111
i=64=4=A
1 1+1+1+1
6321
1+2+3+6 3
12
i2=1 131 14=3A;
6+3+2+1
1
i3=1 121 14=1A
6+3+2+1
i=14=2A 4 1+1+1+1
i=14=2A 4 1+1+1+1
P3.5-2
632
P3.5-3
(a) 1=1+1+1=1⇒R=2Ω
R 6 12 4 2
(b) v=6⋅2=12V (c) p = 6⋅12 = 72 W
i = 8 or R1 = 8 R1 i
8=R2(2−i) ⇒ i=2− 8 or R2 = 8 R2 2−i
(b) v=6⋅2=12V (c) p = 6⋅12 = 72 W
i = 8 or R1 = 8 R1 i
8=R2(2−i) ⇒ i=2− 8 or R2 = 8 R2 2−i
(a) i=2− 8 =4 A ; R1 = 8 =6Ω
123 43
(b)i=8=2A;R2= 8 =6Ω 12 3 2−2
3
(b)i=8=2A;R2= 8 =6Ω 12 3 2−2
3
3-13
(c)R1 =R2
willcause i=12= 1A. ThecurrentinbothR1 and R2 willbe 1A.
2
2 ⋅
R1 R2
R1 +R2
= 8;
R1 =R2
⇒ 2 ⋅ 1R1 = 8 ⇒ R1 = 8 ∴R1 = R2 = 8Ω
2
P3.5-4
Current division:
i= 8 (−6)=−2A
i= 8 (−6)=−2A
P3.5-5
1
i2= 8 (−6)=−3A
i2= 8 (−6)=−3A
16+8
8+8
i =i −i =+1A
12
R
current division: i2 = 1 is and
R+R
12
Ohm's Law: vo = i2 R2 yields
v R +R
is = o 1 2
R R
2 1
pluggingin R = 4Ω, v >9V gives i > 3.15A
1os
andR =6Ω, v <13Vgivesi <3.47A 1os
Soany 3.15A < is < 3.47A keeps 9V<vo<13V.
andR =6Ω, v <13Vgivesi <3.47A 1os
Soany 3.15A < is < 3.47A keeps 9V<vo<13V.
3-14
P3.5-6
c) 0.4= R
R+12
Section 3-7 Circuit Analysis P3.7-1
Section 3-7 Circuit Analysis P3.7-1
⇒ (0.4)(12)=0.6R ⇒ R=8 Ω
a) 24 1.8=1.2A
12+24
b) R 2=1.6 ⇒ 2R=1.6R+1.6(12) ⇒ R=48 Ω
b) R 2=1.6 ⇒ 2R=1.6R+1.6(12) ⇒ R=48 Ω
R +12
(a)R=16+48⋅24 =32Ω
48+24
32⋅32
(b)v= 32+3224=16V;
(b)v= 32+3224=16V;
8+ 32⋅32
32+32
i = 16 = 1 A
32 2
(c)i2= 48⋅1=1A 48+242 3
(c)i2= 48⋅1=1A 48+242 3
3-15
P3.7-2
(a)R1=4+3⋅6 =6Ω
3+6
(b) 1 = 1+1+1⇒Rp=2.4Ω then R2 =8+Rp =10.4Ω Rp 1266
(c)KCL:i+2=i and −24+6i+Ri=0 21221
⇒ −24+6 (i −2)+10.4i = 0 11
⇒ i1= 36 =2.2A ⇒ v1=i1 R2=2.2(10.4)=22.88V 16.4
(b) 1 = 1+1+1⇒Rp=2.4Ω then R2 =8+Rp =10.4Ω Rp 1266
(c)KCL:i+2=i and −24+6i+Ri=0 21221
⇒ −24+6 (i −2)+10.4i = 0 11
⇒ i1= 36 =2.2A ⇒ v1=i1 R2=2.2(10.4)=22.88V 16.4
1
(d) i2 = 6 (2.2)=0.878A,
(d) i2 = 6 (2.2)=0.878A,
1+1+ 1
6 6 12
v2 =(0.878)(6)=5.3V
(e)i= 6 i=0.585A ⇒ P=3i2=1.03W 3 3+62 3
v2 =(0.878)(6)=5.3V
(e)i= 6 i=0.585A ⇒ P=3i2=1.03W 3 3+62 3
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:
i = 1+1 (1.5)=0.75 A 1 2+(1+1)
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:
i = 1+1 (1.5)=0.75 A 1 2+(1+1)
3-17
P3.7-4
(a)
(b)
(a)
(b)
1 = 1 + 1 +1 ⇒ R =4Ω and R = (10+8)⋅9 =6Ω
R2 24 12 8 2 1 b10+8g+9
First,applyKVLtotheleftmeshtoget −27+6ia +3ia =0 ⇒ ia =3A.Next, apply KVL to the left mesh to get 4 ib − 3ia = 0 ⇒ ib = 2.25 A .
First,applyKVLtotheleftmeshtoget −27+6ia +3ia =0 ⇒ ia =3A.Next, apply KVL to the left mesh to get 4 ib − 3ia = 0 ⇒ ib = 2.25 A .
(c)
i =
2
1
8
1 +1+ 1 24 8 12
1 +1+ 1 24 8 12
2.25=1.125A
bgL9 O
and v =−10MNb g 3PQ=−10V
and v =−10MNb g 3PQ=−10V
1 10+8 +9
3-18
P3.7-5
30 v1=6 ⇒ v1=8V
10+30
R2 12=8 ⇒ R2 =20Ω R2 +10
20= R1b10+30g ⇒ R =40Ω R1 +b10+30g 1
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
R2 12=8 ⇒ R2 =20Ω R2 +10
20= R1b10+30g ⇒ R =40Ω R1 +b10+30g 1
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
P3.7-8
1×10−3 =
12×103 =R = p
12×103 =R = p
⇒ R =12×103 =12 kΩ
p
24
12×103 +R
p
(21×103 ) R (21×103 )+ R
(21×103 ) R (21×103 )+ R
⇒ R=28kΩ
Voltage division ⇒ v = 50 130 500 = 15.963 V
130 500+200+20
∴v =v 100 =(15.963)10=12.279V
h 100+30 13
∴ ih = vh = .12279 A
100
3-20
P3.7-9
3-21
P3.7-10
R = 15(20+10) =10 Ω
eq 15+(20+10)
i=−60=−6A, i= 30 60=4A, v= 20 (−60)=−40V a bc
P3.7-11
i=−60=−6A, i= 30 60=4A, v= 20 (−60)=−40V a bc
P3.7-11
R
eq
30+15R
eq
20+10
a)
Req =24 12= (24)(12) =8Ω
24 + 12
b)
from voltage division:
v =40 20 =100V∴i = 3 =5A
v =40 20 =100V∴i = 3 =5A
100
x 20+4 3 x 20 3
fromcurrentdivision: i= i 8 = 5 A
x 8+8 6
3-22
P3.7-12
9+10+17=36 Ω
a.) 36(18)=12 Ω
36+18
P3.7-13
b.) 36R =18 ⇒ 18R=(18)(36) ⇒ R=36 Ω
36+R
Req =2R(R)=2R 2R+R 3
=v2 =240=1920W 2
Req =2R(R)=2R 2R+R 3
=v2 =240=1920W 2
P
deliv.
to ckt Req 3R
Thus R=45 Ω
P3.7-14
Req=2+1+ (612)+(2 2)= 3+4+1=8Ω
∴i = 40 = 40 = 5 A
Req 8
615
i=i =(5)(3)= 3A fromcurrentdivision
i=i =(5)(3)= 3A fromcurrentdivision
1 6+12
2
i2 =i2+2=(5)(12)=52 A
3-23
Verification Problems
VP3-1
VP3-2
KCLatnodea: i =i +i
312
−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:
6i3+3i2+v +12 =0
yields v=−4.0 V not v=−2.0 V
−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:
6i3+3i2+v +12 =0
yields v=−4.0 V not v=−2.0 V
VP3-3
reduce circuit:
5+5=10 in parallel with 20 Ω gives 6.67Ω
by current division: i = 6.67 5 = 1.25 A
20+6.67
∴Reported value was correct.
v = 320 (24)=6.4V
o 320+650+230
∴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 KCLatleftnode: 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 =−6.49A
∴ Reported values were incorrect.
KVL right loop: − 12 + 0.05iB + 1.2iH = 0 KCLatleftnode: 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 =−6.49A
∴ Reported values were incorrect.
VP3-5
Topmesh: 0=4i +4i +2i +1−i =10(−0.5)+1−2(−2)
aaa2b
Lower left mesh: vs =10+2(ia +0.5−ib )=10+2(2)=14 V
Lowerrightmesh: vs +4ia =12 ⇒ vs =12−4(−0.5)=14V The KVL equations are satisfied so the analysis is correct.
Lower left mesh: vs =10+2(ia +0.5−ib )=10+2(2)=14 V
Lowerrightmesh: vs +4ia =12 ⇒ vs =12−4(−0.5)=14V The KVL equations are satisfied so the analysis is correct.
3-25
VP3-6
Apply KCL at nodes b and c to get:
Apply KCL at nodes b and c to get:
KCL equations:
Nodee: −1+6=0.5+4.5
Nodea: 0.5+ic =−1 ⇒ ic =−1.5mA
Noded: ic +4=4.5 ⇒ ic =0.5mA That's a contradiction. The given values of ia
Nodee: −1+6=0.5+4.5
Nodea: 0.5+ic =−1 ⇒ ic =−1.5mA
Noded: ic +4=4.5 ⇒ ic =0.5mA That's a contradiction. The given values of ia
Design Problems
DP3-1
and ib are not correct.
Using voltage division:
v= R2+aRp
m R +(1−a)R +R +aR
Using voltage division:
v= R2+aRp
m R +(1−a)R +R +aR
1p2p12p
vm =8Vwhena=0 ⇒
R2 =1
R1 +R2 +Rp 3
vm =12Vwhena=1 ⇒
R2 +Rp =1
vm =12Vwhena=1 ⇒
R2 +Rp =1
24= R2+aRp 24
R +R +R
R1 +R2 +Rp 2
The specification on the power of the voltage source indicates
242 ≤1 ⇒ R1+R2+Rp≥1152Ω R1 +R2 +Rp 2
Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives 3R2 = R1 + R2 + 2000 and 2(R2 + 2000)= R1 + R2 + 2000 . Solving these equations gives
R1 =6000Ω andR2 =4000Ω.
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.
242 ≤1 ⇒ R1+R2+Rp≥1152Ω R1 +R2 +Rp 2
Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives 3R2 = R1 + R2 + 2000 and 2(R2 + 2000)= R1 + R2 + 2000 . Solving these equations gives
R1 =6000Ω andR2 =4000Ω.
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
division, 200 12 = 4 ⇒ R2 = 400 Ω . The power required to be dissipated by R2
R2 +200
is 82 = 0.16 W < 1 W . To reduce the voltage across any one resistor, let’s implement R2 as the
is 82 = 0.16 W < 1 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
series combination of two 200 Ω resistors. The power required to be dissipated by each of these
resistorsis 42 =0.08W<1W.
200 8
Now let’s check the voltage:
11.88
Hence, vo = 4 V ± 8% and the design is complete. DP3-3
Now let’s check the voltage:
11.88
Hence, vo = 4 V ± 8% and the design is complete. DP3-3
210
210 + 380
190 < v0 < 12.12
190 + 420
3.700<v0 <4.314
4−7.5%<v0 <4+7.85%
Vab ≅ 200 mV
v= 10 120Vab = 10 (120)(0.2)
v= 10 120Vab = 10 (120)(0.2)
10+R 10+R
letv=16 = 240 ⇒ R=5Ω
10+R
∴P=10 =25.6W
162
DP3-4
NN11
i=G v= v whereG = ∑ =N
∴ N = iR = (9)(12) =18 bulbs v6
∴ N = iR = (9)(12) =18 bulbs v6
T R T n=1Rn R
3-27
28
Chapter 4 – Methods of Analysis of Resistive Circuits
Exercises
Ex. 4.3-1
Ex. 4.3-2
Ex. 4.3-2
va va−vb
KCLata: 3 + 2 +3=0 ⇒ 5va−3vb=−18
vb−va
KCL at b: 2 − 3 −1 = 0 ⇒ v b − v a = 8
Solving these equations gives:
va =3Vandvb =11V
KCL at a:
va+va−vb+3=0 ⇒ 3va−2vb=−12 42
KCLata: 3 + 2 +3=0 ⇒ 5va−3vb=−18
vb−va
KCL at b: 2 − 3 −1 = 0 ⇒ v b − v a = 8
Solving these equations gives:
va =3Vandvb =11V
KCL at a:
va+va−vb+3=0 ⇒ 3va−2vb=−12 42
Ex. 4.4-1
Apply KCL to the supernode to get
2+vb+10+vb =5 20 30
Solving:
vb = 30 V and va = vb +10 = 40 V
Apply KCL to the supernode to get
2+vb+10+vb =5 20 30
Solving:
vb = 30 V and va = vb +10 = 40 V
KCL at a:
Solving:
vb−va−vb−4=0
3 2
⇒ −3va+5vb=24 va =−4/3Vandvb =4V
⇒ −3va+5vb=24 va =−4/3Vandvb =4V
4-1
Ex. 4.4-2
Ex. 4.5-1
Ex. 4.5-1
(vb+8)−(−12)+vb=3 ⇒ vb=8Vandva=16V
10 40
Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into
vb =4ia andsolveforvb .
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.
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.
6 vb 9+vb
+ =i ⇒ v =4i =4
⇒ v =4.5V
812a ba12 b
va−6+va−4va=0 ⇒ va=−2V
20 15
Ex. 4.6-1
Mesh equations:
Solving these equations gives:
−12+6i +3i −i −8=0 ⇒ 9i −3i =20
112 12
8−3i −i +6i =0 ⇒ −3i +9i =−8
8−3i −i +6i =0 ⇒ −3i +9i =−8
122 12
i1=13 A and i2=−1 A
66
The voltage measured by the meter is 6 i2 = −1 V.
4-2
Ex. 4.7-1
Meshequation: 9+3i+2i+4i+3=0 ⇒ (3+2+4)i=−9−3 ⇒ i=−12 A
4 9
The voltmeter measures 3 i = −4 V
Ex. 4.7-2
Meshequation: 15+3i+6(i+3)=0 ⇒ (3+6)i=−15−6(3) ⇒ i=−33=−32 A 93
Ex. 4.7-3
Express the current source current in terms of the mesh currents: 3 = i1 − i2 ⇒ i1 = 3 + i2 . 44
ApplyKVLtothesupermesh: −9+4i +3i +2i =0 ⇒ 43+i +5i =9 ⇒ 9i =6 1224222
so i2 = 2 Aand the voltmeter reading is 2i2 = 4 V 33
Meshequation: 15+3i+6(i+3)=0 ⇒ (3+6)i=−15−6(3) ⇒ i=−33=−32 A 93
Ex. 4.7-3
Express the current source current in terms of the mesh currents: 3 = i1 − i2 ⇒ i1 = 3 + i2 . 44
ApplyKVLtothesupermesh: −9+4i +3i +2i =0 ⇒ 43+i +5i =9 ⇒ 9i =6 1224222
so i2 = 2 Aand the voltmeter reading is 2i2 = 4 V 33
4-3
Ex. 4.7-4
Expressthecurrentsourcecurrentintermsofthemeshcurrents: 3=i1 −i2 ⇒ i1 =3+i2. ApplyKVLtothesupermesh: −15+6i1 +3i2 =0 ⇒ 6(3+i2)+3i2 =15 ⇒ 9i2 =−3
Finally, i 2 = − 1 A is the current measured by the ammeter. 3
Problems
Section 4-3 Node Voltage Analysis of Circuits with Current Sources
Expressthecurrentsourcecurrentintermsofthemeshcurrents: 3=i1 −i2 ⇒ i1 =3+i2. ApplyKVLtothesupermesh: −15+6i1 +3i2 =0 ⇒ 6(3+i2)+3i2 =15 ⇒ 9i2 =−3
Finally, i 2 = − 1 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:
0=v1+v1−v2+i=−4+−4−2+i=−1.5+i ⇒ i=1.5A 8686
(checked using LNAP 8/13/02)
0=v1+v1−v2+i=−4+−4−2+i=−1.5+i ⇒ i=1.5A 8686
(checked using LNAP 8/13/02)
4-4
P4.3-2
P4.3-3
KCL at node 1:
v1−v2+v1+1=0 ⇒ 5v1−v2=−20 20 5
KCL at node 2:
v1−v2+2=v2−v3 ⇒ −v1+3v2−2v3=40 20 10
KCL at node 3:
v2−v3+1=v3 ⇒ −3v2+5v3=30 10 15
Solvinggivesv1 =2V,v2 =30Vandv3 =24V. (checked using LNAP 8/13/02)
KCL at node 1:
v1−v2+v1=i1 ⇒ i1=4−15+ 4 =−2A 520 520
KCL at node 2:
v1 − v 2 + i 2 = v 2 − v 3
v1−v2+v1+1=0 ⇒ 5v1−v2=−20 20 5
KCL at node 2:
v1−v2+2=v2−v3 ⇒ −v1+3v2−2v3=40 20 10
KCL at node 3:
v2−v3+1=v3 ⇒ −3v2+5v3=30 10 15
Solvinggivesv1 =2V,v2 =30Vandv3 =24V. (checked using LNAP 8/13/02)
KCL at node 1:
v1−v2+v1=i1 ⇒ i1=4−15+ 4 =−2A 520 520
KCL at node 2:
v1 − v 2 + i 2 = v 2 − v 3
5 15
⇒ i =−4−15+15−18=2A
⇒ i =−4−15+15−18=2A
2 5 15
(checked using LNAP 8/13/02)
4-5
P4.3-4
Node equations:
Whenv1 =1V,v2 =2V
Node equations:
Whenv1 =1V,v2 =2V
−.003+v1 +v1−v2 =0
R 500
1
−v1−v2+v2 −.005=0
−v1−v2+v2 −.005=0
500 R2
−.003+1+−1 =0⇒R = 1
R 500
1
.003+ 1
500
1
=200Ω
2 =667Ω
− −1 + 2 −.005=0⇒R2 =
500 R2
.005− 1
500
P4.3-5
Node equations:
(checked using LNAP 8/13/02)
v1 +v1−v2 +v1−v3 =0
500 125 250
−v1−v2 −.001+v2−v3 =0 125 250
−v2−v3−v1−v3+v3 =0 250 250 500
Solving gives:
v1 =0.261V, v2 =0.337 V, v3 =0.239 V
Finally, v=v1−v3 =0.022V
(checked using LNAP 8/13/02)
−v1−v2 −.001+v2−v3 =0 125 250
−v2−v3−v1−v3+v3 =0 250 250 500
Solving gives:
v1 =0.261V, v2 =0.337 V, v3 =0.239 V
Finally, v=v1−v3 =0.022V
(checked using LNAP 8/13/02)
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 =−6V
KCL at node b:
va −vb +2=vb −vc ⇒ −6−vb +2=vb −vc ⇒ −1−vb +2=vb −vc ⇒ 30=8vb −3vc 6 10 6 10 6 10
Express the branch voltage of the voltage source in terms of its node voltages: 0−va =6 ⇒ va =−6V
KCL at node b:
va −vb +2=vb −vc ⇒ −6−vb +2=vb −vc ⇒ −1−vb +2=vb −vc ⇒ 30=8vb −3vc 6 10 6 10 6 10
KCLatnodec:
Finally:
P4.4-2
P4.4-2
vb −vc =vc ⇒ 4vb −4vc =5vc ⇒ vb =9vc
108 4
30=89v−3v ⇒ v=2V
4ccc
(checked using LNAP 8/13/02)
Express the branch voltage of each voltage source in terms of its node voltages to get:
va =−12V, vb =vc =vd +8
4-7
KCL at node b:
vb −va =0.002+i ⇒ vb −(−12)=0.002+i ⇒ v +12=8+4000i
vb −va =0.002+i ⇒ vb −(−12)=0.002+i ⇒ v +12=8+4000i
b
0.001= vd +i ⇒ 4=vd +4000i 4000
so vb +4=4−vd ⇒ (vd +8)+4=4−vd ⇒ vd =−4V Consequently vb =vc =vd +8=4Vandi=4−vd =2mA
0.001= vd +i ⇒ 4=vd +4000i 4000
so vb +4=4−vd ⇒ (vd +8)+4=4−vd ⇒ vd =−4V Consequently vb =vc =vd +8=4Vandi=4−vd =2mA
4000 4000
KCL at the supernode corresponding to the 8 V source:
KCL at the supernode corresponding to the 8 V source:
P4.4-3
Apply KCL to the supernode:
Apply KCL to the supernode:
4000
(checked using LNAP 8/13/02)
P4.4-4
va−10+va +va−8−.03=0 ⇒ va=7V
100 100 100
(checked using LNAP 8/13/02)
Apply KCL to the supernode:
va +8+(va +8)−12+va −12+ va =0 500 125 250 500
Solving yields
va =4V
(checked using LNAP 8/13/02)
(checked using LNAP 8/13/02)
Apply KCL to the supernode:
va +8+(va +8)−12+va −12+ va =0 500 125 250 500
Solving yields
va =4V
(checked using LNAP 8/13/02)
4-8
P4.4-5
The power supplied by the voltage source is
(checked using LNAP 8/13/02)
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−vm+vm +0.002+vm−8=0 6000 R 3000
The power supplied by the voltage source is
(checked using LNAP 8/13/02)
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−vm+vm +0.002+vm−8=0 6000 R 3000
v (i +i )=v va −vb +va −vc =1212−9.882+12−5.294
a12a4646
= 12(0.5295 + 1.118) = 12(1.648) = 19.76 W
= 12(0.5295 + 1.118) = 12(1.648) = 19.76 W
That is
3+6000v =16 ⇒ R= 6000
3+6000v =16 ⇒ R= 6000
Rm 16
−3
(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.
vm
4-9
Section 4-5 Node Voltage Analysis with Dependent Sources
P4.5-1
P4.5-2
Express the resistor currents in terms of the
node voltages:
i1=va −vc =8.667−10=−1.333A and 1
i2=vb−vc =2−10=−4A 22
Apply KCL at node c:
i1 +i2 = Ai1 ⇒ −1.333+(−4)= A(−1.333)
⇒ A=−5.333=4 −1.333
(checked using LNAP 8/13/02)
Write and solve a node equation:
va−6+va +va−4va =0⇒va=12V 1000 2000 3000
ib =va−4va =−12mA 3000
(checked using LNAP 8/13/02)
First express the controlling current in terms of
the node voltages:
i = 2−vb a 4000
i1=va −vc =8.667−10=−1.333A and 1
i2=vb−vc =2−10=−4A 22
Apply KCL at node c:
i1 +i2 = Ai1 ⇒ −1.333+(−4)= A(−1.333)
⇒ A=−5.333=4 −1.333
(checked using LNAP 8/13/02)
Write and solve a node equation:
va−6+va +va−4va =0⇒va=12V 1000 2000 3000
ib =va−4va =−12mA 3000
(checked using LNAP 8/13/02)
First express the controlling current in terms of
the node voltages:
i = 2−vb a 4000
P4.5-3
Write and solve a node equation:
−2−vb+ vb −52−vb=0 ⇒v =1.5V
−2−vb+ vb −52−vb=0 ⇒v =1.5V
4000 2000 4000 b
(checked using LNAP 8/14/02)
4-10
P4.5-4
Apply KCL to the supernode of the CCVS to get
12−10+14−10−1+ib =0 ⇒ ib =−2A 422
12−10+14−10−1+ib =0 ⇒ ib =−2A 422
Next
i =10−12=−1
i =10−12=−1
a 4 2 ⇒ r=−2=4V
r i = 1 2 − 1 4 − 1 A
P4.5-5
a2
(checked using LNAP 8/14/02)
(checked using LNAP 8/14/02)
First, express the controlling current of the CCVS in
terms of the node voltages: ix = v2 2
Next, express the controlled voltage in terms of the node voltages:
12−v2 =3ix =3v2 ⇒v2 =24 V 25
terms of the node voltages: ix = v2 2
Next, express the controlled voltage in terms of the node voltages:
12−v2 =3ix =3v2 ⇒v2 =24 V 25
soix =12/5A=2.4A.
(checked using ELab 9/5/02)
4-11
Section 4-6 Mesh Current Analysis with Independent Voltage Sources
P 4.6-1
2i1 +9(i1 −i3)+3(i1 −i2)=0
15−3(i1 −i2)+6(i2 −i3)=0 −6(i2 −i3)−9(i1 −i3)−21=0
or
so
i1 =3A, i2 =2Aandi3 =4A.
(checked using LNAP 8/14/02) Top mesh:
4 (2 − 3) + R(2) + 10 (2 − 4) = 0 so R = 12 Ω.
Bottom, right mesh: 8(4−3)+10(4−2)+v2 =0
sov2 =−28V. Bottom left mesh
−v1 +4(3−2)+8(3−4)=0 sov1 =−4V.
(checked using LNAP 8/14/02)
15−3(i1 −i2)+6(i2 −i3)=0 −6(i2 −i3)−9(i1 −i3)−21=0
or
so
i1 =3A, i2 =2Aandi3 =4A.
(checked using LNAP 8/14/02) Top mesh:
4 (2 − 3) + R(2) + 10 (2 − 4) = 0 so R = 12 Ω.
Bottom, right mesh: 8(4−3)+10(4−2)+v2 =0
sov2 =−28V. Bottom left mesh
−v1 +4(3−2)+8(3−4)=0 sov1 =−4V.
(checked using LNAP 8/14/02)
P 4.6-2
14i1 −3i2 −9i3 =0
−3i1 +9i2 −6i3 =−15
−9i1 −6i2 +15i3 =21
4-12
P 4.6-3
Ohm’sLaw: i2 =−6=−0.75A
8
KVL for loop 1:
Ri1 +4(i1 −i2)+3+18=0
KVL for loop 2 +(−6)−3−4(i1 −i2)=0
⇒ −9−4(i1 −(−0.75))=0
⇒ i1=−3A R(−3)+4(−3−(−0.75))+21=0 ⇒ R=4Ω
(checked using LNAP 8/14/02)
KVL loop 1:
25ia −2+250ia +75ia +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 −100ia+500ib =−4
⇒ ia = −6.5mA, ib = −9.3mA
(checked using LNAP 8/14/02)
Mesh Equations:
mesh1: 2i1 +2(i1−i2)+10=0 mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0 mesh3:−10+4(i3−i2)+6i3 =0
KVL for loop 1:
Ri1 +4(i1 −i2)+3+18=0
KVL for loop 2 +(−6)−3−4(i1 −i2)=0
⇒ −9−4(i1 −(−0.75))=0
⇒ i1=−3A R(−3)+4(−3−(−0.75))+21=0 ⇒ R=4Ω
(checked using LNAP 8/14/02)
KVL loop 1:
25ia −2+250ia +75ia +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 −100ia+500ib =−4
⇒ ia = −6.5mA, ib = −9.3mA
(checked using LNAP 8/14/02)
Mesh Equations:
mesh1: 2i1 +2(i1−i2)+10=0 mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0 mesh3:−10+4(i3−i2)+6i3 =0
P4.6-4
P4.6-5
Solving:
i=i2 ⇒ i=− 5 =−0.294A
17
(checked using LNAP 8/14/02)
(checked using LNAP 8/14/02)
4-13
Section 4-7 Mesh Current Analysis with Voltage and Current Sources
P4.7-1
P4.7-2
mesh 1: i1 = 1 A
2
mesh2: 75i2+10+25i2 =0 ⇒i2 =−0.1A
ib =i1−i2=0.6A
(checked using LNAP 8/14/02)
mesha: ia =−0.25A meshb: ib =−0.4A
vc = 100(ia − ib ) = 100(0.15) =15 V
(checked using LNAP 8/14/02)
mesh2: 75i2+10+25i2 =0 ⇒i2 =−0.1A
ib =i1−i2=0.6A
(checked using LNAP 8/14/02)
mesha: ia =−0.25A meshb: ib =−0.4A
vc = 100(ia − ib ) = 100(0.15) =15 V
(checked using LNAP 8/14/02)
P4.7-3
Express the current source current as a function of the mesh currents:
i1−i2 =−0.5⇒i1 =i2−0.5
Apply KVL to the supermesh:
30i1+20i2+10=0 ⇒ 30(i2−0.5)+20i2 =−10
50i2 −15 = −10 ⇒ i2 = 5 = .1A 50
i =−.4A and v =20i =2V 122
(checked using LNAP 8/14/02) 4-14
Apply KVL to the supermesh:
30i1+20i2+10=0 ⇒ 30(i2−0.5)+20i2 =−10
50i2 −15 = −10 ⇒ i2 = 5 = .1A 50
i =−.4A and v =20i =2V 122
(checked using LNAP 8/14/02) 4-14
P4.7-4
Express the current source current in terms of the mesh currents:
ib = ia −0.02 Apply KVL to the supermesh:
250 ia+100 (ia −0.02)+9 = 0 ∴ia =−.02A=−20mA
vc = 100(ia −0.02) = −4 V
P4.7-5
Express the current source current in terms of the mesh currents:
ib = ia −0.02 Apply KVL to the supermesh:
250 ia+100 (ia −0.02)+9 = 0 ∴ia =−.02A=−20mA
vc = 100(ia −0.02) = −4 V
P4.7-5
(checked using LNAP 8/14/02)
Express the current source current in terms of the mesh currents:
i3−i1=2 ⇒ i1=i3−2
Supermesh: 6i1 +3i3 −5(i2 −i3)−8=0 ⇒ 6i1 −5i2 +8i3 =8 Lower,leftmesh: −12+8+5(i2 −i3)=0 ⇒ 5i2 =4+5i3
Supermesh: 6i1 +3i3 −5(i2 −i3)−8=0 ⇒ 6i1 −5i2 +8i3 =8 Lower,leftmesh: −12+8+5(i2 −i3)=0 ⇒ 5i2 =4+5i3
Eliminating i1 and i2 from the supermesh equation:
6(i3 −2)−(4+5i3)+8i3 =8 ⇒ 9i3 =24
6(i3 −2)−(4+5i3)+8i3 =8 ⇒ 9i3 =24
Thevoltagemeasuredbythemeteris: 3i =324=8V
3 9
(checked using LNAP 8/14/02)
4-15
P4.7-6
Mesh equation for right mesh:
4(i−2)+2i+6(i+3)=0 ⇒ 12i−8+18=0 ⇒ i=−10 A=−5 A 12 6
Mesh equation for right mesh:
4(i−2)+2i+6(i+3)=0 ⇒ 12i−8+18=0 ⇒ i=−10 A=−5 A 12 6
P 4.7-7
(checked using LNAP 8/14/02)
i2 =−3A
i1−i2=5 ⇒ i1−(−3)=5
⇒ i1=2A
2(i3 −i1)+4i3 +R(i3 −i2)=0
⇒ 2(−1−2)+4(−1)+R(−1−(−3))=0 ⇒R=5Ω
(checked using LNAP 8/14/02)
i2 =−3A
i1−i2=5 ⇒ i1−(−3)=5
⇒ i1=2A
2(i3 −i1)+4i3 +R(i3 −i2)=0
⇒ 2(−1−2)+4(−1)+R(−1−(−3))=0 ⇒R=5Ω
(checked using LNAP 8/14/02)
4-16
P 4.7-8
Express the controlling voltage of the dependent source as a function of the mesh current
v2 =50i1 Apply KVL to the right mesh:
Express the controlling voltage of the dependent source as a function of the mesh current
v2 =50i1 Apply KVL to the right mesh:
P 4.7-9
(checked using LNAP 8/14/02)
ib =4ib−ia ⇒ib = 1ia 3
ib =4ib−ia ⇒ib = 1ia 3
−100(0.04(50i)−i)+50i+10=0⇒i =0.2A
1111
v2 =50i1 =10V
−1001i +200i +8 = 0
3a a
⇒ia =−0.048A
⇒ia =−0.048A
P4.7-10
(checked using LNAP 8/14/02)
Express the controlling current of the dependent source as a function of the mesh current:
ib = .06−ia
Express the controlling current of the dependent source as a function of the mesh current:
ib = .06−ia
Apply KVL to the right mesh:
−100(0.06−ia)+50(0.06−ia)+250ia =0 ⇒ ia =10mA
−100(0.06−ia)+50(0.06−ia)+250ia =0 ⇒ ia =10mA
Finally:
vo =50ib =50(0.06−0.01) =2.5V
(checked using LNAP 8/14/02)
(checked using LNAP 8/14/02)
4-17
P4.7-11
Apply KVL to the right mesh:
−100(.006−ia)+3[100(.006−ia)]+250ia = 0 ⇒ ia =−24mA
(checked using LNAP 8/14/02)
(checked using LNAP 8/14/02)
Express the controlling voltage of
the dependent source as a function
of the mesh current:
vb =100(.006−ia)
vb =100(.006−ia)
P4.7-12
applyKVLtoleftmesh: −3+10×103 i +20×103 (i −i )= 0⇒30×103 i −20×103 i =3 (1)
11212
applyKVLtorightmesh: 5×103 i +100×103 i +20×103 (i −i )=0⇒ i = 8i (2) 122112
applyKVLtorightmesh: 5×103 i +100×103 i +20×103 (i −i )=0⇒ i = 8i (2) 122112
63
Solving 1 & 2 simultaneously ⇒ i = 55 mA, i = 220 mA
()()
Power delevered to cathode = (5 i1 ) (i2 ) + 100 (i2 )2
Power delevered to cathode = (5 i1 ) (i2 ) + 100 (i2 )2
12
2
= 5(655)(3220)+100(3220) = 0.026 mW
( −5 ) (3600s ) ∴Energy in24hr. = Pt = 2.6×10 W (24hr) hr
= 2.25J
= 5(655)(3220)+100(3220) = 0.026 mW
( −5 ) (3600s ) ∴Energy in24hr. = Pt = 2.6×10 W (24hr) hr
= 2.25J
4-18
P4.7-13
(a) vo =−gRLv andv= R2 vi ⇒vo =−g RLR2
R1 +R2 vi R1 +R2
v
(b) ∴ o =−g
vi
(b) ∴ o =−g
vi
5×10 10
( 3)(3)
1.1×103
( 3)(3)
1.1×103
=−170 ⇒g=0.0374S
PSpice Problems
SP 4-1
4-19
SP 4-2
From the PSpice output file:
VOLTAGE SOURCE CURRENTS
NAME
V_V1 V_V2
V_V1 V_V2
CURRENT
-3.000E+00 -2.250E+00
V_V3 -7.500E-01
-3.000E+00 -2.250E+00
V_V3 -7.500E-01
The voltage source labeled V3 is a short circuit used to measure the mesh current. The mesh
currents are i1 = −3 A (the current in the voltage source labeled V1) and i2 = −0.75 A (the current
in the voltage source labeled V3).
SP 4-3
The PSpice schematic after running the simulation:
The PSpice output file:
**** INCLUDING sp4_2-SCHEMATIC1.net **** * source SP4_2
V_V4 0 N01588 12Vdc
SP 4-3
The PSpice schematic after running the simulation:
The PSpice output file:
**** INCLUDING sp4_2-SCHEMATIC1.net **** * source SP4_2
V_V4 0 N01588 12Vdc
4-20
The PSpice output file:
R_R4
V_V5
R_R5
V_V6
I_I1
I_I2
N01588 N01565 4k
N01542 N01565 0Vdc
0 N01516 4k
N01542 N01516 8Vdc
0 N01565 DC 2mAdc 0 N01542 DC 1mAdc
0 N01516 4k
N01542 N01516 8Vdc
0 N01565 DC 2mAdc 0 N01542 DC 1mAdc
VOLTAGE SOURCE CURRENTS
NAME
V_V4 V_V5 V_V6
V_V4 V_V5 V_V6
CURRENT
-4.000E-03 2.000E-03 -1.000E-03
-4.000E-03 2.000E-03 -1.000E-03
FromthePSpiceschematic:va =−12V,vb =vc =4V,vd =−4V.Fromtheoutputfile: i=2mA.
SP 4-4
The PSpice schematic after running the simulation:
The PSpice schematic after running the simulation:
VOLTAGE SOURCE CURRENTS
NAME
V_V7 V_V8
V_V7 V_V8
CURRENT
-5.613E-01 -6.008E-01
-5.613E-01 -6.008E-01
The current of the voltage source labeled V7 is also the current of the 2 Ω resistor at the top of
the circuit. However this current is directed from right to left in the 2 Ω resistor while the current
i is directed from left to right. Consequently, i = +5.613 A.
4-21
Verification Problems
VP 4-1
VP 4-2
Apply KCL at node b:
vb −va − 1 + vb −vc = 0 425
−4.8−5.2 −1+−4.8−3.0≠0 425
The given voltages do not satisfy the KCL equation at node b. They are not correct.
Apply KCL at node a:
−vb −va −2+va = 0
vb −va − 1 + vb −vc = 0 425
−4.8−5.2 −1+−4.8−3.0≠0 425
The given voltages do not satisfy the KCL equation at node b. They are not correct.
Apply KCL at node a:
−vb −va −2+va = 0
4 2
−20−4−2+4 = −4≠0
4 2
The given voltages do not satisfy the KCL
equation at node a. They are not correct.
4-22
VP 4-3
VP 4-4
KCL at node 1:
0=v1−v2+v1+1 ⇒ −8−(−20)+−8+1=0 205 205
KCL at node 2:
v1 −v2 =2+v2 −v3 ⇒ −8−(−20)=2+−20−(−6)
20 10 20 10 ⇒ 12=6
20 10
KCLatnode3: v2 −v3 +1=v3 ⇒ −20−(−6)+1=−6 ⇒ −4=−6 10 15 10 15 1015
KCL is satisfied at all of the nodes so the computer analysis is correct.
VP 4-4
KCL at node 1:
0=v1−v2+v1+1 ⇒ −8−(−20)+−8+1=0 205 205
KCL at node 2:
v1 −v2 =2+v2 −v3 ⇒ −8−(−20)=2+−20−(−6)
20 10 20 10 ⇒ 12=6
20 10
KCLatnode3: v2 −v3 +1=v3 ⇒ −20−(−6)+1=−6 ⇒ −4=−6 10 15 10 15 1015
KCL is satisfied at all of the nodes so the computer analysis is correct.
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