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:

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).

Exercises Ex. 1.3-1
Ex. 1.3-3
Ex. 1.3-4
i(t)= dq(t) d t
Ex. 1.4-1

Chapter 1 – Electric Circuit Variables

i(t) = 8t2 −4t A
q(t) = ∫tidτ+q(0)= ∫t(8τ2 −4τ)dτ+0=8τ3−2τ2 t =8t3−2t2 C 00303

q(t)=∫ti(τ)dτ+q(0)=∫t4sin3τdτ+0=−4cos3τt =−4cos3t+4 C 003033

0 i(t) = 2

t<0 0<t<2 t > 2

 − 2 e − 2 ( t − 2 ) 

i1 =45μA=45×10-6 A<i2 =0.03mA=.03×10-3 A=3×10-5 A< i3 =25×10-4 A Ex. 1.4-2

Ex. 1.4-3
Ex. 1.4-4

∆q = i∆t = (4000 A)(0.001 s) = 4 C
i=∆q=45×10−9 =9×10−6=9μA ∆t 5×10−3
C = 10×109 electron1.602×10−19 electron  s 

i = 10billion electron1.602×10−19
 s  

C  electron

= 1010 ×1.602×10−19 electron C

=1.602×10−9 C = 1.602 s

nA

s

electron

1-1

Ex. 1.6-1

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.
   
2. (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.
   
3. (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.
   
4. (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.
   
5. (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
P1.3-2
P1.3-3

i(t)= 4 1−e−5t =20e−5t A d()

dt
q(t)= i(τ)dτ+q(0)= 41−e−5τ dτ+0= 4dτ− 4e−5τ dτ=4t+ e−5t− C

∫t ∫t()∫t ∫t 44

0000

55

q(t)=∫t i(τ)dτ=∫t 0dτ=0 C fort≤2soq(2)=0. −∞ −∞
q(t)=∫ti(τ)dτ+q(2)=∫t2dτ=2τt =2t−4 C for2≤t≤4.Inparticular,q(4)=4C. 222
q(t)=∫ti(τ)dτ+q(4)=∫t−1dτ+4=−τt +4=8−t C for4≤t≤8.Inparticular,q(8)=0C. 444
q(t)=∫ti(τ)dτ+q(8)=∫t0dτ+0=0 C for8≤t. 88

1-2

P1.3-4
Section 1-6 Power and Energy P1.6-1
a.) q = ∫idt = i∆t = (10A)(2hrs)(3600s/hr)= 7.2×104 C b.) P=vi=(110V)(10 A)=1100 W
c.)Cost= 0.06$×1.1kW×2hrs=0.132$ kWhr
P1.6-2
P1.6-3

i = 600 A = 600 C s
Silver deposited = 600C×20min×60 s ×1.118 mg= 8.05×105 mg=805 g s min C

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

P 0.06 W

for0≤t≤10s: v=30V and i=30t=2tA ∴P=30(2t)=60tW 15
for10 ≤ t ≤ 15s: v(t)=−25t+b ⇒ v(10)=30V ⇒ b=80V 5

v(t)=−5t+80 and i(t) = 2t A ⇒ P=(2t)(−5t+80)= −10t2+160t W for15≤t≤25s: v=5V and i(t)=−30t+bA

10 i(25)=0 ⇒ b=75 ⇒ i(t)=−3t+75A

∴P = (5)(−3t+75) = −15t+375 W

1-3

∫∫10 ∫15()∫25
Energy= Pdt = 60tdt+ 160t−10t2 dt+ (375−15t)dt
=30t2 10 +80t2 −10t3 15 +375t−15t2 25 =5833.3J 0 3 10 2 15

0 10 15

P1.6-4

a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully charged).

w=∫Pdt = ∫ vidτ =∫

211+ dτ =22t+  3600

t 5(3600)  0.5τ 

0.5 5(3600)

0

τ2
= 441×103 J = 441 kJ

0

3600

0

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

P1.6-5

p(t)= 1(cos3t)(sin3t)= 1sin6t 36
p (0.5) = 1 sin 3 = 0.0235 W 6
p (1) = 1 sin 6 = −0.0466 W 6

1-4

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

t0=0;
tf=2;
dt=0.02;
t=t0:dt:tf;

v=4*cos(3*t);
i=(1/12)*sin(3*t);

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

end

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

% initial time
% final time
% time increment
% time

% device voltage
% device current

% power

P1.6-6

t0=0;
tf=2;
dt=0.02;
t=t0:dt:tf;

v=8*sin(3*t);
i=2*sin(3*t);

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

end

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

% initial time
% final time
% time increment
% time

% device voltage
% device current

% power

p(t)=16(sin3t)(sin3t)=8(cos0−cos6t)=8−8cos6t W

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

1-5

P1.6-7

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

t0=0;
tf=2;
dt=0.02;
t=t0:dt:tf;

v=4*(1-exp(-2*t));
i=2*exp(-2*t);

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

end

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

% initial time
% final time
% time increment
% time

% device voltage
% device current

% power

P1.6-8

P =VI=3×0.2=0.6W
w =P⋅t= 0.6×5×60=180J

p(t)=4(1−e−2t )×2e−2t =8(1−e−2t )e−2t

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

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

t0=0;
tf=1;
dt=0.02;
t=t0:dt:tf;

v=20*(1-exp(-8*t));
i=.030*exp(-8*t);

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

end

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

Here is the plot:

% initial time
% final time
% time increment
% time

 % device voltage
   % device current

% power

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

The circuit element must be able to absorb 0.15 W.

1-8

Exercises Ex. 2.3-1
Ex. 2.3-2
Ex. 2.5-1
Ex. 2.5-2
Ex. 2.8-1

m (i1 + i 2 ) = mi1 + mi 2 ⇒ superposition is satisfied m (a i1 ) = a (mi1 ) ⇒ homogeneity is satisfied
Therefore the element is linear.
m(i1 +i2)+b=mi1 +mi2 +b≠(mi1 +b)+(mi2 +b)⇒superpositionisnotsatisfied Therefore the element is not linear.

Chapter 2 - Circuit Elements

v2 (10)2
P= R = 100 =1W

P=v2 =(10cost)2 =10cos2tW R 10

ic=−1.2 A,vd=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

Ex. 2.8-3

vc =−2 V, id =4vc =−8 A andvd =2.2 V id and vd adhere to the passive convention so
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.

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

2-2

Ex. 2.9-1
Ex. 2.9-2
Ex. 2.10-1 Ex. 2.10.2
At t=6stheswitchisinthedownposition,sov=0V. 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
(c)Whenv=4 V, i= 4 =33 A=33A. 0.12

θ=45°,I=2 mA,Rp =20 kΩ

a=θ ⇒ aRp=45(20kΩ)=2.5kΩ 360 360

vm =(2×10−3)(2.5×103)=5V

v=10V,i=280μA, k=1μA forAD590 °K

i  °K
i=kT ⇒ T= =(280μA)1 =280 K

At t=4sbothswitchesareopen,soi=0A.
At t=4stheswitchisintheupposition,so v =iR=(2mA)(3kΩ)=6V.

k  μA 

°

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
(c)Whenv=12 V, i= 12 =0.04678 A=46.78mA. 256.5
P2.3-4
Leti=1A,thenv=3i+ 5=8V.Next2i =2A but16=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
P2.5-2
P2.5-3

i = is = 3 A and v = R i = 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.

i=is =3 mAandv=24 V R=v= 24 =8000=8kΩ

i .003
P = (3×10−3 )×24 = 72×10−3 = 72 mW

v=vs=10V and R=5Ω i = v = 10=2 A

R5
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
P2.5-5
v2 andi2 donotadheretothepassiveconventionso i2 =−v2 =−150= −6 A R2 25
ThepowerabsorbedbyR isP = v i =150⋅3=450W 1111
ThepowerabsorbedbyR2 isP2 = −v2i2 =−150(−6)= 900W P2.5-6
v2 andi2 doadheretothepassiveconventionsov2 = R2 i2 = 8⋅2 = 16V. ThepowerabsorbedbyR2 isP2 =v2i2 =16⋅2=32 W.
P2.5-7

v=vs =24Vandi=2A R = v = 24 = 12 Ω

i2
p=vi =24⋅2=48W

v1 = v2 = vs =150V;
R1 =50Ω;R2 =25Ω
v1 andi1 adheretothepassiveconventionso
i1 =v1 =150=3A R1 50

i1=i2=is=2A; R=4Ω and R =8Ω

12
v1 andi1 donotadheretothepassiveconventionso

v1=−R1 i1=−4⋅2=−8V. ThepowerabsorbedbyR1 is
P1=−v1i1 =−(−8)(2)=16W.

Model the heater as a resistor, then
witha 250 Vsource: P=v2 ⇒ R = v2 = (250)2 = 62.5Ω
with a 210 V source: P= v2 = (210)2 = 705.6 W R 62.5

R P 1000

2-5

P2.5-8

The current required by the mine lights is: i = P = 5000 = 125 A v 120 3
Power loss in the wire is : i2 R
Thus the maximum resistance of the copper wire allowed is
R = 0.05P = 0.05×5000 = 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 = ρL / A with ρ = 1.7×10−6Ω⋅cm from Table 2.5−1
A = ρ L = 1.7×10−6 ×20,000 = 0.236 cm2 R 0.144

Section 2-6 Independent Sources P2.6-1
(a)
(b) i and P do not depend on is .
Thevaluesofi andP are3Aand45W,bothwhenis
(a)

= 3Aandwhenis =5A.

i =vs =15=3A and P = Ri2 = 5(3)2 = 45W R5

P2.6-2

v=Ris =5⋅2=10 V and P =v2 =102 =20W R5

(b) v and P do not depend on v s . ThevaluesofvandPare10Vand20Wbothwhenvs =10Vandwhenvs =5V

2-6

P2.6-3

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

P2.6-5
(a)
(b)

P=vi=(2 cos t) (10 cos t)=20 cos2 t mW

∫1 ∫1 2 111
w= Pdt= 20costdt=202t+4sin2t =10+5sin2mJ

000

2-7

Section 2-7 Voltmeters and Ammeters P2.7-1

P2.7-2

(a)R=v= 5 =10Ω 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.
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=2v ⇒ v=20V

2-8

Section 2-8 Dependent Sources P2.8-1
P2.8-2
P2.8-3
P2.8-4
Section 2-9 Transducers P2.9-1
P2.9-2

a= θ , θ=360vm 360 Rp I

r = vb = 8 = 4 Ω ia2

vb =8V; gvb =ia =2A; g =ia =2 =0.25 A vb 8 V

i b = 8 A ; d i b = i a = 32A ; d = i a = 32 = 4 A ib8A

va = 2 V ; b va = vb = 8 V ; b = vb = 8 = 4 V va 2 V

θ =

(360)(23V) (100 kΩ)(1.1 mA)

= 75.27°

AD590:k=1μA , °K
v=20 V (voltage condition satisfied)

4 μ A < i < 1 3 μ A 

⇒ 4°K<T<13°K k 

T=i 

2-9

Section 2-10 Switches P2.10-1

At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is 15 V.
P2.10-2
Verification Problems VP2-1
VP2-2

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

i = v = 10 = 2 mA R 5×103

i = v = 15 = 3 mA R 5×103

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.

vo=40V and is=−(−2)=2A.(Noticethattheammetermeasures −is ratherthanis.) So vo =40=20V

is2A Your lab partner is wrong.

We expect the resistor current to be i = vs = 12 = 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
1.)10>0.04 ⇒ R<10=250Ω R 0.04
2.)102 <1 ⇒ R>200Ω R2
Therefore 200 < R < 250 Ω. For example, R = 225 Ω. DP2-2
1.)2R>40 ⇒ R>20Ω 2.)22R<15 ⇒ R<15=3.75Ω
DP2-3

4
Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously.

P =(30mA)2 ⋅(1000Ω)=(.03)2 (1000)=0.9W<1W 1
P =(30mA)2 ⋅(2000Ω)=(.03)2 (2000)=1.8W<2W 2
P =(30mA)2 ⋅(4000Ω)=(.03)2 (4000)=3.6W<4W 3

2-11

Exercises Ex 3.3-1

Chapter 3 – Resistive Circuits

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

2+1+i3 =0⇒ i3 =-3A 2+1=i4 ⇒ i4 =3A
i3 +i6 =1⇒ -3+i6 =1⇒ i6 =4A

-v2 –3=0⇒ v2 =-3V
Apply KVL to the loop consisting of elements C, E, D, and A to get
3+6+v4 –3=0⇒ v4 =-6V Apply KVL to the loop consisting of elements E and F to get
v6 –6=0⇒ v6 =6V 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

Ex 3.3-3
Ex 3.3-4 Ex 3.4-1

im = 9 A ⇒ va = 18 = 6 V and vm = 4 va = 24 V

−18 + 0 −12 − va = 0 −va −10 + 4va − 8 = 0

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
⇒ va = −30 V and im = 2 va + 3 ⇒ 5

3

From voltage division v =12 3  =3V

3 3+9 

then
i = v3 = 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

Ex 3.4-3
Ex 3.4-4 Ex. 3.5-1

P=6W and R1 =6Ω i2=P = 6 =1or i=1A
R6 1
v0 =iR1=(1)(6)=6V
from KVL: −vs + i (2+4+6+2) = 0
⇒ vs=14i=14V From voltage division ⇒ vm = 25 (8) = 2 V
25+75
From voltage division ⇒ vm = 25 (−8) = −2 V 25+75
1 = 1+1+1+1=4 ⇒ Req=103=1kΩ Req 103 103 103 103 103 4 4
By current division, the current in each resistor = 1 (10-3 ) = 1 mA 44
From current division ⇒ im = 10 (−5) = −1 A 10+40

Ex 3.5-2

3-3

Problems
Section 3-3 Kirchoff’s Laws P3.3-1
ApplyKCLatnodeatoget 2+1=i+4⇒ i=-1A
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=13V
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.
Check: The sum of the power supplied by all branches is
-(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0

3-4

P3.3-2

ApplyKCLatnodeatoget 2=i2 +6=0⇒ i2 =-4A ApplyKCLatnodebtoget 3=i4 +6⇒ i4 =-3A Apply KVL to the loop consisting of elements A and B to get
-v2 –6=0⇒ v2 =-6V Apply KVL to the loop consisting of elements C, D, and A to get
-v3 –(-2)–6=0⇒ v4 =-4V Apply KVL to the loop consisting of elements E, F and D to get
4–v6 +(-2)=0⇒ v6 =2V Check: The sum of the power supplied by all branches is
-(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−R2(3)+v=0 (outsideloop) v=12+3R2 orR2=v−12

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

i+12−3 =0 (topnode) R1
i=3−12 or R1 = 12 R1 3−i

3

(a)
(b)
(c)

R2=2−12=−10Ω;R1= 12 =8Ω 3 3 3−1.5

(checked using LNAP 8/16/02) 24 = −12 i, because 12 and i adhere to the passive convention.
∴i=−2AandR1=12 =2.4Ω 3+2
9 = 3v, because 3 and v do not adhere to the passive convention ∴v=3V and R2=3−12=−3Ω

3
The situations described in (b) and (c) cannot occur if R1 and R2 are required to be nonnegative.

3-6

P3.3-4

i =12=2A

1

6
i2 = 20 = 5A 4
i3 = 3−i2 = −2A i4 = i2 + i3 = 3 A

Power absorbed by the 4 Ω resistor = 4⋅i2 = 100 W 2
Power absorbed by the 6 Ω resistor = 6⋅i2 = 24 W 1
Power absorbed by the 8 Ω resistor = 8⋅i2 = 72 W 4

P3.3-5

(checked using LNAP 8/16/02)
v1 = 8 V
v2 =−8+8+12 = 12V v3 =2⋅4=8V

P3.3-6

(checked using LNAP 8/16/02)

v2
4Ω: P = 3 =16W
4
6Ω:P=v2 =24W 6
8Ω: P=v12 =8W 8


P2mA =−3×(2×10−3)=−6×10−3 =−6mW


P =−−7×(1×10−3)=7×10−3 =7mW
(checked using LNAP 8/16/02)

1mA

3-7

P3.3-7

P3.3-8


P2V =+2×(1×10−3)=2×10−3 =2mW

P3V =+3×(−2×10−3)=−6×10−3 =−6mW
(checked using LNAP 8/16/02)
KCL: iR =2+1 ⇒ iR =3A KVL:vR+0−12=0 ⇒ vR=12V
∴ R=vR =12=4Ω iR 3
(checked using LNAP 8/16/02)
KVL: vR +56+24=0 ⇒ vR =−80V KCL:iR+8=0 ⇒ iR=−8A
∴ R=vR =−80=10Ω iR −8
(checked using LNAP 8/16/02)

P3.3-9

3-8

P3.3-10

KCL at node b:
KCLatnodea:

5.61=3.71−5.61+12−5.61 ⇒ 0.801=−1.9+1.278 7R1 5 R1
⇒ R1 = 1.9 =3.983≈4Ω 1.278 − 0.801
3.71+3.71−5.61+3.71−12=0 ⇒ 1.855+(−0.475)+−8.29=0 2 4 R2 R2
⇒ R2 = 8.29 =6.007≈6Ω 1.855 − 0.475
(checked using LNAP 8/16/02)

3-9

Section 3-4 A Single-Loop Circuit – The Voltage Divider P3.4-1
P3.4-2

v = 6 12= 6 12=4V

1

6+3+5+4 18
v2= 3 12=2V;v3= 5 12=10 V

18 183

v4 = 4 12 = 8 V 18 3
(checked using LNAP 8/16/02)

(a) R=6+3+2+4=15Ω (b) i=28=28=1.867A

R 15
(c) p = 28⋅i =28(1.867)=52.27 W

(28 V and i do not adhere to the passive convention.)
(checked using LNAP 8/16/02)

3-10

P3.4-3

(a)i=8=8 ;R1=4=4⋅100=50Ω R2 100 i 8
(b)i=4=4 ;R2=8=8⋅100=200Ω R1 100 i 4
(c)1.2=12i⇒i=0.1A;R=4=40Ω;R =8=80Ω 1i 2i
P3.4-4
P3.4-5

(checked using LNAP 8/16/02)

iR2 =v=8V
12 = i R + v = i R + 8

11

⇒4=iR 1

Voltage division
v = 16 12 = 8 V

1
v3 = 4 12 = 4 V

16+8

4+8
KVL:v −v−v = 0

31 v = −4V

(checked using LNAP 8/16/02)

 100  vs 

usingvoltagedivider:v0= vs ⇒ R=50 −1

100+2R v  o

w i t h v s = 2 0 V a n d v 0 > 9 V , R < 6 1 . 1 Ω  R = 6 0 Ω
withv =28Vandv <13V,R>57.7Ω s 0 

3-11

P3.4-6

a.)  240 18=12V 120+240


b.) 18 18 =0.9W

120+240 

c.)  R 18=2 ⇒ 18R=2R+2(120) ⇒ R=15 Ω  R +120 



d.) 0.2= R ⇒ (0.2)(120)=0.8R ⇒ R=30 Ω R+120 

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Filed Under: event on Wednesday, February 7, 2018