s
P 0.06 W

P1.6-3
30
for 0 ≤ t ≤ 10 s: v = 30 V and i = t = 2t A ∴ P = 30(2t ) = 60t W
15
25
for 10 ≤ t ≤ 15 s: v ( t ) = − t + b ⇒ v (10 ) = 30 V ⇒ b = 80 V
5
v(t ) = −5t + 80 and i (t ) = 2t A ⇒ P = ( 2t )( −5t +80 ) = −10t 2 +160t W
30
for 15 ≤ t ≤ 25 s: v = 5 V and i (t ) = − t +b A
10
i (25) = 0 ⇒ b = 75 ⇒ i (t ) = −3t + 75 A
∴ P = ( 5 )( −3t + 75 ) = −15t + 375 W

1-3

60t dt + ∫10 (160t −10t 2 ) dt + ∫15 ( 375−15t ) dt
10 15 25
Energy = ∫ P dt = ∫0
15 25
+ 80t 2 − 10 t 3 + 375t − 15 t 2 = 5833.3 J
10
= 30t 2
0 3 10 2 15

P1.6-4
a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully
charged).
5( 3600 )
t 5 ( 3600 )  0.5 τ  0.5 2
w = ∫ Pdt = ∫0 vi dτ = ∫0 2 11 +  dτ = 22 t + 3600 τ
 3600  0

= 441× 103 J = 441 kJ

1 hr 10¢
b.) Cost = 441kJ × × = 1.23¢
3600s kWhr

P1.6-5
1 1
p (t ) = ( cos 3 t )( sin 3 t ) = sin 6 t
3 6
1
p ( 0.5 ) = sin 3 = 0.0235 W
6
1
p (1) = sin 6 = −0.0466 W
6

1-4

Here is a MATLAB program to plot p(t):

clear

t0=0; % initial time
tf=2; % final time
dt=0.02; % time increment
t=t0:dt:tf; % time

v=4*cos(3*t); % device voltage
i=(1/12)*sin(3*t); % device current

for k=1:length(t)
p(k)=v(k)*i(k); % power
end

plot(t,p)
xlabel('time, s');
ylabel('power, W')

P1.6-6
p ( t ) = 16 ( sin 3 t )( sin 3 t ) = 8 ( cos 0 − cos 6 t ) = 8 − 8cos 6 t W


clear

tf=2; % final time
t=t0:dt:tf; % time

v=8*sin(3*t); % device voltage
i=2*sin(3*t); % device current

for k=1:length(t)
end

plot(t,p)
xlabel('time, s');
ylabel('power, W')

1-5

P1.6-7
( ) ( )
p ( t ) = 4 1 − e −2 t × 2 e −2 t = 8 1 − e−2 t e −2t


clear

tf=2; % final time
t=t0:dt:tf; % time

v=4*(1-exp(-2*t)); % device voltage
i=2*exp(-2*t); % device current

for k=1:length(t)
end

plot(t,p)
xlabel('time, s');
ylabel('power, W')

P1.6-8
P = V I =3 × 0.2=0.6 W
w = P ⋅ t = 0.6 × 5 × 60=180 J

1-6

Verification Problems

VP 1-1
Notice that the element voltage and current of each branch adhere to the passive convention. The
sum of the powers absorbed by each branch are:

(-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 10 W + 9 W -20 W + 5 W
=0W
The element voltages and currents satisfy conservation of energy and may be correct.

VP 1-2
Notice that the element voltage and current of some branches do not adhere to the passive
convention. The sum of the powers absorbed by each branch are:

-(3 V)(3 A)+(3 V)(2 A)+ (3 V)(2 A)+(4 V)(3 A)+(-3 V)(-3 A)+(4 V)(-3 A)
= -9 W + 6 W + 6 W + 12 W + 9 W -12 W
≠0W

The element voltages and currents do not satisfy conservation of energy and cannot be correct.

Design Problems

DP 1-1
The voltage may be as large as 20(1.25) = 25 V and the current may be as large as (0.008)(1.25)
= 0.01 A. The element needs to be able to absorb (25 V)(0.01 A) = 0.25 W continuously. A
Grade B element is adequate, but without margin for error. Specify a Grade B device if you trust
the estimates of the maximum voltage and current and a Grade A device otherwise.

1-7

DP1-2
( ) ( )
p ( t ) = 20 1 − e −8 t × 0.03 e −8 t = 0.6 1 − e−8t e−8t


clear

tf=1; % final time
t=t0:dt:tf; % time

v=20*(1-exp(-8*t)); % device voltage
i=.030*exp(-8*t); % device current

for k=1:length(t)
end

plot(t,p)
xlabel('time, s');
ylabel('power, W')

Here is the plot:

The circuit element must be able to absorb 0.15 W.

1-8

Chapter 2 - Circuit Elements
Exercises
Ex. 2.3-1
m ( i1 + i 2 ) = mi1 + mi 2 ⇒ superposition is satisfied
m ( a i1 ) = a ( mi1 ) ⇒ homogeneity is satisfied
Therefore the element is linear.

Ex. 2.3-2
m ( i1 + i 2 ) + b = mi1 + mi 2 + b ≠ ( mi1 + b ) + ( mi 2 + b ) ⇒ superposition is not satisfied
Therefore the element is not linear.

Ex. 2.5-1
v 2 (10 )
2

P= = =1 W
R 100
Ex. 2.5-2
v 2 (10 cos t ) 2
P= = = 10 cos 2 t W
R 10
Ex. 2.8-1

ic = − 1.2 A, v d = 24 V
id = 4 ( − 1.2) = − 4.8 A
id and vd adhere to the passive convention so
P = vd id = (24) (−4.8) = −115.2 W
is the power received by the dependent source

2-1

Ex. 2.8-2

vc = −2 V, id = 4 vc = −8 A and vd = 2.2 V
P = vd id = (2.2) (−8) = −17.6 W
is the power received by the dependent source. The power supplied by the
dependent source is 17.6 W.

Ex. 2.8-3

ic = 1.25 A, vd = 2 ic = 2.5 V and id = 1.75 A
P = vd id = (2.5) (1.75) = 4.375 W
is the power received by the dependent source.

2-2

Ex. 2.9-1

θ = 45° , I = 2 mA, R p = 20 kΩ
θ 45
a= ⇒ aR = (20 kΩ) = 2.5 kΩ
360 p 360

vm = (2 ×10−3 )(2.5 ×103 ) = 5 V

Ex. 2.9-2
µA
v = 10 V, i = 280 µA, k = 1 for AD590
°K

i  °K 
i = kT ⇒ T = = (280µA)1  = 280° K
k  µA 
 

Ex. 2.10-1
At t = 4 s both switches are open, so i = 0 A.

Ex. 2.10.2
At t = 4 s the switch is in the up position, so v = i R = (2 mA)(3 kΩ) = 6V .

At t = 6 s the switch is in the down position, so v = 0 V.

Problems
Section 2-3 Engineering and Linear Models

P2.3-1
The element is not linear. For example, doubling the current from 2 A to 4 A does not double the
voltage. Hence, the property of homogeneity is not satisfied.

P2.3-2
(a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and
the line passes through the origin so the equation of the line is v = 0.12 i . The element is indeed
linear.
(b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV
4
(c) When v = 4 V, i = = 33 A = 33 A.
0.12

2-3

P2.3-3
(a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and
the line passes through the origin so the equation of the line is v = 256.5i . The element is indeed
linear.
(b) When i = 4 mA, v = (256.5 V/A)×(4 mA) = (256.5 V/A)×(0.004 A) = 1.026 V
12
(c) When v = 12 V, i = = 0.04678 A = 46.78 mA.
256.5

P2.3-4
Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence,
the property of homogeneity is not satisfied. The element is not linear.

Section 2-5 Resistors

P2.5-1
i = is = 3 A and v = Ri = 7 × 3 = 21 V
v and i adhere to the passive convention
∴ P = v i = 21 × 3 = 63 W
is the power absorbed by the resistor.

P2.5-2
i = is = 3 mA and v = 24 V
v 24
R = = = 8000 = 8 k Ω
i .003
P = (3×10 −3 )× 24 = 72×10 −3 = 72 mW

P2.5-3
v = vs =10 V and R = 5 Ω
v 10
i = = =2 A
R 5
v and i adhere to the passive convention
∴ p = v i = 2⋅10 = 20 W
is the power absorbed by the resistor

2-4

P2.5-4
v = vs = 24 V and i = 2 A
v 24
R= = = 12 Ω
i 2
p = vi = 24⋅2 = 48 W

P2.5-5
v1 = v 2 = vs = 150 V;
R1 = 50 Ω; R2 = 25 Ω
v 1 and i1 adhere to the passive convention so
v 1 150
i1 = = =3 A
R 1 50
v 150
v2 and i 2 do not adhere to the passive convention so i 2 = − 2 = − = −6 A
R2 25
The power absorbed by R1 is P = v1 i1 = 150 ⋅ 3 = 450 W
1

The power absorbed by R 2 is P 2 = − v 2i 2 = −150(−6) = 900 W

P2.5-6
i1 = i 2 = is = 2 A ;
R1 =4 Ω and R2 = 8 Ω
v 1 and i 1 do not adhere to the passive convention so
v 1 =− R 1 i 1 =−4⋅2=−8 V.
The power absorbed by R 1 is
P1 =−v 1i 1 =−(−8)(2) = 16 W.

v2 and i 2 do adhere to the passive convention so v2 = R 2 i 2 = 8 ⋅ 2 = 16 V .
The power absorbed by R 2 is P 2 = v 2i 2 = 16 ⋅ 2 = 32 W.

P2.5-7
Model the heater as a resistor, then
v2 v2 (250) 2
with a 250 V source: P = ⇒ R = = = 62.5 Ω
R P 1000
v 2 (210) 2
with a 210 V source: P = = = 705.6 W
R 62.5

2-5

P2.5-8
P 5000 125
The current required by the mine lights is: i = = = A
v 120 3
Power loss in the wire is : i 2 R
Thus the maximum resistance of the copper wire allowed is
0.05P 0.05×5000
R= = = 0.144 Ω
i2 (125/3) 2
now since the length of the wire is L = 2×100 = 200 m = 20,000 cm
thus R = ρ L / A with ρ = 1.7×10−6 Ω⋅ cm from Table 2.5−1
ρL 1.7×10−6 ×20,000
A= = = 0.236 cm 2
R 0.144

Section 2-6 Independent Sources

P2.6-1
v s 15
= 3 A and P = R i 2 = 5 ( 3 ) = 45 W
2
(a) i = =
R 5
(b) i and P do not depend on is .
The values of i and P are 3 A and 45 W, both when i s = 3 A and when i s = 5 A.

P2.6-2
v 2 102
(a) v = R i s = 5 ⋅ 2 = 10 V and P = = = 20 W
R 5
(b) v and P do not depend on v s .
The values of v and P are 10V and 20 W both when v s = 10 V and when v s = 5 V

2-6

P2.6-3
Consider the current source:
i s and v s do not adhere to the passive convention,
so Pcs =i s v s =3⋅12 = 36 W
is the power supplied by the current source.

Consider the voltage source:
i s and v s do adhere to the passive convention,
so Pvs = i s vs =3 ⋅12 = 36 W
is the power absorbed by the voltage source.
∴ The voltage source supplies −36 W.

P2.6-4
Consider the current source:
i s and vs adhere to the passive convention
so Pcs = i s vs =3 ⋅12 = 36 W
is the power absorbed by the current source.
Current source supplies − 36 W.

Consider the voltage source:
i s and vs do not adhere to the passive convention
so Pvs = i s vs = 3 ⋅12 =36 W
is the power supplied by the voltage source.

P2.6-5
(a) P = v i = (2 cos t ) (10 cos t ) = 20 cos 2 t mW
1
1 1 1 1 
(b) w = ∫0 P dt = ∫0 20 cos t dt = 20 t + sin 2t  = 10 + 5 sin 2 mJ
2

2 4 0

2-7

Section 2-7 Voltmeters and Ammeters

P2.7-1
v 5
(a) R = = = 10 Ω
i 0.5

(b) The voltage, 12 V, and the
current, 0.5 A, of the voltage
source adhere to the passive
convention so the power

P = 12 (0.5) = 6 W

is the power received by the
source. The voltage source
delivers -6 W.

P2.7-2
The voltmeter current is zero
so the ammeter current is
equal to the current source
current except for the
reference direction:

i = -2 A

The voltage v is the voltage of
the current source. The power
supplied by the current source
is 40 W so

40 = 2 v ⇒ v = 20 V

2-8

Section 2-8 Dependent Sources

P2.8-1

vb 8
r = = =4 Ω
ia 2

P2.8-2
ia 2 A
vb = 8 V ; g v b = i a = 2 A ; g = = = 0.25
vb 8 V

P2.8-3
i a 32 A
i b = 8 A ; d i b = i a = 32A ; d = = =4
ib 8 A

P2.8-4
vb 8 V
va = 2 V ; b va = vb = 8 V ; b = = =4
va 2 V

Section 2-9 Transducers

P2.9-1
θ 360 vm
a= , θ =
360 Rp I
(360)(23V)
θ = = 75.27°
(100 kΩ)(1.1 mA)

P2.9-2
µA
AD590 : k =1 °
,
K
v =20 V (voltage condition satisfied)

4 µ A < i < 13 µ A 

i  ⇒ 4 ° K< T <13° K
T = 
k 

2-9

Section 2-10 Switches

P2.10-1
At t = 1 s the left switch is open and the
right switch is closed so the voltage
across the resistor is 10 V.

v 10
i= = = 2 mA
R 5×103

At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is
15 V.
v 15
i= = = 3 mA
R 5×103

P2.10-2
At t = 1 s the current in the resistor
is 3 mA so v = 15 V.

At t = 4 s the current in the resistor
is 0 A so v = 0 V.


VP2-1
vo =40 V and i s = − (−2) = 2 A. (Notice that the ammeter measures − i s rather than i s .)
vo 40 V
So = = 20
is 2 A
Your lab partner is wrong.

VP2-2
vs 12
We expect the resistor current to be i = = = 0.48 A. The power absorbed by
R 25
this resistor will be P = i vs = (0.48) (12) = 5.76 W.
A half watt resistor can't absorb this much power. You should not try another resistor.

2-10

Design Problems

DP2-1
10 10
1.) > 0.04 ⇒ R < = 250 Ω
R 0.04

102 1
2.) < ⇒ R > 200 Ω
R 2

Therefore 200 < R < 250 Ω. For example, R = 225 Ω.

DP2-2
1.) 2 R > 40 ⇒ R > 20 Ω
15
2.) 2 2 R < 15 ⇒ R < = 3.75 Ω
4
Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously.

DP2-3
P = ( 30 mA ) ⋅ (1000 Ω ) = (.03) (1000 ) = 0.9 W < 1 W
2 2
1

P2 = ( 30 mA ) ⋅ ( 2000 Ω ) = (.03) ( 2000 ) = 1.8 W < 2 W
2 2

P3 = ( 30 mA ) ⋅ ( 4000 Ω ) = (.03) ( 4000 ) = 3.6 W < 4 W
2 2

2-11

Chapter 3 – Resistive Circuits
Exercises

Ex 3.3-1

Apply KCL at node a to get 2 + 1 + i3 = 0 ⇒ i3 = -3 A

Apply KCL at node c to get 2 + 1 = i4 ⇒ i4 = 3 A

Apply KCL at node b to get i3 + i6 = 1 ⇒ -3 + i6 = 1 ⇒ i6 = 4 A

Apply KVL to the loop consisting of elements A and B to get

-v2 – 3 = 0 ⇒ v2 = -3 V

Apply KVL to the loop consisting of elements C, E, D, and A to get

3 + 6 + v4 – 3 = 0 ⇒ v4 = -6 V

Apply KVL to the loop consisting of elements E and F to get

v6 – 6 = 0 ⇒ v6 = 6 V

Check: The sum of the power supplied by all branches is

-(3)(2) + (-3)(1) – (3)(-3) + (-6)(3) – (6)(1) + (6)(4) = -6 - 3 + 9 - 18 - 6 + 24 = 0

3-1

Ex 3.3-2
Apply KCL at node a to
determine the current in the
horizontal resistor as shown.

Apply KVL to the loop
consisting of the voltages source
and the two resistors to get

-4(2-i) + 4(i) - 24 = 0 ⇒ i = 4 A

2
Ex 3.3-3 −18 + 0 − 12 − va = 0 ⇒ va = −30 V and im = va + 3 ⇒ im = 9 A
5

18
Ex 3.3-4 −va − 10 + 4va − 8 = 0 ⇒ va = = 6 V and vm = 4 va = 24 V
3

Ex 3.4-1
From voltage division
 3 
v3 = 12   = 3V
 3+9 
then
v
i = 3 = 1A
3

The power absorbed by the resistors is: (12 ) ( 6 ) + (12 ) ( 3) + (12 ) ( 3) = 12 W
The power supplied by the source is (12)(1) = 12 W.

3-2

Ex 3.4-2
P = 6 W and R1 = 6 Ω

P 6
i2 = = = 1 or i =1 A
R1 6

v0 = i R1 =(1) (6)=6V

from KVL: − v+ i (2 + 4 + 6 + 2) = 0
s
⇒ v = 14 i = 14 V
s

25
Ex 3.4-3 From voltage division ⇒ v =
m 25+75
(8) = 2 V

25
Ex 3.4-4 From voltage division ⇒ v =
m 25+75
( −8 ) = −2 V

Ex. 3.5-1

1 1 1 1 1 4 103 1
= + 3+ 3+ 3= 3 ⇒ R = = kΩ
R 3
10 10 10 10 10 eq 4 4
eq
1 -3 1
By current division, the current in each resistor = (10 ) = mA
4 4

Ex 3.5-2
10
From current division ⇒ i =
m 10+40
( −5 ) = − 1 A

3-3

Problems

Section 3-3 Kirchoff’s Laws

P3.3-1

Apply KCL at node a to get 2 + 1 = i + 4 ⇒ i = -1 A

The current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W is
power received by element B. The power supplied by element B is 12 W.

Apply KVL to the loop consisting of elements D, F, E, and C to get

4 + v + (-5) – 12 = 0 ⇒ v = 13 V

The current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 W
is the power supplied by element F.


-(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0

3-4

P3.3-2

Apply KCL at node a to get 2 = i2 + 6 = 0 ⇒ i2 = -4 A

Apply KCL at node b to get 3 = i4 + 6 ⇒ i4 = -3 A

Apply KVL to the loop consisting of elements A and B to get

-v2 – 6 = 0 ⇒ v2 = -6 V

Apply KVL to the loop consisting of elements C, D, and A to get

-v3 – (-2) – 6 = 0 ⇒ v4 = -4 V

Apply KVL to the loop consisting of elements E, F and D to get

4 – v6 + (-2) = 0 ⇒ v6 = 2 V


-(6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0

3-5

P3.3-3
KVL : −12 − R 2 (3) + v = 0 (outside loop)
v − 12
v = 12 + 3R 2 or R 2 =
3
12
KCL i+ − 3 = 0 (top node)
R1
12 12
i = 3− or R1 =
R1 3−i

(a)
v = 12 + 3 ( 3) = 21 V
12
i = 3− =1 A
6

(b)
2 − 12 10 12
R2 = = − Ω ; R1 = =8Ω
3 3 3 − 1.5

(checked using LNAP 8/16/02)

(c)
24 = − 12 i, because 12 and i adhere to the passive convention.
12
∴ i = − 2 A and R1 = = 2.4 Ω
3+ 2
9 = 3v, because 3 and v do not adhere to the passive convention
3 − 12
∴ v = 3V and R 2 = = −3 Ω
3

The situations described in (b) and (c) cannot occur if R1 and R2 are required to be nonnegative.

3-6

P3.3-4
12
i = =2A
1 6
20
i = = 5A
2 4
i = 3−i = − 2 A
3 2
i = i +i = 3A
4 2 3

Power absorbed by the 4 Ω resistor = 4 ⋅ i 2 = 100 W
2
1
4 (checked using LNAP 8/16/02)

P3.3-5
v1 = 8 V
v2 = −8 + 8 + 12 = 12 V
v3 = 2⋅ 4 = 8 V
v2
4Ω : P = 3 = 16 W
4
2
v2
6Ω : P = = 24 W
6
v2
8Ω : P = 1 = 8 W
(checked using LNAP 8/16/02) 8

P3.3-6

P2 mA = − 3 × ( 2 ×10−3 )  = −6 × 10−3 = −6 mW
 

P1 mA = −  −7 × (1× 10−3 )  = 7 × 10−3 = 7 mW
 


3-7

P3.3-7

P2 V = +  2 × (1× 10−3 )  = 2 × 10−3 = 2 mW
 
P3 V = + 3 × ( −2 × 10 )  = −6 × 10−3 = −6 mW

−3



P3.3-8

KCL: iR = 2 + 1 ⇒ iR = 3 A
KVL: vR + 0 − 12 = 0 ⇒ vR = 12 V
vR 12
∴ R= = =4Ω
iR 3


P3.3-9

KVL: vR + 56 + 24 = 0 ⇒ vR = −80 V
KCL: iR + 8 = 0 ⇒ iR = −8 A
vR −80
∴ R= = = 10 Ω
iR −8


3-8

P3.3-10

5.61 3.71 − 5.61 12 − 5.61 −1.9
KCL at node b: = + ⇒ 0.801 = + 1.278
7 R1 5 R1
1.9
⇒ R1 = = 3.983 ≈ 4 Ω
1.278 − 0.801

3.71 3.71 − 5.61 3.71 − 12 −8.29
KCL at node a: + + = 0 ⇒ 1.855 + ( −0.475 ) + =0
2 4 R2 R2
8.29
⇒ R2 = = 6.007 ≈ 6 Ω
1.855 − 0.475


3-9

Section 3-4 A Single-Loop Circuit – The Voltage Divider

P3.4-1
6 6
v = 12 = 12 = 4 V
1 6+3+5+ 4 18
3 5 10
v = 12 = 2 V ; v = 12 = V
2 18 3 18 3
4 8
v = 12 = V
4 18 3


P3.4-2

(a) R = 6 + 3 + 2 + 4 = 15 Ω
28 28
(b) i = = = 1.867 A
R 15
( c ) p = 28 ⋅ i =28(1.867)=52.27 W
(28 V and i do not adhere
to the passive convention.)


3-10

P3.4-3

i R2 = v = 8 V
12 = i R1 + v = i R1 + 8
⇒ 4 = i R1

8 8 4 4 ⋅ 100
(a) i= = ; R1 = = = 50 Ω
R 2 100 i 8
4 4 8 8 ⋅ 100
(b) i = = ; R2 = = = 200 Ω
R1 100 i 4
4 8
( c ) 1.2 = 12 i ⇒ i = 0.1 A ; R1 = = 40 Ω; R2 = = 80 Ω
i i

P3.4-4
Voltage division
16
v1 = 12 = 8 V
16 + 8
4
v3 = 12 = 4 V
4+8

KVL: v3 − v − v1 = 0
v = −4 V


P3.4-5
 100  v 
using voltage divider: v =  ⇒ R = 50  s − 1
v
0  100 + 2 R  s
 v 
 o 
with v = 20 V and v > 9 V, R < 61.1 Ω 
s 0 
 R = 60 Ω
with v = 28 V and v < 13 V, R > 57.7 Ω 
s 0 

3-11

P3.4-6

 240 
a.)   18 = 12 V
 120 + 240 
 18 
b.) 18   = 0.9 W
 120 + 240 
 R 
c.)   18 = 2 ⇒ 18 R = 2 R + 2 (120 ) ⇒ R = 15 Ω
 R + 120 
R
d.) 0.2 = ⇒ ( 0.2 )(120 ) = 0.8 R ⇒ R = 30 Ω
R + 120


3-12

Section 3-5 Parallel Resistors and Current Division

P3.5-1
1
6 1 1
i = 4= 4= A
1 1 + 1 + 1 +1 1+ 2 + 3 + 6 3
6 3 2 1
1
3 2
i = 4 = A;
2 1 + 1 + 1 +1 3
6 3 2 1
1
i = 2 4 =1 A
3 1 + 1 + 1 +1
6 3 2 1
1
i = 4=2 A
4 1 + 1 + 1 +1
6 3 2

P3.5-2

1 1 1 1 1
(a) = + + = ⇒ R = 2Ω
R 6 12 4 2
(b) v = 6 ⋅ 2 = 12 V
(c) p = 6 ⋅12 = 72 W

P3.5-3

8 8
i= or R1 =
R1 i
8 8
8 = R 2 (2 − i ) ⇒ i = 2 − or R 2 =
R2 2−i

8 4 8
(a) i = 2− = A ; R1 =
4
=6Ω
12 3
3
8 2 8
(b) i = = A ; R2 =
2
=6Ω
12 3 2−
3

3-13

1
( c ) R1 = R 2 will cause i= 2 = 1 A. The current in both R1 and R 2 will be 1 A.
2
R1 R 2 1
2 ⋅ = 8 ; R1 = R 2 ⇒ 2 ⋅ R1 = 8 ⇒ R1 = 8 ∴ R1 = R 2 = 8 Ω
R1 + R 2 2

P3.5-4
Current division:

8
1 16 + 8 ( )
i = −6 = −2 A

8
2 8+8( )
i = −6 = −3 A

i = i −i = +1 A
1 2

P3.5-5
 R 
current division: i =  1  i and
2 R + R  s
 1 2

Ohm's Law: v = i R yields
o 2 2
 v  R + R 
i =  o  1 2
s  R  R 
 2  1 
plugging in R = 4Ω, v > 9 V gives i > 3.15 A
1 o s
and R = 6Ω, v < 13 V gives i < 3.47 A
1 o s
So any 3.15 A < i < 3.47 A keeps 9 V < v < 13 V.
s o

3-14

P3.5-6

 24 
a)   1.8 = 1.2 A
 12 + 24 
 R 
b)   2 = 1.6 ⇒ 2 R = 1.6 R + 1.6 (12 ) ⇒ R = 48 Ω
 R + 12 
R
c) 0.4 = ⇒ ( 0.4 )(12 ) = 0.6 R ⇒ R = 8 Ω
R + 12

Section 3-7 Circuit Analysis

P3.7-1

48 ⋅ 24
(a) R = 16 + = 32 Ω
48 + 24
32 ⋅ 32
(b) v = 32 + 32 24 = 16 V ;
32 ⋅ 32
8+
32 + 32
16 1
i= = A
32 2
48 1 1
(c) i2 = ⋅ = A
48 + 24 2 3

3-15

P3.7-2

3⋅ 6
(a) R1 = 4 + =6Ω
3+ 6
1 1 1 1
(b) = + + ⇒ R p = 2.4 Ω then R 2 = 8 + R p = 10.4 Ω
Rp 12 6 6
(c) KCL: i2 + 2 = i1 and − 24 + 6 i2 + R 2i1 = 0
⇒ −24+6 (i1 −2)+10.4i1 = 0
36
⇒ i1 = =2.2 A ⇒ v1 =i1 R 2 =2.2 (10.4)=22.88 V
16.4

1
(d ) i2 = 6 ( 2.2 ) = 0.878 A,
1 1 1
+ +
6 6 12
v2 = ( 0.878 ) (6) = 5.3 V
6 2
(e) i3 = i2 = 0.585 A ⇒ P = 3 i3 = 1.03 W
3+ 6

3-16

P3.7-3
Reduce the circuit from the right side by repeatedly replacing series 1 Ω resistors in parallel with
a 2 Ω resistor by the equivalent 1 Ω resistor

This circuit has become small enough to be easily analyzed. The vertical 1 Ω resistor is
equivalent to a 2 Ω resistor connected in parallel with series 1 Ω resistors:

1+1
i1 = (1.5 ) = 0.75 A
2 + (1 + 1)

3-17

P3.7-4
(a) 1 1 1 1 (10 + 8) ⋅ 9
= + + ⇒ R2 = 4 Ω and R1 = = 6Ω
R2 24 12 8 b g
10 + 8 + 9

(b)

First, apply KVL to the left mesh to get −27 + 6 ia + 3 ia = 0 ⇒ ia = 3 A . Next,
apply KVL to the left mesh to get 4 ib − 3 ia = 0 ⇒ ib = 2.25 A .
(c)

1
i2 = 8 2.25 = 1125 A
. b gLM b10 +98g + 9 3OP = −10 V
and v1 = − 10
1 1 1
+ +
24 8 12
N Q

3-18

P3.7-5

30
v1 = 6 ⇒ v1 = 8 V
10 + 30

R2
12 = 8 ⇒ R2 = 20 Ω
R2 + 10

20 =
b
R1 10 + 30 g ⇒ R1 = 40 Ω
b
R1 + 10 + 30 g
Alternate values that can be used to change the numbers in this problem:

meter reading, V Right-most resistor, Ω R1, Ω
6 30 40
4 30 10
4 20 15
4.8 20 30

3-19

P3.7-6

P3.7-7
24
1× 10−3 = ⇒ R p = 12 ×103 = 12 kΩ
12 ×103 + R p

12 × 10 = R p
3
=
( 21×10 ) R3

⇒ R = 28 kΩ
( 21×10 ) + R
3

P3.7-8

 130 500 
Voltage division ⇒ v = 50  = 15.963 V
 130 500 + 200 + 20 
 
 100   10 
∴v = v 
h  = (15.963)   = 12.279 V
 100 + 30   13 
v
∴ i = h = .12279 A
h 100

3-20

P3.7-10

15 ( 20 + 10 )
Req = = 10 Ω
15 + ( 20 + 10 )
60  30   60   20 
ia = − = −6 A, ib =  R  
= 4 A, vc =   ( −60 ) = −40 V
Req  30 + 15   eq   20 + 10 

P3.7-11 a)

(24)(12)
Req = 24 12 = =8Ω
24 + 12

b) from voltage division:
100
 20  100 5
v = 40  = V∴ i = 3 = A
x  20 + 4  3 x 20 3

 8  5
from current division: i = i = A
x 8+8
  6

3-22

P3.7-12
9 + 10 + 17 = 36 Ω
36 (18 )
a.) = 12 Ω
36+18

36 R
b.) = 18 ⇒ 18 R = (18 )( 36 ) ⇒ R = 36 Ω
36+R

P3.7-13
2 R( R ) 2
Req = = R
2R + R 3
v 2 240
Pdeliv. = = =1920 W
to ckt Req 2 R
3
Thus R =45 Ω

P3.7-14

R = 2 + 1 + ( 6 12 ) + ( 2 2 ) = 3 + 4 + 1 = 8 Ω
eq
40 40
∴i = = =5 A
Req 8
 6 
i1 = i 
 6 + 12 
( )
 = ( 5) 3 = 3 A
1 5 from current division

i2 = i 
 2 
 2+2
( )
 = ( 5) 2 = 2 A
1 5

3-23


VP3-1

KCL at node a: i = i + i
3 1 2
− 1.167 = − 0.833 + ( −0.333)
− 1.167= − 1.166 OK
KVL loop consisting of the vertical
6 Ω resistor, the 3 Ω and4Ω resistors,
and the voltage source:
6i + 3i + v + 12 = 0
3 2
yields v = −4.0 V not v = −2.0 V

VP3-2

reduce circuit: 5+5=10 in parallel with 20 Ω gives 6.67Ω

 6.67 
by current division: i =   5 = 1.25 A
 20 + 6.67 
∴Reported value was correct.

VP3-3
 320 
v =
o  320 + 650 + 230 
( 24 ) = 6.4 V ∴Reported value was incorrect.


3-24

VP3-4

KVL bottom loop: − 14 + 0.1iA + 1.2iH = 0
KVL right loop: − 12 + 0.05iB + 1.2iH = 0
KCL at left node: iA + iB = iH
This alone shows the reported results were incorrect.
Solving the three above equations yields:
iA = 16.8 A iH = 10.3 A
iB = −6.49 A
∴ Reported values were incorrect.

VP3-5

 1 
Top mesh: 0 = 4 i a + 4 i a + 2  i a + − i b  = 10 ( −0.5 ) + 1 − 2 ( −2 )
 2 

Lower left mesh: vs = 10 + 2 ( i a + 0.5 − i b ) = 10 + 2 ( 2 ) = 14 V

Lower right mesh: vs + 4 i a = 12 ⇒ vs = 12 − 4 (−0.5) = 14 V

The KVL equations are satisfied so the analysis is correct.

3-25

VP3-6
Apply KCL at nodes b and c to get:

KCL equations:

Node e: −1 + 6 = 0.5 + 4.5

Node a: 0.5 + i c = −1 ⇒ i c = −1.5 mA

Node d: i c + 4 = 4.5 ⇒ i c = 0.5 mA

That's a contradiction. The given values of ia
and ib are not correct.

Design Problems

DP3-1
Using voltage division:

R 2 + aR p R 2 + aR p
vm = 24 = 24
R1 + (1 − a ) R p + R 2 + aR p R1 + R 2 + R p

vm = 8 V when a = 0 ⇒
R2 1
=
R1 + R 2 + R p 3
vm = 12 V when a = 1 ⇒
R2 + R p 1
=
R1 + R 2 + R p 2
The specification on the power of the voltage source indicates
242 1
≤ ⇒ R1 + R 2 + R p ≥ 1152 Ω
R1 + R 2 + R p 2
Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives
3R 2 = R1 + R 2 + 2000 and 2 ( R 2 + 2000 ) = R1 + R 2 + 2000 . Solving these equations gives
R1 = 6000 Ω and R 2 = 4000 Ω .

With these resistance values, the voltage source supplies 48 mW while R1, R2 and Rp dissipate
24 mW, 16 mW and 8 mW respectively. Therefore the design is complete.

3-26

DP3-2
Try R1 = ∞. That is, R1 is an open circuit. From KVL, 8 V will appear across R2. Using voltage
200
division, 12 = 4 ⇒ R 2 = 400 Ω . The power required to be dissipated by R2
R 2 + 200
82 1
is = 0.16 W < W . To reduce the voltage across any one resistor, let’s implement R2 as the
400 8
series combination of two 200 Ω resistors. The power required to be dissipated by each of these
42 1
resistors is = 0.08 W < W .
200 8
Now let’s check the voltage:
190 210
11.88 < v < 12.12
190 + 420 0 210 + 380

3.700 < v0 < 4.314

4 − 7.5% < v0 < 4 + 7.85%

Hence, vo = 4 V ± 8% and the design is complete.

DP3-3
Vab ≅ 200 mV
10 10
v= 120 Vab = (120) (0.2)
10 + R 10 + R
240
let v = 16 = ⇒ R=5Ω
10 + R
162
∴P= = 25.6W
10

DP3-4
N N 1 1
i = G v = v where G = ∑ = N 
T R T
n = 1 Rn  R
iR ( 9 )(12 )
∴N= = = 18 bulbs
v 6

3-27

Chapter 4 – Methods of Analysis of Resistive Circuits
Exercises

Ex. 4.3-1

v v −v
a a b
KCL at a: + + 3 = 0 ⇒ 5 v − 3 v = −18
3 2 a b

v −v
b a
KCL at b: − 3 −1 = 0 ⇒ v − v = 8
2 b a

Solving these equations gives:

va = 3 V and vb = 11 V

Ex. 4.3-2
KCL at a:

v v −v
a a b
+ + 3 = 0 ⇒ 3 v − 2 v = −12
4 2 a b

v v −v
b a b
− −4=0
KCL at a: 3 2
⇒ − 3 v + 5 v = 24
a b
Solving:
va = −4/3 V and vb = 4 V

Ex. 4.4-1
Apply KCL to the supernode to get

v + 10 v
2+ b + b =5
20 30
Solving:

v = 30 V and v = v + 10 = 40 V
b a b

4-1

Ex. 4.4-2

( vb + 8) − ( −12) + vb = 3 ⇒ v = 8 V and v = 16 V
10 40 b a

Ex. 4.5-1
Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into
vb = 4 ia and solve for vb .
6 vb  9 + vb 
+ =i ⇒ v = 4i = 4   ⇒ v = 4.5 V
8 12 a b a  12  b
 

Ex. 4.5-2
The controlling voltage of the dependent source is a node voltage so it is already expressed as a
function of the node voltages. Apply KCL at node a.

v −6 v −4v
a + a a = 0 ⇒ v = −2 V
20 15 a

Ex. 4.6-1

Mesh equations:
−12 + 6 i + 3  i − i  − 8 = 0 ⇒ 9 i − 3 i = 20
 1 2
1   1 2

8 − 3  i − i  + 6 i = 0 ⇒ − 3 i + 9 i = −8
 1 2
  2 1 2

Solving these equations gives:
13 1
i = A and i = − A
1 6 2 6

The voltage measured by the meter is 6 i2 = −1 V.

4-2

Ex. 4.7-1

 3 −12
Mesh equation: 9 + 3 i + 2 i + 4  i +  = 0 ⇒ ( 3 + 2 + 4 ) i = −9 − 3 ⇒ i= A
 4 9
The voltmeter measures 3 i = −4 V

Ex. 4.7-2

−33 2
Mesh equation: 15 + 3 i + 6 ( i + 3) = 0 ⇒ ( 3 + 6 ) i = −15 − 6 ( 3) ⇒ i= = −3 A
9 3

Ex. 4.7-3

3 3
Express the current source current in terms of the mesh currents: = i1 − i 2 ⇒ i1 = + i 2 .
4 4
3 
Apply KVL to the supermesh: −9 + 4i1 + 3 i 2 + 2 i 2 = 0 ⇒ 4  + i 2  + 5 i 2 = 9 ⇒ 9 i 2 = 6
4 
2 4
so i 2 = A and the voltmeter reading is 2 i 2 = V
3 3

4-3

Ex. 4.7-4

Express the current source current in terms of the mesh currents: 3 = i1 − i 2 ⇒ i1 = 3 + i 2 .
Apply KVL to the supermesh: −15 + 6 i1 + 3 i 2 = 0 ⇒ 6 ( 3 + i 2 ) + 3 i 2 = 15 ⇒ 9 i 2 = −3
1
Finally, i 2 = − A is the current measured by the ammeter.
3

Problems

Section 4-3 Node Voltage Analysis of Circuits with Current Sources

P4.3-1

KCL at node 1:

v v −v
1 1 2 −4 − 4 − 2
0= + +i = + + i = −1.5 + i ⇒ i = 1.5 A
8 6 8 6


4-4

P4.3-2
KCL at node 1:
v −v v
1 2 1
+ + 1 = 0 ⇒ 5 v − v = −20
20 5 1 2
KCL at node 2:
v −v v −v
1 2 2 3
+2= ⇒ − v + 3 v − 2 v = 40
20 10 1 2 3
KCL at node 3:
v −v v
2 3 3
+1 = ⇒ − 3 v + 5 v = 30
10 15 2 3
Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V.


P4.3-3

KCL at node 1:

v −v v
1 2 1 4 − 15 4
+ =i ⇒ i = + = −2 A
5 20 1 1 5 20

KCL at node 2:

v −v v −v
1 2 2 3
+i =
5 2 15
 4 − 15  15 − 18
⇒ i = − + =2A
2  5  15


4-5

P4.3-4

Node equations:
v1 v1 − v2
−.003 + + =0
R1 500
v1 − v2 v2
− + − .005 = 0
500 R2
When v1 = 1 V, v2 = 2 V
1 −1 1
−.003 + + = 0 ⇒ R1 = = 200 Ω
R1 500 1
.003 +
500
−1 2 2
− + − .005 = 0 ⇒ R2 = = 667 Ω
500 R2 1
.005 −
500


P4.3-5
Node equations:
v1 v − v 2 v1 − v3
+ 1 + =0
500 125 250
v − v2 v − v3
− 1 − .001 + 2 =0
125 250
v − v3 v1 − v3 v3
− 2 − + =0
250 250 500
Solving gives:

v1 = 0.261 V, v2 = 0.337 V, v3 = 0.239 V

Finally, v = v1 − v3 = 0.022 V


4-6

Section 4-4 Node Voltage Analysis of Circuits with Current and Voltage Sources

P4.4-1

Express the branch voltage of the voltage source in terms of its node voltages:

0 − va = 6 ⇒ va = −6 V
KCL at node b:

va − vb v −v −6 − vb v −v vb v −v
+2= b c ⇒ +2= b c ⇒ −1− +2= b c ⇒ 30 = 8 vb − 3 vc
6 10 6 10 6 10

vb − vc vc 9
KCL at node c: = ⇒ 4 vb − 4 vc = 5 vc ⇒ vb = vc
10 8 4

9 
Finally: 30 = 8  vc  − 3 vc ⇒ vc = 2 V
4 

P4.4-2

Express the branch voltage of each voltage source in terms of its node voltages to get:

va = −12 V, vb = vc = vd + 8

4-7

KCL at node b:
vb − va vb − ( −12 )
= 0.002 + i ⇒ = 0.002 + i ⇒ vb + 12 = 8 + 4000 i
4000 4000

KCL at the supernode corresponding to the 8 V source:
v
0.001 = d + i ⇒ 4 = vd + 4000 i
4000
so vb + 4 = 4 − vd ⇒ ( vd + 8 ) + 4 = 4 − vd ⇒ vd = −4 V
4 − vd
Consequently vb = vc = vd + 8 = 4 V and i = = 2 mA
4000


P4.4-3

Apply KCL to the supernode:
va − 10 va va − 8
+ + − .03 = 0 ⇒ va = 7 V
100 100 100


P4.4-4
Apply KCL to the supernode:

va + 8 ( va + 8 ) − 12 va − 12 va
+ + + =0
500 125 250 500

Solving yields
va = 4 V


4-8

P4.4-5

The power supplied by the voltage source is

v −v v −v   12 − 9.882 12 − 5.294 
va ( i1 + i 2 ) = va  a b + a c  = 12  + 
 4 6   4 6 

= 12(0.5295 + 1.118) = 12(1.648) = 19.76 W


P4.4-6
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
 6000  R 3000

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.

4-9

Section 4-5 Node Voltage Analysis with Dependent Sources

P4.5-1
Express the resistor currents in terms of the
node voltages:

va − vc
i 1= = 8.667 − 10 = −1.333 A and
1
v −v 2 − 10
i 2= b c = = −4 A
2 2

Apply KCL at node c:

i1 + i 2 = A i1 ⇒ − 1.333 + ( −4 ) = A (−1.333)
−5.333
⇒ A= =4
−1.333


P4.5-2
Write and solve a node equation:

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

va − 4va
ib = = −12 mA
3000


P4.5-3
First express the controlling current in terms of
the node voltages:
2 − vb
i =
a 4000
Write and solve a node equation:

2 − vb v  2 − vb 
− + b − 5  = 0 ⇒ vb = 1.5 V
4000 2000  4000 


4-10

P4.5-4
Apply KCL to the supernode of the CCVS to get

12 − 10 14 − 10 1
+ − + i b = 0 ⇒ i b = −2 A
4 2 2

Next
10 − 12 1
ia = =−  −2 V
4 2 ⇒ r = =4
1 A
r i a = 12 − 14 
 −
2


P4.5-5
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
12 − v 2 = 3 i x = 3 ⇒ v2 = V
2 5

so ix = 12/5 A = 2.4 A.

(checked using ELab 9/5/02)

4-11

Section 4-6 Mesh Current Analysis with Independent Voltage Sources

P 4.6-1
2 i1 + 9 (i1 − i 3 ) + 3(i1 − i 2 ) = 0
15 − 3 (i1 − i 2 ) + 6 (i 2 − i 3 ) = 0
−6 (i 2 − i 3 ) − 9 (i1 − i 3 ) − 21 = 0
or
14 i1 − 3 i 2 − 9 i 3 = 0
−3 i1 + 9 i 2 − 6 i 3 = −15
−9 i1 − 6 i 2 + 15 i 3 = 21
so
i1 = 3 A, i2 = 2 A and i3 = 4 A.


P 4.6-2
Top mesh:
4 (2 − 3) + R(2) + 10 (2 − 4) = 0
so R = 12 Ω.

Bottom, right mesh:
8 (4 − 3) + 10 (4 − 2) + v 2 = 0
so v2 = −28 V.

Bottom left mesh
−v1 + 4 (3 − 2) + 8 (3 − 4) = 0
so v1 = −4 V.


4-12

P 4.6-3
−6
Ohm’s Law: i 2 = = −0.75 A
8
KVL for loop 1:
R i1 + 4 ( i1 − i 2 ) + 3 + 18 = 0

KVL for loop 2
+ (−6) − 3 − 4 ( i1 − i 2 ) = 0
⇒ − 9 − 4 ( i1 − ( −0.75 ) ) = 0
⇒ i 1 = −3 A
R ( −3) + 4 ( −3 − ( −0.75 ) ) + 21 = 0 ⇒ R = 4 Ω

P4.6-4

KVL loop 1:

25 ia − 2 + 250 ia + 75 ia + 4 + 100 (ia − ib ) = 0
450 ia −100 ib = −2

KVL loop 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


P4.6-5
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


4-13

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