Linear circuit analysis - solution manuel (R. A. DeCarlo and P. Lin) (z-lib.org).pdf

PROBLEM SOLUTIONS CHAPTER 1.
Solution 1.1. (a) Charge on one electron: -1.6019×10-19
C. This means that charge on 10
13
electrons
is: -1.6019×10
-6
C. Net charge on sphere is: 1.6019×10-6
C (POSITIVE).
Solution 1.2. (a) 1 atom ≡ -4.646×10
-18
C. By proportionality, 64g ≡ NA atoms.
3.1g ≡ ? atoms ⇒ 3.1g ≡
3.1NA
64
atoms.
Total Charge = −4.646 ×10
−18 C
atom
×
3.1× 6.023 ×1023
64
atoms =−1.3554 ×10
5
C
(b) Total charge per atom is -4.646×10
-18
C. Total charge per electron is –1.6019×10-19
C. Therefore,
there are 29 electrons per atom of copper.
(c) 0.91 A ≡ 0.91 C/s. i =
∆Q
∆t
⇒ ∆t =
∆Q
i
=
1.36 ×105
0.91
=1.49 ×10
5
sec.
(d) We know there are
3.1NA
64
= 2.9174 ×10
22
atoms in the penny. Removing 1 electron from
0.05 ×
3.1NA
64
atoms means removing 0.05 ×
3.1NA
64
electrons. Therefore,
Net charge = 0.05 ×
3.1NA
64
×1.6019 ×10
−19
= 234C
Solution 1.3 (a) 7.573 ×10
17
× −1.6019 ×10
−19
( )= −0.1213C
(b) Current =
0.1213
10
−3 = 121.3A flowing from right to left.
(c) Again, use proportionality:
10 A =
x ×1.6019 ×10−19
60sec
⇒ x =
10 × 60
1.6019 ×10
−19 = 3.75 ×10
21
(d) i t
( )=
dq
dt
= 1− e
−5t
A. This is an exponential evolution with an initial value of 0, a final value of 1,
and a time-constant of 1/5 (signal reaches ~63% of it’s final value in one time-constant).

0.2 time in sec
i(t)
1
(e) Current is the slope of the charge waveform. Therefore, by inspection:
Solution 1.4 (a) 6.023×10
23
×(-1.6019×10
-19
) = –9.65×10
4
C.
(b) Current flows from right to left (opposite electrons), and:
I =
9.65 ×104
10
−3 = 9.65 ×10
7
A

(c) Using proportionality:
5A =
x × 1.6019 ×10−19
60sec
⇒ x =
5 × 60
1.6019 × 10
−19 =1.87 ×10
21
(d) i t
( )=
dq
dt
= 1+ 0.5πcos πt
( ) ⇒ i 1sec
( )= 1− 1.57 = −0.57A. Current flows from left to right.
Solution 1.5 (a) i t
( )= 1− 4e
−2t
+ 3e
−3t
t ≥ 0. Then
q t
( )= i t
( )dt
0
t
∫ = 1 − 4e
−2τ
+ 3e
−3τ
( )d
0
t
∫ τ = τ]−∞
t
− 4 e
−2τ
dτ
0
t
∫ + 3 e
−3τ
dτ
0
t
∫
= t − 4 −0.5e
−2τ
[ ]0
t
+ 3 −0.333e
−3τ
[ ]0
t
= t + 2e
−2t
− e
−3t
−1
(b) By inspection:
(c) q t
( )= 120cos 120πt
( ). Hence
i t
( )=
dq
dt
= −120π× 120sin 120πt
( ) =−14400πsin 120πt
( ) A
Solution 1.6. (a) i(t) = 1 − cos(πt) A. Hence
q(t) = i(τ)dτ
0
t
∫ = 1 − cos(πτ)
( )dτ
0
t
∫ = t −
1
π
sin(πτ)



0
t
= t −
1
π
sin(πt) C
(b) Charge is integral of current. Graphically, the charge at time t is the area under the current curve up to
time t: (note the quadratic nature between 2 and 4 seconds)

Solution 1.7
Again, Q is the running area under the current curve. Between 0 and 3 seconds, current decreases linearly
until zero. So, Qtot = 7.5 C. From 0 to 6: Qtot = 7.5 + Q3_6 = 7.5 -1/1×0.5 + -1/1×0.5 + -1×1 = 5.5 C,
where the curve from 3 to 6 was divided into two triangular sections and one rectangular one.
Solution 1.8 Charge is the area under the current curve. Thus, Q = 0.1*4 – 0.1*2 = 0.2 C.
Solution 1.9 Calculate the change in energy for the electron: ∆E = Q ∆V = 3.218×10
-15
.
Equate this to kinetic energy:
3.218 ×10
−15
=
1
2
mv
2
⇒ v = 8.4 ×10
7
m / s
where the mass of an electron, 9.1066×10
-31
has been substituted.
Solution 1.10 P = VI. Hence I = P/V = 2×10
3
/120 = 16.6667 A
PROBLEM Solution 1.11 (a) It is necessary to integrate the i(t) curve to obtain q(t). We do this
interval by interval:
(i) 0 ≤ t ≤ 1 ms, q(t) = 0 + dτ
0
t
∫ = t µC
(ii) 1 ms ≤ t ≤ 2 ms, q(t) = 1 − 2 dτ
1
t
∫ = 3 − 2t µC
(iii) 2 ms ≤ t ≤ 3 ms, q(t) = −1+ dτ
2
t
∫ = −3 + t µC

(iv) 3 ms ≤ t ≤ 5 ms, q(t) = 0 + 8 − 2τ
( )dτ
3
t
∫ = 8t − t
2
−15 µC
(v) 5 ms ≤ t, q(t) = 0 µC
0 1 2 3 4 5 6
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Charge
in
micro
C
TextEnd
Time in ms
(b) Voltage is the ratio of the power and current curve. In this case, the division can be done graphically
by inspection. Note that the ratio of a quadratic function and a linear function is a linear function:

0 1 2 3 4 5 6
-1
-0.5
0
0.5
1
1.5
2
Voltage
in
V
TextEnd
Time in ms
Solution 1.12 (a) VA = P/I = 20/4 = 5 V
(b) PB = VI = 2×7 = 14 W
(c) VC = P/I = -3W/3A = -1V
(d) ID = P/V = -27W/3V = -9A
(e) IE = P/V = 2/1 = 2A
(f) PF = VI = -4×5 = -20W
In all of the above, note that the direction of the current flow relative to the polarity of the voltage across a
device determines whether power is delivered or absorbed. Power is absorbed when current flows from the
positive terminal of the device to the negative one.
Solution 1.13 (a) By inspection: Circuit Element (CE) 1 absorbs –5W, and CE 2 absorbs 6W.
(b) Compute power absorbed by all elements including independent sources:
I3A: -15
CE1: -5
V3V: -12
CE2: +6
V5V: 10
I2A: 16
----------

Sum: 0 (Verifies conservation of power.)
Solution 1.14 (a) Compute power absorbed:
I5A: -85
CE1: 98
V3V: 33
CE2: 16
V7V: -42
I2A: -20
-------------
Sum: 0
(b) Add all terms:
I-source: Pabsorbed = −3 1− e
−t
( )= −3 + 3e
−t
watts
V-source: Pabsorbed = −2 3e
−t
−1
( )= −6e
−t
+ 2 watts
CE1: Pabsorbed = 3e
−t
× 3 1− e
−t
( )= 9e
−t
− 9e
−2t
watts
CE2: Pabsorbed = 3e
−t
−1
( )3e
−t
−1
( )= 9e
−2t
− 6e
−t
+1 watts
Simple algebraic manipulation of the the sum of all the above terms reveals that the result is zero.
Solution 1.15 (a) When IL = 1, P = VLIL = (16-4)×1 = 12 W. When IL = 2, P = VLIL = (16-16)×1 = 0.
(b) P = (16-4IL
2
)IL. Differentiate this w.r.t. IL and set to zero: 16 – 12IL
2
= 0. Therefore, IL = 1.155A.
Solution 1.16 (a) When IL = 2, P = (16-4)×2 = 24W. When IL = 3, P = (16 - 9)×3 = 21 W.
(b) Maximum occurs in the interval from 0 to 4: P = (16 - IL
2
) IL
Differentiate w.r.t. IL and set to zero: 16 – 3IL
2
= 0.
Therefore, IL = 2.31 A.
Solution 1.17 (a) Power is the product of the current and voltage. We can compute the product
graphically:

0 0.5 1 1.5 2 2.5 3
0
2
4
6
8
10
Power
in
Watts
TextEnd
Time in s
(b)
W t
( ) = p t
( )dt
0
t
∫ = 10 −10e
−7τ
( )dτ
0
t
∫ = 10τ]0
t
− −10
7 e
−7τ
[ ]0
t
=10t +
10
7
e
−7t
−
10
7
This can be used as an aid to plot the work function:
0 0.5 1 1.5 2
0
1
2
3
4
5
6
7
8
9
Time in s
Energy
in
J
TextEnd
Solution 1.18 (a) Since, i t
( )= 115 − 23t mA ,

q 7
( )= i τ
( )dτ
0
7
∫ = 115t −
23t2
2



0
7
×10
−3
= 0.2415 C
(b) Energy is the integral of power:
E = p t
( )dt
0
7
∫ = v t
( )× i t
( )dt
0
7
∫ = 25 i t
( )dt
0
7
∫
= 25 × 0.2415 = 6.0375 C
Solution 1.19 (a) ∆t = 100o
F, Rate of temp. increase is 2.5 Wh/o
F per gallon:
Energy = 2.5Wh/o
F/gallon×100o
F×30gallons = 7500 Wh = 2.7×107
J.
(b) Heater generates P = 120×10 = 1200 W. We want 7500 Wh. Therefore, the total number of hours
needed is 7500Wh/1200W = 6.25 h.
Solution 1.20 First compute the change in temperature required, in o
F:
∆t = 80-25 = 55o
C = 55×9/5 o
F= 99o
F
Next, compute the energy spent every hour, which means on 40 gallons of water:
E = 2.5 Wh/o
F/gallon×99o
F×40gallons = 9900 Wh
Since the heater is not 100 % efficient, we spend more energy than is actually needed to heat the water:
E_spent = 9900 Wh/0.9 = 11000 Wh
So, far, this was the energy spent every hour. Over six hours, the total energy spent is:
E6h = 11000×6 = 66,000 Wh
Finally, the total energy spent per month is Em = 66,000×30 = 1980 kWh
and the bill is 1980 kWh×0.14$/kWh = $277.2
Solution 1.21
Energy = 120 W × 6 h = 720 Wh = 0.72 kWh
Therefore, cost per day = 0.72 kWh × 8 = 5.76 cents, and cost per month is 5.76×31 = $1.785.
Solution 1.22

We need to compute the difference between the inner diameter of the tube and the outer one in order to get
the cross-sectional area:
area = πRout
2
−πRin
2
= π0.003
2
− π0.0018
2
=1.81 ×10
−3
m
2
Then, R = 1.7×10-5
×(12/1.81) = 11.3 mΩ.
Solution 1.23 L = 20 m, W = 0.015m, H = 0.001 m. Thus, A = W×H, and R = 5.1× copper×L/A
= 0.116 Ω.
Solution 1.24. (a) 500 ft, 20 gauge wire: 10.35 Ω/1000 ft from table 1.3. This implies that
R = 5.175 Ω.
(b) 55 ft, 20 gauge, nickel wire:
R = 5.1×
10.35
1000
× 55 = 2.9 Ω
(c) Rtot = 2.9 + 5.175 = 8.08 Ω.
Solution 1.25. R T
( ) = R 20
( )1 +α T − 20
( )
[ ]. Substituting at T = –10 yields:
21 = R 20
( ) 1+ 0.0039 −30
( )
[ ] or R 20
( )= 23.78 Ω
Evaluating at T = +10 yields,
R 10
( ) = 23.78 + 23.78 × 0.0039 × −10
( ) or R 10
( ) = 22.85Ω
Solution 1.26. For tungsten, we know that α = 0.0045. Therefore:
R 150
( ) = R 20
( )1 +α T − 20
( )
[ ] = 200 1 + 0.0045 150 − 20
( )
[ ]
= 317Ω
Rate of change of resistance is (317-200)/(150-20) = 0.9 Ω/o
C.

Solution 1.27. Plug numbers directly into the same formula as problem 1.26:
0.0022 = 0.002 + 0.002×0.0039(T-20)
Rearrange to obtain: T = 45.64o
C.
Solution 1.28. (a) Power in a wire: P = I
2
R. Rearranging, we can express the current as
I = P
R .
Substitute given P and R to obtain I = 0.707 mA.
(b) Use the same formula for current obtained above to get 50 A.
Solution 1.29. Use formula for power: P = V
2
/R. Rearranging, R = V
2
/P = 96 Ω.
Solution 1.30 (a) I = V/R = 12 A, out of the positive terminal of the battery.
(b) Up through the resistor.
(c) Absorbed power by resistor: P = V
2
/R = 14.4 W. Same power is delivered by source.
(d) From table 1.2 and 1.3, 1000 feet of 18 AWG aluminum wire has resistance:
»R1000ft = 1.6*6.51
R1000ft = 1.0416e+01
By proportionality, 1000 × 0.1 = L ×10.416 . Hence,
»L = 100/10.416
L = 9.6006e+00 meters.
Solution 1.31 (a) V = 10 V.
(b) P = V
2
/R, which means that R = V
2
/P = 100/25 = 4 Ω.
(c) I = V/R = 10/4 = 2.5 A. Current flow is downwards through resistor.
(d) Up through resistor.
(e) P = V
2
/R10 = 100/10 = 10 W. Hence, I10 = V/R10 = 1 A. Without applying material from a future
chapter, a legitimate way to obtain Isource is to apply conservation of power first and then compute Isource
from the power formula. Hence, Psource =10 + 25 = 35 watts. Using material from a later chapter, in
particular KCL, we may conclude that, Isrc = 2.5 + 1 = 3.5 A. Thus, Psource = VIsource = 10×3.5 = 35 W.
This approach indicates that power is conserved.

Solution 1.32 (a) From 0 to 1 s, i(t) = 10
-3
t. Thus, i
2
R = 10
−6
t
2
R is the power absorbed during this
interval. Integrating this expression for the power from 0 to 1 s gives us the total energy used:
10−6
t3
R
3
0
1
= 5000
10−6
3
= 0.001667 J.
Finally, we need to multiply this by 2 to account for the interval from 1 to 2 seconds. Thus, the total energy
spent is 3.33 mJ.
(b) The same charge that got transported in one direction during the interval from 0 to 1 is being
transported back in the interval from 1 to 2 (by symmetry). Therefore, total charge transfer is zero.
Solution 1.33. (a) 60 W + 120 W = 180 W.
(b) P = IV þ I = P/V = 180/12 = 15 A.
(c) P = Energy/Time þ Time = 1.2 MJ/180 W = 6.67×10
4
sec = 1.85 h.
Solution 1.34. P = I
2
R. Therefore, 325 = 25×(5+4+2R). Solving for R, yields R = 2 Ω.
Solution 1.35. (a) Use definition of power and substitute given power:
V2 = P × R = 98 × 2 = 14 V
Similarly, I3 =
P
R
=
12
3
= 2 A, V4 = P× R = 16 × 4 = 8 V, I5 =
768.8
5
= 12.4 A, and
V6 = 486 × 6 = 54 V.
(b) Ptot = Pdissipated = 98 + 12 + 16 + 768.8 + 486 = 1380.8 W.
(c) Vin = V2 + V6 = 68 V. Iin + I3 = I5 + I4 and I4 =
V4
4
. Thus, Iin = I5 + I4 – I3 = 12.4 A.
Solution 1.36. (a) Sources are the top, right-most, and bottom left. The reason is that current flows out
of the positive terminal of the device.
(b) The 32/16 element is a 2 Ω resistor. The 54.5/18.167 element is a 3Ω resistor. The 13/2.167 element
is a 6 Ω resistor. The 93/2.833 element is a 32.827 Ω resistor. The 24/5 element is a 4.8 Ω resistor.

Solution 1.37. Power: 12 = Ix
2
R, which means that R = 12/Ix
2
. Now, analyze the loop: 16 = Ix(R+4).
Substitute the value of R into this expression: 16 = Ix
12
Ix
2 + 4Ix . Hence: Ix
2
− 4Ix + 3 = 0 .
This equation has two solutions: one is at Ix = 1 A or R = 12 Ω. The other is at Ix = 3 A or R = 4/3 Ω.
Solution 1.38. (a) Conservation of power:
16Ix = 4Ix
2
+12 + 10 − 6
Hence
0 = Ix
2
− 4Ix + 4 = Ix − 2
( )2
Thus, Ix = 2 A.
(b) 32Ix = 4Ix
2
+ 28 ⇒ Ix
2
−8Ix + 7 = Ix − 7
( ) Ix −1
( )= 0 . Hence, Ix = 7A or Ix = 1A.
Solution 1.39. (a)
(i) AA: I = 36/12 = 3A
BB: I = 24/12 = 2A
CC: I = 14.4/12 = 1.2A
(ii) Sum = 6.2A
(iii) P = VI = 6.2×12 = 74.4W. This is equal to the sum of the powers absorbed by the bulbs.
(iv) R = V/I
AA: R = 12/3 = 4Ω
BB: R = 12/2 = 6Ω
CC: R = 12/1.2 = 10Ω
(b) Each AA bulb draws 3 A. Thus, up to five bulbs can be connected without blowing the fuse (5×3=15).
So, 6 or more would blow the fuse.
(c) Similar analysis suggests that 13 or more bulbs would blow the fuse. Intuitively, the bulbs draw less
current, so more of them can be used.
Solution 1.40. (a)
p(t) = i
2
t
( )R = 20cos 2πt
( )
[ ]2
×10 = 4000
1 + cos 4πt
( )
2
= 2000 1 + cos 4πt
( )
( ) W

(b)
W t
( ) = p t
( )dt
0
t
∫ = 2000t + 2000 cos 4πt
( )dt
0
t
∫ = 2000t +
2000sin 4πt
( )
4π
J
0 0.5 1 1.5 2 2.5 3
0
1000
2000
3000
4000
5000
6000
Energy
in
J
TextEnd
Time in s
Solution. 1.41. When the switch is closed, a constant current of 5/10000 = 0.5 mA flows through the
circuit. When the switch is open, no current flows. So, 50% of the time, a 0.5 mA current flows, and the
other 50% no current flows. The average current is therefore 0.25 mA.

Solution. 1.42 When the switch is at A, the current is 5/5000 = 1 mA. When the switch is at B, the
current is 5/10000 = 0.5 mA. Now, the switch is at position A 20% of the time (1ms out of a 5ms period,
after which the events repeat). So, the average current is 0.2*1 mA + 0.8*0.5 mA = 0.6 mA.
Solution. 1.43 The current in the load resistor is 2 A. So, the power is 2
2
×RL = 8 W.
Solution 1.44. Vin = IinR1 è Iout = µVin/R2 = µIinR1/R2.
Solution 1.45 (a) I1 = Vin/R1. Hence, Vout = αVinR2/R1.
(b)
Vout
Vin
=
αR2
R1
=
100 ×10
R1
= 5 ⇒ R1 = 200 Ω
(c)
Power − gain =
α
2
Vin
2
R1
2 R2
Vin
2
R
1
=
α2
R2
R1
= 500
Solution 1.46 (a) V1 = 200 mA × 5 Ω = 1 V implies V2 = 0.8×8 = 6.4 V. Hence
Vout = 5×6.4 = 32 V and Iout = 32/64 = 0.5 A.
(b) Current Gain = 0.5/0.2 = 2.5.
(c) Power values for the 5, 8, and 64 Ω resistors are, respectively, P5 = 0.2 W, P8 = 5.12 W, P64 = 16 W.
Solution 1.47 (a) I1 = 5A è I2 = 3×5/3 = 5 A è Iout = 25 A, Vout = 50 V.
(b) Voltage Gain = 5.
(c) Pin = 5×5×2 = 50 W, P1 = 5×5×3 = 75W, P2 = 25×50 = 1250W.
Solution 1.48 I1 = Vin/10 = 0.1 A, VR = 10×(Vin/10)×R = R; Vout = 5R×10 = 50R =50RVin
Vout/Vin = 50R. If we want Vout/Vin to be 150, R has to be 3 Ω.

SOLUTIONS CHAPTER 2
SOLUTION 2.1. Using KCL at the center node of each circuit:
(a) I3 = I2 − I1 = −1 − 2 = −3A
(b) I3 = I1 + I2 − I4 = 2 −1 − 0.5 = 0.5A
SOLUTION 2.2. KCL at the bottom node gives I1 = −7 − 8 = −15A , and at the right node
I4 =−6 − 8 = −14 . From these, KCL at the top node gives I3 = I4 − 5 = −19A, and finally at the central
node gives I2 = 6 + I3 − 7 = −20 A,
SOLUTION 2.3. Use a gaussian surface on the top triangle. Performing KCL around this surface yields
1A − 2A + 3A + 4 A− 5A = I = 1A.
SOLUTION 2.4. Use a gaussian surface around the bottom rectangle. KCL yields
I1 = 2A +10A + 3A = 15A.
SOLUTION 2.5. Using KVL, V1 = 55V −15V + 105V −100V − 30V = 15V.
SOLUTION 2.6. Using KVL, Vx = 5V −1V −1V −1V +1V −1V = 2V.
SOLUTION 2.7. Using KVL once again.
v1 = 7 + 6 + 5 = 18V
v2 = 6 + 7 −8 = 5V
v3 = −5 − 6 = −11V
v4 = 8 − 7 =1V
SOLUTION 2.8. KVL is used to find the voltage across each current source, and KCL to find the current
through each voltage source.
I3V = 6A − 7A = −1A
I4V = I3V + 8A = 7A
I5V =−8A− 6A = −14A
V7 A = 4V + 3V = 7V
V8A = −4V + 5V = 1V
V6 A = V8 A − 3V = −2V

Chap 2 Probs P2 - 2 © R. A. DeCarlo, P. M. Lin
SOLUTION 2.9. Using the same method as before, the current and voltages are found through and across
each sources.
I5V = 9 +8 − 7 = 10 ⇒ P = 50W
I4V = −6 − I5V = −16 ⇒ P =−64W
I2V = 6 − 7 = −1A ⇒ P = −2W
I3V = −I2V − 9 = −8A ⇒ P =−24W
V8A = 4 − 5 = −1V ⇒ P =−8W
V9 A = 3 + V8 A = 2V ⇒ P = 18W
V7 A = 2 − V9 A = 0 ⇒ P = 0W
V6 A = 5 − V7A = 5 ⇒ P = 30W
Summing all the power give 0W, hence conservation of power.
SOLUTION 2.10. Doing KVL around the right loop does not balance out. Changing 8V to 5V would fix
this.
SOLUTION 2.11. Using KVL to determine the voltages, and KCL to determine the currents:
Vy = 8V
Vx = Vy − 4 = 4V
Ia = 4 A
Iy = 4 −14 + 2Ia = −2A
Ix = Ia − Iy = 6A
SOLUTION 2.12. First Vin = I2 ⋅ 8Ω = 24V. Then I1 = Vin / 3Ω = 8A and I3 = 12A − I1 − I2 =1A.
Therefore RL = Vin / I3 = 24Ω ⇒ P = I3 ⋅ Vin = 24W
SOLUTION 2.13. (a) First, from current division, get
I1 =
1/ 3
1/ 3 +1/ 6 + 1/ RL





 ⋅ 12 − aI1
( )⇒ I1 =
12 / 3
1 + a
( )/ 3 +1/ 6 +1/ RL





 .
(b) Using the previous equation and solving for 1/ RL = 12/ 3I1
( )−1/ 6 − 1+ a
( )/ 3 = 0.5S or RL = 2Ω.
The power P = I3
2
RL =
1/ RL
1/ 3 +1/ 6 +1/ RL





 ⋅ 12 − aI1
( )





 ⋅ RL = 18W
SOLUTION 2.14. For the power delivered by the source to be 60W, the voltage across it should be
V = P / 2A = 30V. Therefore the current through the 20Ω must be I20Ω = 30 / 20 = 1.5A, and by KCL the
current through IRL
= 2 − I20Ω = 0.5A. From this, RL = V / IRL
= 60Ω.

SOLUTION 2.15. Writing KVL around the loop 25V − 4I −15V − 5I − I = 0 ⇒ I =1A, and
P5Ω = I
2
R5Ω = 5W
SOLUTION 2.16. The total power supplied by the source is P = 50V ⋅ 0.5A = 25W . The power absorbed
by the resistor is P60Ω = 0.5A
( )2
⋅ 60Ω = 15W . Therefore by conservation of power, the power absorbed by
X is 10W.
SOLUTION 2.17. (a) As this loop is open, no current flows through it, so IR is 0A. The output voltage is
VOUT = −2V + 3V − 2V = −1V by KVL.
(b) Writing out the KVL equation around the loop 3 − 2 − IRR − 2 − IR2R − IRR = 0 ⇒ −1 = IR 4R.
Therefore IR =−1/ 4R and VOUT = IR R = −1/ 4V .
SOLUTION 2.18. Writing out KVL around the loop 60 − 30I − 30 − 20 + 60 − 40I = 0 ⇒ I = 1A. From
ohm's law R = V / I = 30Ω.
SOLUTION 2.19. (a) Using Ohms law Iin = V2 / 20 +12
( )= 0.75A, and V1 =12 ⋅ Iin = 9V . To find R, write
KCL and get VR = 30 − V2 = 6V . Therefore using Ohms law again, R = VR / IIN = 8Ω.
(b) Writing KVL around the loop, 30 = aV1 + IinR + Iin 20 + V1, and substituting Iin = V1 / 12,
V1 = 30 / R + 32
( ) /12 + a
[ ] is obtained. Next substitute back V1 =12Iin and solve for
R =
30
12IIN
− a





 ⋅12 − 32 = 40Ω
SOLUTION 2.20. (a) i. Using R = Vxy / Ibat the value of each resistors starting with the top one are 2.7Ω,
0.6Ω, and 0.25Ω. Using the same relationship, the resistance for the motor is 1.25Ω.
ii. Using P = Vxy
2
/ R the power dissipated by each resistor is 16.875W, 3.75W, 1.5625W, and for the
motor 7.8125W.
iii. The relative efficiency is = 7.8125/ (12⋅2.5) ⋅100 = 26 %
(b) i. Performing voltage division across each resistor
VAB = 0
VBC =12 ⋅ RBC / (RBC + RCD + Rmotor ) = 3.43V
VCD = 12⋅ RCD / (RBC + RCD + Rmotor ) =1.43V
Vmotor = 12 ⋅ Rmotor / (RBC + RCD + Rmotor ) = 7.14V

ii. Ibat = 12 / (RBC + RCD + Rmotor ) = 5.71A
iii. The relative efficiency is = (Vmotor
2
/ Rmotor)/ (12 ⋅5.71)⋅100 = 59.5 %
(c) i. Repeating the steps from (b), the voltages across the first two resistance are 0, then across the other
and the motor 2V, and 10V
ii. Ibat = 12 / (RCD + Rmotor) = 8A
iii. And the relative efficiency is = (Vmotor
2
/ Rmotor)/ (12 ⋅8)⋅100 = 83.3 %
(d) What is the largest equivalent resistance of the motor that will draw 30A? R = 12/ 30A = 0.4Ω.
SOLUTION 2.21. (a) Observe that i =− IO , thus v = ki
3
= −kI0
3
.
(b) Using KVL and previous equation, vx = (R1 + R2)IO + VO + kIO
3
.
(c) The power is = IOvx = (R1 + R2)IO
2
+ VOIO + kIO
4
SOLUTION 2.22. I100Ω =
0.04
100
= 0.02 A. Therefore V300Ω = 0.02 × (100 + 200) = 6 V. By KCL,
I150Ω = 0.02 +
6
300
= 0.04 A. Req, seen by the source, is 300 Ω. Therefore Vs = 0.04 × 300 = 12 V.
SOLUTION 2.23. Using KCL IR = 5 − 3 = 2A, and KVL VR = 10 + 6 = 16V . Thus
R = VR / IR =16 / 2 = 8Ω.
SOLUTION 2.24. Using KCL, KVL, along with Ohm’s law,
I5Ω
= 6 − 7 = −1A
I4V = 8 − I5Ω = 9A
I2Ω = 8 + 7 =15A
V6 A = 4 + 5I5Ω = −1V
V8A = −4 + 2I2Ω = 26V
V7 A = V8 A − 5I5Ω = 31V
Now, the power delivered or absorbed by each element is calculated:
P6A = I6A ⋅V6 A = −6W
P7A = I7A ⋅V7 A = 217W
P8A = I8A ⋅ V8A = 208W
P4V = I4V ⋅ 4 = 36W
P5Ω = I5Ω
2
⋅5 = 5W
P2Ω = I2Ω
2
⋅ 2 = 450W

Note that for passive elements, when the power is positive it is absorbed, while for independent sources it is
generated when the power is positive.
SOLUTION 2.25. Note that I1 = 6A. Thus by KCL
I3 = 6 − 0.5I1 = 3A
I2 = 2 + 0.2I1 = 3.2A
I4 = 8 − 0.3I1 = 6.2A
And finally using KVL
V2 = 8A⋅1+ 4I4 + 3I3 = 41.8V
V1 = 2I2 −3I3 = −2.6V
SOLUTION 2.26. (a) Using KCL,
I4 = 5 − 4 =1A
I3 = I4 − 2 = −1A
I2 = 3 − 2 = 1A
I1 = −I2 − 5 = −6A
(b) Using KVL and Ohm’s law,
V
I
V
V
I
V
V
I
V
V
I
V
4
4
10
10
12
12
18
3
4
4
3
3
2
2
1
1
=
=
−
=
=
=
=
−
=
=
(c)
W
V
V
V
P
W
V
V
P
W
V
V
V
P
W
V
V
A
P
A
A
A
A
60
)
(
5
24
)
(
4
40
)
(
2
90
)
(
3
1
3
4
5
4
3
4
3
2
1
2
1
2
3
=
−
+
=
=
−
−
=
−
=
−
−
=
=
−
=

SOLUTION 2.27. Write KVL around the outside loop, 40 = 500Ix + (400 + 200)i. And write KCL
equation i = Ix − 2Ix . Solving yields Ix = −0.4A. The dependent source delivers 2Ix ⋅(−600i) = 192W , and
the independent 40Ix = −16W . Finally the resistors absorb 500Ix
2
+ 400i
2
+ 200i
2
=176W verifying the
conservation of energy since the source generate 192W-16W=176W.
SOLUTION 2.28. By voltage division V2 =
[((90 ||180) + 60)||40]
[((90||180) + 60)||40] +160
⋅
60
60 + (90||180)
⋅ Vs = 1/14 ⋅Vs .
Therefore Vs = 14V2 = 280V .
SOLUTION 2.29. By voltage division
vx = 9V ⋅
18 + 3
(18 + 3) + 6
= 7V
SOLUTION 2.30. By voltage division we get the following two equations in order to solve for the two
unknowns.
V2 = V1 ⋅
R1
R1 + R2
V1 =100V ⋅
R1 + R2
R1 + R2 + 60
Solving yields R1 = 40Ω, and R2 = 100Ω.
SOLUTION 2.31. Dividing 1400 in four gives 350. If we only need 1/4 and 2/4, the resistor string can be
made of three resistances: 350Ω, 350Ω, and 700Ω.
SOLUTION 2.32. Using voltage division, at t=0 vR = 15⋅
2R
3R
= 10V , and t = 5 s vR = 10V , and at t = 10
the voltage goes back to 0V.

0 5 10 15
0
1
2
3
4
5
6
7
8
9
10
time (sec)
Volts
Vb =
Rb
Ra + Rb
Vin and Vd =
Rd
Rc + Rd
Vin
By KVL, if Vout = 0, then
0 = Vout = Vb − Vd =
Rb
Ra + Rb
−
Rd
Rc + Rd





Vin
For arbitrary Vin, this requires that
Rb
Ra + Rb
=
Rd
Rc + Rd
or equivalently that Rb Rc = Ra Rd .
SOLUTION 2.34. First Geq = 1m +1.5m + 2m + 3m = 7.5mS . By current division
I2 =100mA⋅
1.5m
Geq
= 20mA, P = 100mA ⋅ I2 /1.5mS =1.33W .
SOLUTION 2.35. By current division, for I1 to be 2A then 160 + R = 300 ||600 for an even split. Thus
R = 40Ω.
SOLUTION 2.36. By current division, i1 = 0.4A⋅
1/10
1/10 + 1/ 40
= 0.32 A. Therefore using KVL
vd =10i1 − 0.25i1 = 3.12V .
SOLUTION 2.37. (a)Req = (8k ||2k) + (9k ||1k) = 2.5kΩ

(b) Req = 2k ||[(2k ||2k) + (2k ||2k)] = 1kΩ
SOLUTION 2.38. (a) Req = 2 + 15 +10 + 10 + 40 + 30 + 20 + 8 =135Ω.
(b) Four of the resistors are shorted, thus Req = 2 + 15 +10 + 8 = 35Ω.
(c) Lumping the series resistance together Req = 8 +[50 ||(50||25)] + 2 = 22.5Ω
SOLUTION 2.39. (a) Req = [2R + (4R ||4R)]||[2R+ (4R||4R)] = 2R
(b) Req = 2R||2R + (4R ||4R ||4R||4 R) = 2R
SOLUTION 2.40. (a) First Req = 150 + [375||(250 + 500)] = 400Ω. Next Iin = 14V / Req = 35mA. The
power delivered by the source is then = 14Iin = 0.49W .
(b) Req = 150 + [375||(250 + 500)||1k] = 350Ω, and Iin = 14 / Req = 40mA . The power delivered by the
source is = 14Iin = 0.56W .
As the equivalent resistance decreases, more of it gets dissipated by it.
SOLUTION 2.41. Req1 > Req2. Without going into a detailed analysis using methods of Chapter 3, we
present the following intuitive argument. First note that the points a and b represent points on an
unbalanced bridge circuit meaning that the voltage between a and b would not be zero. Also note that when
two resistors are placed in parallel, the equivalent resistance becomes smaller than either resistance. The
addition of the resistor R in circuit 2 essentially creates an internal parallel resistance resulting in an Req2
lower than Req1.
SOLUTION 2.42. Req1 = Req2. As was the case in the previous problem, this is a balanced bridge circuit.
Hence no voltage appears between a and b making the additional resistor irrelevent.
SOLUTION 2.43. (a)Rin = [(20||20) +10]||(1/ 0.12)||(1/ 0.08) = 4Ω.
(b) Rin = 6R ||[(R ||R ||0.5R) + 0.75R + (2R||2R)] = 1.5R
SOLUTION 2.44. (a) The infinite resistance are essentially open circuits, thus
Req = 1 + 2 + 3 + 4 + 2 + 4 + 3 + 2 +1 = 22Ω
(b) 0 resistances are short circuits. Labeling one branch x and the other y, it can be seen that the circuit is a
set of 3 resistor strings in parallel to each other between x and y, then added in series to the two 1 Ohm
resistor. Thus Req = [(2 + 3)||(4 + 2 + 4)||(2 + 3)]+1 +1 = 4Ω.

(c) Writing out Req = 1 +[Rx ||(2 + 3 + 4 + 2 + 4)] + 3 + 2 +1 = 7 +[Rx ||15], and solving for Rx = 3.75Ω.
(d) No, it requires methods to be covered in the next chapter.
SOLUTION 2.45. Using the formulas for parallel resistances, the circuit of figure 2.45 reduces to
(a) RAC = (2 +1) / /6
( )+ 8
[ ]/ /10 = 5 Ω
(b) RAB cannot be calculated by series parallel formulas, but RBC can be done.
RBC = (8 +10) / /6
( )+ 2
[ ] / /1 = 0.86667 Ω
SOLUTION 2.46. (a) Req = 300 + (R ||5.6k), thus R = 800Ω.
(b) Req = R+ (R ||1.2K) , the following quadratic equation must be solved R
2
+1.4k ⋅ R −1.2M = 0 . This
yields R = 600Ω.
(c) Req = 500 + 300 + (800||400 || R). Solving for R yields 800Ω.
SOLUTION 2.47.
(a) Using the fact that the resistance seen into terminal a-b is the same as that seen in terminal c-d, we can
obtain the following relationship. eq
eq R
R
R
R ||
+
= . This produces a quadratic equation whose solution is
Req = 1.618R.
(b) Using the previous argument )
5
(
||
10
5 eq
eq R
R +
+
= . Solving for Ω
= 18
.
11
eq
R .
SOLUTION 2.48. By current division Ix =
1/ 18k
1/18k +1/ 9k





 ⋅
1
6k +(9k ||18k)
1
6k + (9k ||18k)
+
1
4k










⋅36m = 3mA

SOLUTION 2.49. The 500 Ω resistor has no effect on the current entering the circuit to its right.
0.15 =
30
R
+
30
600
=
30
R
+ 0.05
Hence R = 30/0.1 = 300 Ω.
SOLUTION 2.50. (a) First, express the total current as I =
120
0.5 + (20 ||30||40 || RL2 )
. Next, find RL2 that
will cause I to be 15A. Thus RL2 = 40Ω or less will cause the fuse to blow as this will cause the current to
be 15A or more.
(b) Repeating the previous procedure, RL2 = 20Ω.
(c) RL2 = 120Ω.
SOLUTION 2.51. At time 0, all switches are open and Vout =
260
260 + 40
⋅ 220 =190.7V .
Then at t = 5s, switch one closes and
Vout =
260||260
(260 ||260) + 40
⋅ 220 = 168.2V .
At t = 10s,
Vout =
130 ||260||260
(130 ||260||260) + 40
⋅ 220 =136.2V .
Finally at t =15 s,
Vout =
65||130 ||260||260
(65||130 ||260||260) + 40
⋅ 220 = 98.5V .
0 2 4 6 8 10 12 14 16 18 20
80
100
120
140
160
180
200
Time (sec)
Volts

SOLUTION 2.52. (a) Lumping the two sources together and the resistors into an equivalent resistor gives
i1(t) =
9cos(2t) − 3cos(2t)
7k + 9k + 8k + (2k ||3k ||6k)
= 0.24cos(2t)mA .
(b) By current division i2 (t) =
1/ 6k
1/ 2k +1/ 3k +1/ 6k





 ⋅i1(t) = 40cos(2t) A.
SOLUTION 2.53. (a) Starting with,
Req1 = 5||(10 + 10) = 4Ω
Req 2 = 10||(6 + Req1) = 5Ω
Req 3 = 5 + Req2 = 10Ω
(b) Using the values R just obtained,
Va = 100⋅
Req 2
Req3
= 50V
Vb = Va ⋅
Req1
6 + Req1
= 20V
Vc = Vb ⋅
10
10 +10
= 10V
(c) Finally,
Iin =
100
Req3
=10A
Id =
Va
6 + Req1
= 5A
Ie =
Vb
10 +10
= 1A
.
SOLUTION 2.54. (a) Circuit a: Using voltage division,
vout (t) = vin (t) ⋅
300||(20 + 30 + 50)
[300 ||(20 +30 + 50)] + 5





 ⋅
30
30 + 20 + 50




= 33.75sin(377t)V, and Ohm’s law
iout(t) = vout(t)/ 30 = 1.125sin(377t)A. The instantaneous power is then
P(t) = iout (t)⋅ vout (t) = 37.969sin
2
(377t)W .
Circuit b: By current division
iout(t) = iin (t) ⋅
1/ (20 + 30 + 50)
1/ (20 + 30 + 50) +1/ 300 +1/ (50 +100)





 = 60sin(377t)A, and from Ohm’s law
vout (t) = 50 ⋅ iout(t) = 3000sin(377t)V . The instantaneous power is P(t) = 180sin
2
(377t)kW .
(b) No, since the current source forces the amount of current in the circuit.

SOLUTION 2.55. (a) Noting that i2 = v1 /10 = 6A, then we can write KCL at the top left node,
isource = i2 + v1 / 6 + (v1 − 5i2 )/ 5 = 22A. Thus P = 60⋅ 22 = 1.32kW .
(b) First, determine the current through each resistor:
i2 = 60 /10 = 6A
i2.5Ω =
60
2.5 + (5||5)





 = 12A
i5Ω = 1/ 2 ⋅i2.5Ω = 6A
Then calculate the power absorbed by each resistor:
P
10Ω = 10i2
2
= 360W
P2.5Ω = 2.5i2.5Ω
2
= 360W
P5Ω = 5i5Ω
2
= 180W
SOLUTION 2.56. From Ohm’s law I1 = 100m / 200 = 0.5mA. By current division
IRL =
20k
20k + 200




⋅150I1 = 75.257mA, and PRL = 200IRL
2
= 1.103W .
SOLUTION 2.57. First, using voltage division, Vx = Vs
2
2 +1




= (2 / 3)Vs . Then using KCL and the
previous equation, Is = (Vs / 3) − Vx = −(1/ 3)Vs . Finally using Ohm’s law Req = Vs / Is =−3Ω.
SOLUTION 2.58. Observing the following relationship, V1 = Vin , the following nodal equation can written:
Iin = Vin / 3 + Vin / 6 − 2Vin = −1.5Vin .
SOLUTION 2.59. Step 1. From voltage division
V1 =
18
18 + 4 + 2
Vs = 0.75Vs and Vin =
22
24
Vs =
11
12
Vs
Hence
Pin =
Vin
2
22
=
11×11
22 ×144
Vs
2
=
11
288
Vs
2
Step 2. For the load, by current division

I2Ω =
6
6 + 2
AV1 =
3
4
A
3
4
Vs




=
9A
16
Vs
Therefore
P2Ω = 2 × I2Ω
2
= 2
81A2
256
Vs
2
=
81A2
128
Vs
2
Step 3. P2Ω = 10 × Pin implies that
81A2
128
Vs
2
=10
11
288
Vs
2
Hence
A =
128 ×110
81 × 288
= 0.7769
SOLUTION 2.60. By voltage division V1 =
6
6 + 2
Vin = (3/ 4)Vin . By current division, and substituting the
previous equation I2 =
3
3 + 6
4V1 = Vin . Using voltage division and Ohm’s law, and substituting the previous
equation,
Vout = 4.5I2
10
10 + 5
= 3Vin = 30V
Iout = 4.5I2 / (10 + 5) = 0.3Vin = 3A
Finally, from the previous equations |Vout / Vin |= 3 .
SOLUTION 2.61. Writing out KCL when the switch is closed, ibat = 150A +
Vbat − 0.04ibat
240




. Solving
gives ibat = 150.02A and Vout ≈ 6V . When the switch is open Vout = Vbat
240
240 + 0.04




≈12V. Therefore,
the reason for the radio stopping is insufficient supply voltage.
SOLUTION 2.62. (a) Using the following relationship P = V
2
/ R, the resistance of each headlight on low
beam is R = V
2
/ P = 4.11Ω.
(b) Using the same relationship R = 2.22Ω.
(c) By voltage division, Vout =14.7
240
240 + 0.04
= 14.698V .
(d) Using voltage division, Vout = 14.7
240 ||4.11||4.11
(240||4.11||4.11) + 0.04
= 14.417V

(e) Using voltage division, Vout =14.7
240||2.22||2.22
(240 ||2.22||2.22) + 0.04
= 14.186V
11.96 =
15
15 + R0
12 =
180
15 + R0
Therefore
R0 =
180 −15 ×11.96
11.96
= 0.050167 Ω
SOLUTION 2.64. (a) Using KVL Vt = 102 − 0.05⋅80 = 98 V.
(b) Using KVL Vt = 102 + 0.05⋅ 50 =104.5 V.
(c) P = Vt ⋅50 = 5.225. kW
SOLUTION 2.65. Minimum load means the minimum load resistance that the system can handle.
»MaxPwr = 0.8*50e6
MaxPwr = 40000000
»Vs = 750e3;
»Iline = MaxPwr/Vs
Iline = 5.3333e+01
»Rmin = Vs/Iline
Rmin = 1.4062e+04, i.e., Rmin = 14.062 kΩ.
SOLUTION 2.66. (a) Using the following general form for a non-ideal voltage source: vout = −Rsiout + Vs ,
one sees that for zero current vout = Vs = 40V . The slope of the line is
−40
1000
= −Rs = −0.04Ω, thus
Rs = 0.04Ω.
(b) This curve represents a resistor’s I-v characteristic, thus the slope
60
1
1
3
= R = 45Ω.
(c) The general form for a non-ideal current source is iout = −
1
Rs
Vout + Is . When the voltage is zero,
iout = Is = 5A. From the slope of the line,
−4000
5
, Rs = 4000 / 5 = 800 Ω.

SOLUTION 2.67. Using the following formula:
T
n
nI
Cn





 = 1, solve for T, and get 0.625 hrs, or 37.5 min.
SOLUTION 2.68. Using the same equation as before and solving for Cn = nI
T
n




1/
, with n=10, and
T=55/60 hrs, the capacity obtained is 20 Ah.
SOLUTION 2.69. C20 = 50 Ah
(a) In eq. 9, solving for I with n=20, and T=10, I=4.2A
(b) Calculate the capacity for n=10 and T=10, this yields 42 Ah.
SOLUTION 2.70. (a) Using a sequence of voltage division,
V1 = 50mV
48
50
= 48mV
V2 = 50V1
195
200
= 2.34V
Vload = 2.5V2 = 5.85V
And the power is RRL
= Vload
2
/ RL = 2.278W .
(b) Following is the graph, and the script used to generate it.
0 10 20 30 40 50 60 70
0
1
2
3
4
5
Power
in
Watts
0 10 20 30 40 50 60 70
0
200
400
600
800
Resistance in Ohms
Current
in
mA
%Script for Question 70 in chapter 2
RL=8:1:64;
V2=2.34;
IL=2.5*V2 ./ RL;
%Note the use of the ".*" which means that the division
%is performed for each value of RL.

PL=RL .* (IL .^ 2);
%Plot the Power versus RL
subplot(2,1,1);
plot(RL,PL);
ylabel('Power in Watts');
%Plot the Current versus IL
subplot(2,1,2);
plot(RL,1000.*IL);
xlabel('Resistance in Ohms');
ylabel('Current in mA');
%The use of subplot lets you subdivide the graphing
%window in two halfs.
SOLUTION 2.71. (a) Using the following script:
%Script for problem 2.71
R1=15; R2=4; R3=9; R4=2; R5=8;
R6=18;
Ra= R4+R5;
Ga= 1/Ra;
Gb= Ga+1/R1;
Rb= 1/Gb;
Rc= 1/(1/R6+1/R3)+Rb;
Gc= 1/Rc;
Geq= Gc+1/4;
Req= 1/Geq;
Irc= 20*Gc/Geq;
Vrb= Irc*Rb;
Vout= Irc*(Ga/Gb)*8;
Req
Vout
So (a) Req = 3Ω, and (b) Vout = 24V
SOLUTION 2.72. Using the following script:
R1=1e3; R2=2.2e3; R3=2e3; R4=5e3; R5=3e3; R6=R5;
R7=3.2e3; R8=1.2e3; R9=1.6e3;

Ga=1/R7+1/(R8+R9);
Ra=1/Ga;
Gb=Ga+1/R6;
Rb=1/Gb;
Gc=1/R4+1/(R5+Rb);
Rc=1/Gc;
Gd=1/R2+1/(R3+Rc);
Rd=1/Gd;
Geq=1/R1+Gd;
Req=1/Geq
%Going through the same step to find Vout
Id=200e-3*(Gd/(Geq));
Ic=Id*((1/(R3+Rc))/Gd);
Ib=Ic*((1/(R5+Rb))/Gc);
Ia=Ib*Ga/Gb;
Iout=Ia*((1/(R8+R9))/Ga)
Vout=Iout*R9
The following values are obtained:
Req =
5.9121e+02
Iout =
5.5431e-03
Vout =
8.8689e+00
R1
a
b
c
d
R9
R7
R6
R4
R2
SOLUTION 2.73. Using the following script:
R1=20; R2=40; R3=60; R4=30; R5=10; R6=135;
R7=150; R8=300; R9=130; R10=200; R11=50;
Ga=1/R10+1/R11;
Ra=1/Ga;
Rb=Ra+R9+(1/(1/R7+1/R8));

Gb=1/Rb;
Gc=Gb+1/R6;
Rc=1/Gc;
Rd=Rc+R5+(1/(1/R3+1/R4));
Gd=1/Rd;
Ge=Gd+1/R2;
Re=1/Ge;
Rin=R1+Re
Ie=10/Rin;
Id=Ie*Gd/Ge;
I1=Id*(1/R6)/Gc
Ib=Id*Gb/Gc;
Vout=Ib*Ra
The following values are obtained:
Rin = 50Ω
Vout = 0.667V
I1 = 33.3mA
R1
e d
R2
R11
R10
R9
R8
R7
R6
R5
R3
R4
a
c b
SOLUTION 2.74. An identical procedure to the one followed in the previous problem will yield the
following values:
Rin = 50.53Ω
Iout = 133.8mA

Solution 3.1. Select the bottom node as the reference node, and write a node equation at the positive
terminal of the V1 resistor:
V1 − V0
3R
+
V1
6R
+
V1 − 4V0
6R
= 0
⇒ 2V1 − 2V0 + V1 +V1 − 4V0 = 0
⇒ 4V1 = 6V0
⇒ V1 = 1.5V0
Solution 3.2 Write a node equation at the top node:
0.6 −
Vx
100
−
2Vx
100
−
Vx
50
= 0
⇒ −Vx − 2Vx − 2Vx = −60
⇒ −5Vx = −60
⇒ Vx = 12V
Solution 3.3
0.6 −
Vx
100
+
25Vx
100
−
Vx
50
−
Vx − 0.2Vx
40
= 0
⇒ −
3Vx
100
+
25Vx
100
−
8Vx
400
= −0.6
⇒ Vx = −
0.6 × 400
80
⇒ Vx = −3V
Solution 3.4 (a)
It is evident from the figure that Vc = 20. We need to write two equations in Va and Vb and put them in
matrix form. In this case, we can write the matrix equation by inspection. Note that the resistors are
identified by conductance values.
15m −5m
−5m 35m






Va
Vb





 =
0.5
0.5







(b) Solve the matrix equation by inverting the left-most matrix:
Va
Vb





 =
1
525 − 25
35m 5m
5m 15m






0.5
0.5






=
1
0.5
35 5
5 15






0.5
0.5






=
40
20






(c) Vx = Vab = Va −Vb = 20V , Vda = -Va = -40, Vdb = -Vb = -20.
(d) Pi = 0.5×40 = 20W, Pv = 20×(20-20) = 0. Pdiss = 40×40×10m + 20×20×5m + 20×20×5m = 20 W.
Power delivered equals dissipated power.
Solution 3.6 Write two nodal equations:
Vs1 − V1
3000
= Is3 +
V1 −V2
6000
V2
30000
=
Vs2 − V2
12000
+
V1 − V2
6000
Rewrite equations as:
2Vs1 − 2V1 = 6000Is3 +V1 −V2
2V2 = 5Vs2 − 5V2 +10V1 −10V2
Cast into a matrix equation
−3 1
−10 17






V1
V2





 =
6000Is3 − 2Vs1
5Vs2






Solving the matrix equation yields:
V1
V2





 =
181.46
124.39






Power absorbed by the 6k resistor is (V1-V2)2
/R = 0.5429W .
Similarly, Ps1 = (Vs1-V1)/3000×Vs1 = 4.7W, Ps2 = (Vs2 −V2) /12000 ×Vs2 = −0.32W
Ps3 = Is3 × (Vs2 − V1) = −1.21W
Solution 3.7 (a) Again, the matrix equation can be written by inspection:

G1 + G2 + G4 −G4
−G4 G3 + G4 + Gs






VB
VC





 =
50G1
50G3






(b) Substituting the values of conductances and inverting the above matrix equation yields:
VB = 34.0132V
VC = 33.6842V
(c) Power delivered is 80.7566W . Using the Principle of Conservation of Power:
Pdel = P1 + P2 + P3 + P4 + P5
or,
Pdel = 50 ×
VA −VB
20
+
VA −VC
20





 = 80.7566W
(d) In this part, we take the above matrix equation and solve it for each value of Gs. If we do this, we can
get a feel for the behavior of VB and VC w.r.t. changes in Gs. The following plot is the voltage difference
between the two nodes as a function of Gs, and hence as a function of temperature.
As can be seen, in this figure, the voltage difference between B and C does not change linearly with Rs.
Since this resistance itself changes linearly with temperature, this means that VB-VC does not change
proportionally to temperature.
Solution 3.8 The answer is:

G1 + G2 −G1 −G2 0
−G1 G1 + G3 + G4 + G7 −G4 −G7
−G2 −G4 G4 + G5 + G6 + G2 −G6
0 −G7 −G6 G6 + G8 + G7














V1
V2
V3
V4














=
−Is1
0
0
Is2














Solution 3.9 Write the matrix equation by inspection:
4 /100 −1/100 −1/100
−1/100 4 /100 −1/100
−1/100 −1/100 4 /100










V1
V2
V3










=
Vs1
100
Vs1
100
Is2














Solving the equation in MATLAB, we get: V1 = 7, V2 = 7, V3 = 11, and P = 0.6.
P = Vs1 ×
VS1 − V1
100
+
VS1 −V2
100




= 10 ×
3
10
+
3
10




= 0.6W
Solution 3.10 (a)
Nodal equation for A:
VA −Vs1
10
+
VA −VB
10
+
VA −VC
10
= 0
(b) At node B:
VB
10
+
VB −VA
10
− Is2 = 0
(c) At node C:
Is2 − Is3 +
VC − VA
10
= 0
(d) Manipulate algebraically to cast as the following matrix equation:
3 −1 −1
−1 2 0
−1 0 1










VA
VB
VC










=
Vs1
10Is2
10 Is3 − Is2
( )










VA
VB
VC










=
−13 1
3
131
3
−43 1
3













(e) P = (-10-13.333)/10×(-10) = 23.33W.
Solution 3.12
We are required to write the equations in matrix form. First, write a node equation at VA and Vout:
VA − 5 +
VA
5
+
VA −Vout
10
= 0
Vout −VA
10
− 7.5VA +
Vout
10
= 0
Now group the coefficients for VA and Vout, and write the matrix equation:
1+1/5 +1/10 −1/10
−1/10 + 7.5 2/10






VA
Vout





 =
5
0






VA
Vout





 =
1
−37






where the matrix inversion was performed in MATLAB. The ratio of the output voltage to the input voltage
is -37/5.
Solution 3.13 (a) Nodes A and B are already labeled:
VA − 9
( )0.1 + VA − 0.5VB
( )0.2 + VA − VB
( )0.3 = 0
VB −VA
( )0.3 + VB − 9
( )0.5 + 0.4VB = 0
This can be rearranged Into:
0.1+ 0.2 + 0.3
( )VA − 0.3VB − 0.2 × 0.5
( )VB = 0.9
−0.3VA + 0.3 + 0.5 + 0.4
( )VB = 4.5
The matrix equation can now be easily obtained:
0.6 −(0.5)(0.2) − 0.3
−0.3 1.2






VA
VB





 =
0.9
4.5






VA
VB





 =
4.8
4.95






(b) Iin = -(VA-Vs)0.1 - (VB-Vs)0.5 = 2.445A

(c) Ps = VSIin = 22.005W, Pdep = 0.5 ×Vout × (0.5Vout −VA) × 0.2 = −1.151W .
(d) P = V2
/R = (4.95)×(4.95)×0.4 = 9.801W.
Solution 3.14 (a) We write node equations at VA and VB:
−Is1 +
VA
20k
+ gm1VA +
VA −VB
10k
= 0
VB − VA
10k
− gm2 VA − VB
( ) +
VB
2.5k
+ Is2 − gm1VA = 0
Rearranging, we have:
1
20k
+ gm1 +
1
10k




VA −
VB
10k
= Is1
−1
10k
− gm2 − gm1




VA +
1
10k
+ gm2 +
1
2.5k




VB = −Is2
(b)
1/20k + gm1 +1/10k −1/10k
−1/10k − gm2 − gm1 1/10k + gm2 +1/2.5k






VA
VB





 =
Is1
−Is2






(c) The above matrix is inserted into MATLAB, with all the values substituted, to obtain:
VA
VB





 =
9.722
5.972






(d) Vo = VA-VB = 3.75V
(e) P1 = VAIs1 = 0.0292W, Pgm1 = -Vogm1VA = -0.008W, Pgm2 = VBgm2Vo = 0.0112W, P2 = -VBIs2 = -
0.0119W.
Solution 3.15 I1 = 0.4. Write nodal equations at A and B
VA
100
+
VA − VB
20
+ 0.03 VA − VB
( ) = 0.4
VB − VA
20
+
VB
40
+
VB − 80
VB
40
40
= 0
Rearranging and casting into matrix form:

1/100 +1/20 + 0.03 −1/20 − 0.03
−1/20 1/20 +1/40 −1/40






VA
VB





 =
0.4
0






VA
VB





 =
40
40






It is obvious then, that Vx, the voltage between A and B, is zero.
Solution 3.16
1) VA = 3000ix = 3000 ×
VA −VB
9000
=
VA − VB
3
Equation at node B:
VB − VA
9000
+
VB
6000
+
VB −VD
18000
= 0 is equivalent to:
2) −2VA + 6VB − VD = 0
Equation at node D:
VD − VB
18000
+
VD
9000
+ IS = 0 which can be rewritten as:
3) 3VD −VB = −360 Solving the system formed by equations (1), (2) and (3) we obtain:
VA = 9V , VB = −18V , VD = −126V .
Solution 3.17 (a) Choose E as the reference node
VA = 2ix
At node B
6 = (VB-VC)/3 + (VB-VA)/2
Or 6 = 5/6VB –1/3VC –VA/2
At node C, VC = 2iy
At node D, VD = -12V
iy = (VD-VA)/2 = VD/2 –ix
From here on, the solution involves algebraic manipulations to solve the system of equations. MATLAB or
hand analysis can be performed to obtain:
VA = 48V, VB = 12V, VC = -60V.
(b) P6A = 6(12 = 72W

I12V = 30-8 = 22A ( P12V = 264W.
P2ix
= 2ix × (−iz − iY ) = 2 × 24 × (+18 + 30) = 2304W
P2iy
= 2iy × (−ix +
vc − vD
6
) = 2 × (−30)× (−24 − 8) =1920W
(c) P3 Ω = ix × (VB −VC) =1728W = 1728W
P6 Ω = 6×8×8 = 384W
P2 Ωy = 2×iy×iy = 1800W
P2 Ω = (VB-2ix) (VB-2ix)/2 = 648W.
(c) VD = -12V, VA = 2ix, VC = 2iz.
Substitute the above VA and VC into the node equation for node B:
iz = (VB-2ix)/2 = VB/2 –ix and
ix = (VB-2iz)/3 = VB/3 –2/3iz
Substitute iz into ix to obtain: ix = 0, VA = 0. Then, VB can be deduced to be 12.
Finally, VC = 12V.
Now, compute the powers.
P6A = 72W , P3 Ω = 0, P6 Ω = (VC-VD)2
/6 = 96W, P2 Ωy = 72W, P2 Ωz = 72W.
P2ix = 0W , P2iy = 12 × 4 = 48W
Solution 3.18 The three node equations at A, C, and D are:
−0.8 − 0.3 = 0.015VA + 0.02VA − 0.02VC
Vc = 440
−0.8 + 2.5 = −0.005VD + 0.025VC − 0.025VD
As can be seen, these really reduce to only two equations in two unknowns. These can be solved rather
easily either by hand or by MATLAB to obtain: VA = 220V, VD = 310V . Note that all of these voltages
are already referenced to node B (i.e. VA = VAB, etc).
Solution 3.19 (a) Supernode is BC (50 V source).
(b) Only one node equation needs to be written:
P12V =12 × (
VA − VD
2
+
VC − VD
6
) = 12 × (6 + 4) =120W,

VB
90
−
VA
90
+
VB
10
+
VC
10
−
VA
10
+
VC
90
= 0
with the constraint that VC −VB = 50.
(c) The constraint equation can be substituted into the B node equation to obtain
VB = 125V. Thus, VC = 175V, and ix = (VC-VA)/10 = -12.5A.
(d) (VA-VB)/90 = 1.94A è P300 = 300×(1.94+12.5) = 4332W.
VC/90 = 1.94 è P50 = -528W.
P50 = 50 × (ix +
VC
90
) = −528W
Solution 3.20 (a) VB = VA – 440 and VC = VA – 460.
(b) Supernode is one including A, B, and C.
(c) VC − 40
( )0.15 + 0.05VB + 0.25VA − 25 + 0.2 VA − 40
( )= 0
This can easily be rearranged to get VA = 200V.
(d) power Ps = VA × I = 200 × 25 = 5000W = 5KW
Solution 3.22 (a) VC = Vs2 = 6V
(b) Ix = 0.01VA
(c) Supernode at A,B, encompassing the controlled source. So, we have one equation:
Is1 = 0.01VA + 0.0125VB + 0.1VB − 0.1VC
(d) Substitute the constraint equations, VA –VB = 20Ix = 0.2VA , (equivalently: VA =
VB
0.8
) into the above
equation: VA =
VB
0.8
Is1 = 0.01
VB
0.8
+ 0.0125VB + 0.1VB − 0.1VC
⇒ VB = 6.4V
⇒ VA = 8V
(e) Ix = 0.08A.
(f) P0.0125Ω = VB
2
/ R = 0.512W
(g) P = Is1
×VA = 0.2 × 8 =1.6W

Solution 3.23 (a) VC = Vs2 = 50V.
(b) ix = VA/100.
(c) Supernode A,B:
Is1 = 0.01VA + 0.05VB + 0.05VB − 0.05Vs2 + 0.09VA − 0.09Vs2
VA = VB + 300ix = VB + 3VA
(d) Solving the above two equations yields: VA = -90V, VB = 180V.
(e) ix =
VA
100
=
−90
100
= −0.9A
(f) (VB-VC) (VB-VC)/R = 845W.
(g) PS1
= IS1
× VA = 2 × (−90) = −180W
Solution 3.24 (a) VB − VC = 3Vx = 3VB; VC = −2VB
(b) Supernode at B and C, encompassing controlled source.
(c)
Is2 = (
VB
10
−
VA
10
) +
VB
10
+
VC
10
+ (
VC
10
−
VA
10
); VC = −2VB; 10Is2
= −2(VB +VA)
equivalently: 2VA + 2VB = −10
(d)
(0.1VA − 0.1Vs1) + (0.1VA − 0.1VB) + (0.1VA − 0.1VC ) = 0; 0.3VA + 0.1VB = 0.1VS1
; equivalently:
3VA − VB = −10
(e) Again, any method can be used to simplify and solve the system of two equations. The solution is:
VA = -2.5V, VB = -2.5V.
Solution 3.25 (a) VA = Vs1 = 16V.
(b) Supernode at C and D, encompassing controlled voltage source.
(c) Is2 = (0.75mVD − 0.75mVB ) + (1mVC −1mVA)
(d) VC = 4VB + VD
(e)
0.75mVD + (0.75mVB − 0.75mVD) + (0.25mVB − 0.25mVA) = 0
or 1mVB − 0.25mVA = 0
(f) We now have three equations in VB, VC, and VD. These can be solved using any method. By inspection,
we can immediately deduce VB from VA using the last equation: VB = 4V.
The remaining two equations can be solved to obtain: VC = 20V and VD = 4V.

Solution 3.26 (a) The supernode is the combination of A, C, and the controlled voltage source.
(b) Write node equations starting at the supernode:
(G2VA − G2Vin ) + (G3VA ) + (G4VA − G4VB) + (G6VC ) + (G5VC − G5VB ) = 0
⇒ G2 + G3 + G4
( )VA + −G4 − G5
( )VB + G6 + G5
( )VC = G2Vin
and
(2G6VC ) + (G4VB − G4VA ) + (G5VB − G5VC ) + (G1VB − G1Vin) = 0
⇒ −G4
( )VA + G1 + G4 + G5
( )VB + 2G6 − G5
( )VC = G1Vin
and
VA −VC = 3Vx, VA −VC = 3(Vin − VA ), 4VA − VC = 3Vin
In matrix form:
0.8
−0.4
4





−0.5
0.6
0
0.2
0.1
−1





VA
VB
VC










=
6
6
180










(c) The above system of equations can be solved to obtain: VA = 38.75V, VB = 40V, VC = -25V.
(d) Iin = (G7Vin )+ (G2Vin – G2VA) + (G1Vin – G1VB) = 5A.
è Req = 12Ω and P = 300W.
P = Vin × Iin = 60 × 5 = 300W
(e) Iout = VCG6 = -2.5A è P = 62.5W.
Solution 3.27 (a) Supernode is A,B encompassing controlled voltage source.
(b)
(VA – Vs1) + 0.4VB + (0.2VB – 0.2VC) = 0
è VA + 0.6VB – 0.2VC = Vs1
(c) VA – VB = IB = 0.4 VB è VA = 1.4VB.
(d) Is2 = 0.2VB + (0.2VC – 0.2VB) = 0.2VC.
(e)
In matrix form:
1
1
0





0.6
−1.4
0
−0.2
0
0.2





VA
VB
VC










=
8
0
2










The solution is: VC = 10V,VA = 7V,VB = 5V .

(f) i = (VA-Vs1)1S = -1 è Pccvs = (VA −VB ) × (VA − Vs1
) ×1S = −2W
Pvccs = (Vs1-VC)(0.2VB) = -2W.
Solution 3.28 (a) Supernode at A,C, CCVS.
(b) Node equation at supernode:
Is + 0.25mVA = G1VA + (G2VA –G2VB) + (G5VC) + (G4VC – G4VB)
è Is = (G1 + G2 – 0.25m)VA + (–G2 – G4)VB + (G4+G5)VC
Constraint:
VA – VC = 104
ix = 104
G3VB
è 0 = VA – 104
G3VB – VC
At node B:
G3VB + (G2VB – G2VA) + (G4VB – G4VC) = 0
è –G2VA + (G3+G2+G4) – G4VC = 0
è (c) Matrix equation:
G1 + G2 − 0.25m −G2 − G4 G4 + G5
1 −10
4
G3 −1
−G2 G3 + G2 + G4 −G4










VA
VB
VC










=
Is
0
0










;
0
1000
−0.2





−1
−1000
1.1
1
−1000
−0.8





VA
VB
VC










=
2
0
0










(d) Substitute the values of conductances and solve the above matrix equation in MATLAB to obtain:
VA = -38V, VB = -20V, VC = -18V.
(e) P = V × I = (10
4
G3VB) × −Is + VAG1 + G2(VA − VB)
[ ]= (−20) × −2m −1.9m − 3.6m
[ ]= 0.15W
Solution 3.29
Loop equation: Vin = 2kI1 + 500(I1 + 20m)
è Vin = 2500I1 + 10 è I1 = 20mA.
Pvin = 20m×60 = 1.2W.
PI = 20m×(500I1 + 500×20m) = 0.4W.
P2k = I1×I1×R = 0.8W.
P500 = (I1 + 20m)2
R = 0.8W.
total power absorbed by resistors: PR = 0.8 + 0.8 = 1.6
total power delivered by sources: Ps =1.2 + 0.4 W=1.6W

Conservation of power is verified.
Solution 3.30
Loop equation: 100(I1 − 0.5) + 200I1 + 500 × (I1 + 20m) = 0
è I1 = 0.05A.
P0.5A = I × V100Ω; V100Ω = 100 × (0.5 − 0.05); where V100Ω is the voltage on the 100Ω resistor.
P0.5A = 0.5(0.5×100 – 0.05×100) = 22.5W.
P20m = 20m(I1+20m)500 = 0.7W.
Solution 3.31
Loop equation: 3.3 = 50I1 + (50m + I1)100 + (I1 – 30m)40 + (I1 – 50m)60
è I1 = 0.01A.
The power delivered by the independent voltage source:
P = I1×3.3 = 0.033W.
Solution 3.32
Loop equation: 50 = 300I1 + (I1 – 0.4I1)500
50 = (300 + 500 – 200)I1 è I1 = 0.0833A.
Power absorbed by the 500Ω resistor.
P500 = (I1 – 0.4I1)2
500 = 1.25W.
Solution 3.33
Loop equation: 1000(I1 − Is) + 4000I1 + 5000(I1 − gmVx ) = 0→10000I1 − 2Vx = 50
and
Vx = 1000(Is − I1) →1000I1 + VX = 50 .
Solve the above two equations in I1 and Vx to obtain: I1 = 12.5mA, Vx = 37.5V.
Thus, Req = Vx/Is = 750_,
P = Ivccs ×Vvccs = gm ×Vx × 5000 × (gmVx − I1)=0.1875W

Solution 3.34
Loop equation: Vin = 2Iin + 14Iin – 10V1
V1 = 2Iin
After replacing V1 in the loop equation we obtain:
è Vin = – 4Iin è
R1eg =
Vin
Iin
= −4Ω
Solution 3.35
Loop equation: Vs = 500I1 + 100(I1 + 0.5) + 400(I1 − 0.001Vx ) +100(I1 + 0.005Vy )
Vx = 500I1,Vy = 400I1 - 400×0.001Vx = 400I1 – 200I1 = 200I1
After replacing Vx and Vy in the loop equation we obtain:
Vs – 50 = 1000I1 è I1 = 0.1A
Vy = 200I1 = 20V è P400ohm = Vy
2
/400Ω= 1W
Req = Vs/I1 = 150/0.1 = 1500Ω.
Solution 3.36
Select clockwise loop current I1 in the left loop. Select anti clockwise loop current I2 in the right loop.
The two mesh equations are:
12 = I1 +10(I1 + I2)
and 10(I2 + I1) + 2I2 + 12 = 0
The two simultaneous equations can be solved easily to obtain: I1 = 0.75A, I2 = +0.375A.
P10ohm = (I1 + I2)2
/10 = 0.127W.
Battery 1 supplies more current. (I1 > I2)
Solution 3.37
(a) The equation for the left loop is:
660 = I1R + 1.296(I1 + I2)+ 590 + I1R
The equation for the right loop is:
660 = (0.3 – R) I2+ 1.296 (I1 + I2) + 590 + (0.3 – R) I2

Simplifying the two equations:
70 = 1.596I1 + 1.296I2
70 = 1.296I1 + 1.596I2
The solution of these two equations is: I1 = I2 = 24.2A.
(b) I1 + I2 = 48.4 , voltage across locomotive = 590 + 48.4×1.296 è power = 31592W.
(c) Because the locomotive is 1/3 distance from either station it follows that
R = 1/3×0.3 = 0.1Ω. The two equations become:
70 = I1(2R + 1.296) + 1.296I2
70 = 1.296I1 + (1.296 + 0.6 – 2R)I2
The solution of these two equations is: I1 = 32.64A, I2 = 16.32A.
Current in locomotive motor I1 + I2 = 48.96A.
Voltage across locomotive 590 + (I1 + I2) × 1.296
It follows that:
è P = (I1 + I2)(590 + 49×1.296) = 31993W.
Solution 3.38
(a)
(b) The three loop equations are:
660 – 590 = 0.1I1 + 1.296 (I1 – I2) + 0.1I1
0 = 1.296 ( I2 − I1) + 0.2I2 + 1.296 ( I2 − I3)
–70 = 1.296( I3 − I2) + 0.2I3
These three equations can be solved using any method to obtain:

I1 = 46.8A, I2 = 0, I3 = – 46.8A.
(c) Motor currents are 46.8A each.
(d) Ps = VI = 660×46.8 = 30.9kW. Each source supplies 30.9kW.
Solution 3.39 (a) Define three meshes with three mesh currents. The first, I1, is a clockwise current
around the first mesh. The second, I2, is a clockwise current around the middle loop of the circuit
(through the 10mS, 5ms, and 5ms conductances). The third, I3, is a counterclockwise current through the
right-most loop containing the voltage source.
*current names shown above.
(b) I1 = 0.5A
(
I2
10m
−
I1
10m
) +
I2
5m
+ (
I2
5m
+
I3
5m
) = 0.
(
I3
25m
+
I3
5m
) +
I2
5m
= 20
These are two equations in two unknown currents. After grouping the terms, it can be verified that:
I2 = 0.1A, I3 = 0.
(c) Vx = 20V
Vad = (0.5 – I2)/10m = 40V
Vbd = 20V
(d) P0.5 = Va × 0.5 = 40×0.5 = 20W
P20V = 0W
Presistors = 2× I2
2
/5m + (0.5 – I2)2
/10m = 20W
The conservation of power is verified.
Solution 3.40 (a) We can either write down the equations or evaluate the matrix by inspection:

90 ×(I2 − 4.8m) + 10kI2 + 50 = 0
90kI3 + 10 × (I3 − 4.8m) = 50
OR
100k 0
0 100k






I2
I3





 =
90k × 4.8m − 50
10k × 4.8m + 50






(b) The solution of the above equation is: I2 = 3.82mA, I3 = 0.98mA.
(c) Current source: P = 4.8m×[(4.8m – I2)90k + (4.8m – I3)10k] = 0.61W.
Voltage source: P = 50(I3 – I2) = – 142mW.
Solution 3.41 (a) By inspection:
112k −90k −10k
−90k 100k 0
−10k 0 100k










I1
I2
I3










=
180
−60
60










(b) Using MATLAB:
I1 = 4.4mA, I2 = 3.36mA, I3 = 1.04mA
→ ix = −I2 = −3.36mA
(c) P180 = 180×4.4m = 0.792W, P60 = 60( I3 − I2) = -0.139W.
Solution 3.42 The matrix equation is:
8 −6 3
−6 8 −2
3 −2 4










I1
I2
I3










=
14
0
6










whose solution is: I1 = 4A, I2 = 3A, I3 = 0A
v = (I1 + I3 + I2) ×2 = 2V
Solution 3.43 (a) First, note that two mesh currents are needed. Two clockwise currents are defined: I1
in the middle loop, and I2 in the right-most loop:
Middle loop equation:
100I1 – 100Is1 + 20Ix + 80I1 – 80I2 = 0, where Ix = Is1 – I1
and

Right-most loop equation:
80I2 – 80I1 + 10I2 + Vs2 = 0
These can easily be cast into the following matrix equation:
160
−80



−80
90



I1
I2





 =
16
−6






(b) The solution of this equation is: I1 = 0.12A, I2 = 0.04A.
(c) VA = 100(Is1 – I1) = 8V and VB = 80x(I1 − I2) = 6.4V .
(d) Ps1
= Is1VA = 1.6W.
(e)P0.0125S = (I1 − I2)
2
/0.0125 = 0.512W.
Solution 3.44 (a) Create two clockwise mesh currents in the top loop (I2) and the bottom-left loop (I1) .
The bottom-right loop has an independent current source. Writing the loop equations:
Vs1 = 200 (I1 – I2) + 200 (I1 + Is2)
200 (I2 – I1) + 100I2 + 300Ix + 200 (I2 + Is2) = 0, where Ix = I1 – I2
(b) Solving, we get: I1 = -0.1A, I2 = -0.7A, Ix = 0.6A.
(c) VB = (I1 + Is2)200 = 130V.
(d) Pvs1 = I1Vs1 = –25W, Pis2 = (VB + (Is2 + I2)200)Is2 = 105W, P300ix = (−I2)(300Ix) = 126W.
Solution 3.45 (a) Create two clockwise mesh currents in the top loop (I1) and the middle loop (I2) (all
resistor loop):
Top loop equation:
0.5vx = 500 (I1 - I2,) +500I1 where vx = –500I1
and
Middle loop equation:
600 (I2 – Is1) + 500 (I2 – I1) + 900 (I2 + Is2) = 0
(b) Solving, we get: I1 = 0.015A, I2 = 0.0375A, vx = -7.5V.
(c) Pis1 = Is1[0.5vx + (Is1 – I2)600] = 109.7W
P0.5vx = 0.5vx(I1 – Is1) = 1.63W
Pis2 = (I2 + Is2)×900×Is2 = 53.2W

Solution 3.46 Write the mesh equations in terms of R’s and then substitute the values from the matrix:
Mesh 1 equation:
v1 = R1 (i1 − i3) + R2 (i1 – i2)-25i2
From this equation, and the first row of the matrix equation, we can deduce that
R1 = 5 Ω
and
R2 + 25 = 40 ⇒ R2 = 15Ω.
Similarly:
Mesh 3 equation: R1(i3 − i1) + i3R4 + R3(i3 − i2) =0
From which we can deduce:
R3 = 25 Ω and R4 = 5 Ω.
Solution 3.47
Modified loop 1 equation:
Vs1 = 3MI1 + v + 2MI1
Constraint equation:
– I1 + I2 = Is3
Modified loop 2 equation:
v = 2MI2 + Vs2 + 8MI2
Or in matrix form:
5M 0 1
−1 1 0
0 −10M 1










I1
I2
v










=
Vs1
Is3
Vs2










Solving: I1 = -1.1 A, I2 = -0.95 A.
The power Ps3 = Is3 xv= 7.58 W
Solution 3.50 I2 = 2A, I3 = –7A
Loop 1 equation:
Vs = 3I1 + 3(I1 − I2) + 6(I1 + I3 − I2) + 2vy + 2(I1 + I3) =14I1 + 2vy − 74 ⇒ 14I1 + 2VY = 88
vy = 3(I1 – I2)= 3I1 − 6

Solving the above system, we obtain: I1 = 5A è vy = 9, v can be found from the loop 3 equation
v + (I1 + I3 − I2).6 + 2vy + 2(I3 + I1) = 0 . Solving, we obtain v =10V
Finally, Pvs = Vs×I1 = 70W.
Solution 3.51
Mesh 1 equation:
9kI1 + 3k (I1 – I4) + 6k (I1 – I3) + 12k (I1 – I2) = 0
where we have used the fact that Ix = I1 – I4 (and I4 = – 4mA)
Mesh 3 equation:
6k (I3 – I1) – v2 +2kI3 + v = 0
Mesh 2 equation:
2.4kI2 + 12k (I2 – I1) – v = 0
Constraint equations:
I2 – I3 = 0.5mA
I3 – I4 = 0.5Ix = 0.5I1 – −0.51I4 = 0.51I1 + 2mA ; 0.51I1 − I3 = −2mA
The above five equations need to be put into matrix form:
30k −12k −6k 0 0
−6k 0 8k 1 −1
−12k 14.4k 0 −1 0
0 1 −1 0 0
0.5 0 −1 0 0
















I1
I2
I3
v
v2
















=
−12
0
0
0.5m
2m
















The solution is:
I1
I2
I3

I1
I2
I3
v
v2
















=
−0.002A
−0.0025A
−0.003A
−12V
−24V
















Ix = I1 − I4 = 2mA
Power delivered by 0.5mA current source: P1 = 0.5mxv = −6mW
Power delivered by the dependent current source: P2 = 0.5Ix ⋅ v2 = 1m × (−24) = −24mW
Power delivered by 4mA current source: P3 = 4m × (3k⋅ Ix − V2) = 120mW
Solution 3.52
Write the following equations:
Mesh 2 equation:
100 (I2 – I1) + 150 (I2 – I4 – I6) + v = 0
Mesh 4 equation:
3vx + 30 (I4 – I3) + 150 (I4 - I2 + I6) = 0
Mesh 5 equation:
800I5 – 3vx + 10( I5 − I1 + I6) =0
Mesh 6 equation:
250I6 +150(I6 + I4 − I2) +10(I6 + I5 − I1) = 0
Constraint equation:
I1
I2 I3
I4
I5
I6

Is2
= −I2 + I3
Substituting the values of I1 = Is1
= 1.15 and I3 = Is3 = 0.95 , and noting that
vx = (I1 – I6 – I5)10
we can write:
250 −150 0 −150 1
−150 180 −30 120 0
0 0 840 40 0
−150 150 10 410 0
1 0 0 0 0
















I2
I4
I5
I6
v
















=
100I1
30I3 − 30I1
40I1
10I1
I3 − IS2
















I2
I4
I5
I6
v
















=
0.65
0.45
0.05
0.1
35
















Finally, vx = 10V and v is as given above.
Solution 3.53 (a)
Replace the voltage source by current sources:
At node 1
is1 = (V2 − V1) /12k
At node 2
(V2 − V1) /12k + (V2 − V3) /90k + (V2 − V4) /10k = 0
At node 3
is2 = (V3 – V2) /90k + V3/10k
At node 4
is2 + (V4 −V2) /10k + V4/90k = 0
V1 = 180
V4 – V3 = 60
(b) In matrix form:

−1/12k 1/12k 0 0 −1 0
−1/12k 1/12k +1/90k +1/10k −1/90k −1/10k 0 0
0 −1/90k 1/90k +1/10k 0 0 −1
0 −1/10k 0 1/90k +1/10k 0 1
1 0 0 0 0 0
0 0 −1 1 0 0




















V1
V2
V3
V4
is1
is2


















=
0
0
0
0
180
60


















The solution from MATLAB is
180.0000
127.2000
33.6000
93.6000
-0.0044
0.00232
(c) Power delivered by S1 is:
Ps1 = Vs1 × (−Is1) = 0.792W
Power delivered by S2 is:
Ps2 = Vs2 × (−Is2) = −0.139W
Solution 3.54 (a) Replace the 100 ohm resistor, the controlled voltage source, and vs2 by current source.
Then write the node equations:
Is1 = ix + I1 + 0.09 (V1 – V3)
I1 = V2/20 + 0.05 (V2 – V3)
– I2 = 0.05 (V3 – V2) + 0.09 (V3 – V1)
V3 = Vs2
V1 – V2 = 300ix = 300V1/100
In matrix form:
0.1
0
−0.09
0
2








0
0.1
−0.05
0
1
−0.09
−0.05
0.14
1
0
1
−1
0
0
0
0
0
1
0
0








V1
V2
V3
I1
I2
















=
2
0
0
50
0
















(b) Using MATLAB to solve the above system the solution is:
V1 = -90.0000V
V2 = 180.0000V

V3 = 50.0000V
I1 = 15.5000A
I2 = -6.1000A
(c) Power delivered by the current source is
Ps1 = Is1 × V1 = −180W
Power delivered by the voltage source is:
Ps2 = Vs2 × (−I2) = 305W
Solution 3.55 Modify the circuit so that it looks like the following:
The modified node equations are:
Is1 = - Iy + Ix
Ib = Iy + 0.2 (VC – VB)
Is2 = 0.2 (VC – VB) + Ix
The equations describing the constitutive relationships of elements in the original network are:
VA – VB = Ib
Ix = 0.2Vb
Ib = 0.4Vb
Iy = 8 −VA
These can be cast into a matrix equation and solved easily to obtain the same result as previously arrived
at.
In matrix form:

0
0
0
1
0
0
1












0
−0.2
−0.2
−1
0.2
0.4
0
0
0.2
0.2
0
0
0
0
1
0
1
0
−1
0
0
−1
1
0
0
0
0
1
0
−1
0
−1
0
−1
0
−1
0
0
0
0
0
0












VA
VB
VC
Ix
Iy
Ib
Is1
























=
0
0
2
0
0
0
8
























VA
VB
VC
Ix
Iy
Ib
Is1
























=
7V
5V
10V
1A
1A
2A
0A
























Solution 3.56 Modify the circuit as follows:
The modified node equations are:
At node A: Is = G1VA + G2(VA −VB ) + Ia
At node B: G2(VA −VB ) = Ix + G4(VB −VC)
At node C: 0.25mVA + Ia = G4(VC − VB) + G5VC
The equations describing the constitutive relationships of elements in the original network are:
Ix = G3VB
VA – VC = 104
Ix

These can be cast into a matrix equation that can be solved in MATLAB.
In Matrix form:
0.25m
−0.2m
−0.25m
0
1








−0.2m
1m
−0.8m
−0.1m
0
0
−0.8m
1m
0
−1
1
0
−1
0
0
0
1
0
1
−104








VA
VB
VC
Ia
Ix
















=
2m
0
0
0
0
















The solution is:
VA
VB
VC
Ia
Ix
















=
−38V
−20V
−18V
0.0075A
−0.002A
















We observe that we have obtained the same results as in problem 3.28.
Solution 3.57 Replace dependent source by i35 (from 3 to 5). Also, replace voltage source by i10 (from 1
to 0). Now, write the modified noted equations. The reference node is O:VO = 0V :
At node 1: i10 = (V6 – V1) + (V2 – V1)
At node 2: 2 = (V2 – V1) + (V2 – V3)
At node 3: i35 = (V4 – V3) + (V2 – V3)
At node 4: 2 = 2 + (V4 – V3) + V4
At node 5: i35 = (V5 – V6) + V5
At node 6: 2 = (V5 – V6) – V6
Constraints:
V3 – V5 = 15vx = 15V4
V1 = 5
The following matrix equation is obtained:

−2 1 0 0 0 1 −1 0
−1 2 −1 0 0 0 0 0
0 1 −2 1 0 0 0 −1
0 0 −1 2 0 0 0 0
0 0 0 0 2 −1 0 −1
0 0 0 0 1 −2 0 0
0 0 1 −15 −1 0 0 0
1 0 0 0 0 0 0 0


























V1
V2
V3
V 4
V5
V6
i10
i35


























=
0
2
0
0
0
2
0
5


























The solution of this equation is obtained from MATLAB:
V1 =5.0000V
V2 =3.3571V
V3 =0.2857V
V4 =-0.1429V
V5 =1.8571V
V6 =-0.0714V
i10 =-6.7143A
i35 = 3.7857A
The power delivered by the dependent voltage source connected between nodes 3 and 5:
P35 =15vx (−i35) = 15 ×V4 × (−i35) = 8.115W
The power delivered by the current source connected between nodes 2 and 4:
P24 = (2A) × (V2 −V4 ) = 7W
The power delivered by the current source connected between nodes 4 and 6:
P46 = (2A) × (V4 −V6) = −0.143W
The power delivered by the voltage source connected between nodes 1 and 0:
P10 = 5V × (I12 + I16) = 5x (V1 −V2) + (V1 − V6)
[ ]= 33.57W
Solution 3.59 Using the appropriate element stamps for each element of the circuit, we obtain the
following system:
0.15 + 0.2 −0.15 −0.2 0
−0.15 0.15 + 0.05 0 −1
−0.2 0 0.25 + 0.2 1
0 −1 1 0














VA
VB
VC
Ix














=
−8 − 3
3
25
440















Solution 3.60
1/20k +1/10k + gm1 −1/10k
−1/10k − gm1 − gm2 gm2 +1/10k +1/2.5k






V1
V2





 =
Is1
−Is2






The solution is the same as that of problem 3.14.
Solution 3.62 (a) Because RT (T) can be approximated by a straight line between (250Ω,0
o
C) and
(80Ω,50
o
C) it follows that:
RT (T)= – 3.4T + 250
(b) For T = 25
o
C , RT = 165Ω
(c) The voltage across the RT + RL series combination can be obtained from voltage division:
VT,L =
RT + RL
RT + 2RL + R
⋅12 = 4.7857V
This is the same as the voltage across Rx because the meter is at zero deflection. Thus,
Rx
Rx + R
⋅12 = 4.7857. It follows that Rx =165.84Ω.
(d) We first denote the nodes:
A - the node common to R,Rx and the voltmeter;
B - the node common to R,Rand the voltage source;
C - the node common to RL,RT and RL;
D - the node common to Rx,RL and the voltage source.
The reference node is D:VD = 0. It follows that VB =12V .
We also have: vout = VA − VC
The node equations are:
1
2
0

At node A:
VA −VC
Rm
+
VA −12
R
+
VA
Rx
= 0
Equivalently: VA(RRx + RmRx + RmR) - VC ⋅ RRx = 12RmRx
VA × 4199.86 − VC × 41.46 =19900.8 (1)
At node C:
VC −12
R + RL
+
VC
RT + RL
+
VC − VA
Rm
= 0
Equivalently: (VC −12) × 0.004 +
VC
RT + 2.5
+ (VC −VA) ⋅10
−4
= 0
VC(0.004 ⋅ RT + 0.01 +1+ 0.00025 + RT ×10
−4
) −
−VA(RT ×10
−4
+ 0.00025) = 0.48 × (RT + 2.5)
The last equation can be rewritten as:
VC(0.0041RT +1.01025) - VA(RT ×10
−4
+ 0.00025) = 0.48(RT + 2.5) (2)
From (1) and (2), we obtain:
(0.0041RT +1.01025) ×
−19900.8 + VA × 4199.86
41.46
−
− VA × (RT ×10
−4
+ 0.00025) = 0.48(RT + 2.5)
Equivalently: VA × (RT × 0.415 +102.337) = 2.448 × RT + 486.12
It follows that VA =
2.448 × RT + 486.12
0.415 × RT +102.337
From the equation at node A:
vout = VA − VC = −Rm ×
VA −12
R
+
VA
Rx





 =
2.448 × RT + 486.12
0.415 × RT +102.337
×(−100.3) + 480
At T = 0
o
C: RT = 250Ω. It follows that vout = −54.4415V
At T = 50
o
c: RT = 80Ω. It follows that vout = 80V
(e): The formula has been derived at part d):
T RT vout
0
o
C 250Ω −54.4415V
5
o
C 233Ω −52.4136V
10
o
C 216Ω −50.2368V
15
o
C 199Ω −47.8938V
20
o
C 182Ω −45.3650V
25
o
C 165Ω −42.6273V

30
o
C 148Ω −39.6537V
Solution 3.63
Place a source Vin between C and D, and calculate the current drawn from the source as below:
Loop 1 equation:
I1R1 + (I1 − I2)R2 + (I1 − I3)R3 = 0
Equivalently:
I1(R1 + R2 + R3) − I2R2 − I3R3 = 0
Loop 2 equation:
(I2 − I1)R2 + I2R4 −1 = 0
Equivalently:
−I1R2 + I2(R2 + R4) =1
Loop 3 equation:
1 + I3R5 + (I3 − I1)R3 = 0
Equivalently:
−R3I1 + I3(R3 + R5) = −1
We obtain the following system of equations:
30I1 − 4I2 − 6I3 = 0
−4I1 + 6I2 =1
−6I1 +14I3 = −1





⇒ I1 = 0.0096A, I2 = 0.1731A, I3 = −0.0673A;
Iin = I2 − I3 = 0.2404 A
Reg,CD =
Vin
Iin
=
1
0.2404
= 4.16Ω

Solution 3.64
The node equation at node A is:
VAG1 + (VA − VB)G2 + (VA −VC)G3 = 0
Equivalently:
(G1 + G2 + G3)VA −VBG2 − VCG3 = 0
The supernode is identified by a Gaussian surface enclosing the controlled voltage source. The supernode
equation is:
G2(−VA + VB) − 6 + G4VC + G3(VC − VA ) = 0
Equivalently, we have:
−VA(G2 + G3) + G2VB +VC(G3 + G4 ) = 6
One way of obtaining the solution to the problem is:
We multiply the above two equations by 30.
−30(G2 + G3)VA + 30G2VB + 30(G3 + G4 )VC =180
and
30(G1 + G2 + G3)VA − 30G2VB − 30G3VC = 0
By equating the coefficient of the above two equations with the coefficients of the first and second given
equations, we obtain:
30G2 = 30 ⇒ G2 = 0.1S
30G3 = 2 ⇒ G3 = 0.067S
30(G1 + G2 + G3) = 11⇒ G1 = 0.2S
30(G3 + G4 ) = 32 ⇒ G4 = 0.87S
can be obtained as follows:
VC −VB = VX = (VC − VA )
Equivalently:
VA −VB + (1− )VC = 0
By comparing with the third given equation ⇒ = 3.
Solution of 3.66
(a)
At node A: (VA-VC)/2 + (VA-VB)/2 + (VA-VD)/2 = 14
At node B: (VB-VA)/2 + (VB-VC)/2 + (VB-VD)/2 = 7
At node C: (VC-VA)/2 + (VC-VB)/2 + (VC-VD)/2 + 2VC = 0
At node D: (VD-VA)/2 + (VD-VB)/2 + (VD-VC)/2 + 0.5VD = 0

These can be solved in MATLAB to obtain:
22.0000
18.5000
7.5000
12.0000
(b) Mesh analysis would result in the same voltages
The loops and their current loops are:
A,C,14A:I1
A,C,D:I2
A,B,D:I3
B,C,D:I4
C,D,reference node:I5
B,D,7 A:I6











I1 = 14A,I6 = 7A
Loop ACDequation: 2(I2 − I1) + 2(I2 − I3) + 2(I2 + I4 + I5) = 0
Loop BCD equation: 2(I3 + I6 + I4 ) + 2(I2 + I4 + I5) + 2I4 = 0
Loop ABD equation: 2(I3 + I4 + I6) + 2(I3 − I2) + 2I3 = 0
Loop CDref node equation: 2(I5 + I2 + I4 ) + 2(I5 − I6) + 0.5(I5 + I1) = 0
In matrix form:

6
2
−2
2







−2
2
6
0
2
6
2
2
2
2
0
4.5







I2
I3
I4
I5














=
28
−14
−14
7














;
I2
I3
I4
I5














=
6.75
1.75
−5.5
1














VA = 2(I1 − I2) + 0.5(I1 + I5) = 22V
VB = 2(I4 + I3 + I6) + 2(I6 − I5) = 18.5V
VC = 0.5(I1 + I5) = 7.5V
VD = 2(I6 − I5) =12V
(c) Mesh analysis requires more work.
(d) The removal of the top resistor will result in more node equations than loop equations. The addition of
a resistor between node A and reference node will result in more loop equations than node equations.

Thevenin Probs, 7/11/01 - P4.1 - @R.A. Decarlo & P. M. Lin
PROBLEM SOLUTIONS CHAPTER 4
SOLUTION 4.1. First, find Vout / Vs for each circuit. Then solve for R knowing
Vout = P⋅10 =±14.142V .
(a) Writing KCL at the inverting terminal, 1/1k(v− − vs) =1/ R(Vout − v− ) ⇒ Vout / Vs = −R/ 1k , since
the inverting terminal is a virtual short. Solving for R = −Vout ⋅1k / Vs = 2.828kΩ.
(b) Writing KCL at the inverting terminal, Vs /1.5k = (Vout − Vs)/ R ⇒ Vout / Vs = R / 1.5k +1, solving
for R = 1.5k(Vout / Vs −1) = 2.743kΩ.
(c) From (a) Vout / Vs = −12k / R, thus R = −12k ⋅Vs / Vout = 4.243kΩ.
(d) This is the same circuit as (b) except the output voltage is taken across two resistors. Thus
Vout =
P
10
10 + 6
( ) = 22.627V . Using the general form from (b), R = 400 Vout / Vs −1
( ) = 1.410kΩ
SOLUTION 4.2. (a) First, find the voltage at the non-inverting terminal as v+ = 1/ 2 ⋅Vs. Then write KCL
at the inverting terminal, and make use of the virtual short property,
(Vs / 2)/10 k = (Vout − Vs / 2) / 30k ⇒ Vout / Vs = 30k(1/ 20k +1/ 60k) = 2 .
(b) Relating the output of the amplifier to the output of the circuit, Vout = Vamp(500 / 800). Then writing
KCL at the inverting terminal, Vs / 400 = (Vamp − Vs) /1.2k ⇒ Vamp / Vs = 1.2k / 400 +1 = 4. Therefore
Vout / Vs = (Vamp / Vs )⋅ (Vout / Vamp ) = 2.5.
(c) Note that since no current goes into the non-inverting terminal of the op-amp, the voltage at that node is
–Vs. KCL at the inverting terminal, −Vs / 4k = (Vout + Vs )/ 20k ⇒ Vout / Vs = −6.
SOLUTION 4.3. Write KCL for both terminals,
(V− −Vi )/1k = (Vo − V− ) / 2k
V− /1k = (Vo − V− )/ 3k
Solving and doing the appropriate substitutions, Vo / Vi = −8 .
SOLUTION 4.4. This is essentially the basic inverting configuration, which is defined as
Vo / Vi = −2k / 1k = −2 .

SOLUTION 4.5. (a) By voltage division VL = 1V ⋅
100
200
= 0.5V . Using Ohm’s law
Is = IL =
1
100 +100
= 5mA .
(b) No current flows in the input terminal of an ideal op-amp, thus Is = 0A and VL = 1V . From Ohm’s
law Ia = IL = VL /100 = 10mA.
SOLUTION 4.6. (a) Using voltage division,
V1 = Vs
32||(8 + 24)
[32||(8 + 24)] + 8





 =
2
3
Vs
Vout = V1
24
24 + 8




= 0.5Vs
(b) By voltage division,
V1 = Vs
32
32 + 40




= 0.8Vs
Vout = V1
24
24 + 8




= 0.6Vs
(c) Using voltage division, V1 = Vs
32
32 +8




= 0.8Vs , as no current enters the non-inverting terminal of the
op-amp. Due to the virtual short property, Vout = V1
24
24 + 8




= 0.6V1. This is indeed the same results as
(b), which should be expected because of the isolation provided by the ideal buffers.
SOLUTION 4.7. Write KCL at the inverting terminal,
−Vs1 /1k − Vs2 / 2k = Vout / 4k ⇒ Vout = −4Vs1 − 2Vs2 = 40mV .
SOLUTION 4.8. (a) The voltage at the non-inverting terminal is V+ = 3 / 2V , KCL at the inverting
terminal gives (1.5− 2.5) /10k = (Vout −1.5)/ 30k ⇒ Vout = −1.5V . The power is
P = Vout
2
/ 500 = 4.5mW.
(b) The voltage at the non-inverting terminal is 3V this time, thus KCL
(3 − 2.5)/ 10k = (Vout − 3) / 30k ⇒ Vout = 4.5V . The power is P = Vout
2
/ 500 = 40.5mW.
SOLUTION 4.9. (a) Define the point between the two op-amp as Vint. Observe that the first op-amp is in
the basic inverting configuration, and the second the non-inverting configuration. By inspection,

Vint / Vs = −R1 / 1k
Vout / Vint = (1+ R2 /1k)
Cascading the two stages, (Vint / Vs)(Vout / Vint) = Vout / Vs =− R1 /1k(1+ R2 /1k) . Solving for
R1 = 20 ⋅1k / (1+ R2 /1k) = 5kΩ. The power absorbed is P = (20 ⋅0.5)
2
/ 8 =12.5W .
(b) Using the same equations as (a), solve for R2 = (20 ⋅1k / 2k −1)1k = 9kΩ.
(c) Rewriting the equation obtained in (a), R1
2
+1kR1 − 20M = 0, and solving the quadratic equation yields
R1 = R2 = 4kΩ.
SOLUTION 4.10. This is a cascade of two non-inverting configuration op-amp of the form
Vo / Vs = (1+10k /10k) for each. Therefore 2 ⋅ 2 = 4.
SOLUTION 4.11. This system is made up of a non-inverting stage with a gain of 1+10k/10k, a voltage
divider of gain 8k/(2k+8k), and a second non-inverting stage of gain 1+10k/10k. The product of all three
yields Vout / Vin = (2)(0.8)(2) = 3.2.
SOLUTION 4.12. (a) By inspection, the gain of the first stage is –1. Then write KCL for the second stage
Vs1 / 2R − Vs2 / R = Vout / 2R ⇒ Vout = Vs1 − 2Vs2 = 10V .
(b) The first stage gain is –0.5, thus Vout = 2R(0.5Vs1) / 2R − 2R(Vs2)/ 0.5R =−7.5V, using the same
procedure as in (a).
SOLUTION 4.13. (a) This is a cascade of a summing amplifier with the following transfer characteristic,
Vo = −4Vs1 − 2Vs2 , and an inverting stage of gain –1.5. Thus Vout = 1.5(4Vs1 + 2Vs2 ) = 2.25V .
(b) Notice that the only difference is the gain of the inverting stage, which is now –2. Therefore
Vout = 2(4Vs1 + 2Vs2) = 3V .
SOLUTION 4.14. This circuit is a cascade of two summing amplifier where the output of the first is an
input of the second stage. The transfer function of the first stage is Vo = −2RVs1 / 2R − 2RVs2 / R, which
is substituted in the transfer function of the second stage to obtain
Vout = −R[−2RVs1 / 2R − 2RVs2 / R]/ R− RVs3 / R = Vs1 + 2Vs2 − Vs3 =−2V .
SOLUTION 4.15. Writing KCL at the inverting node, −V1 / R1 − V2 / R2 − V3 / R3 = Vout / Rf , and
solving for Vout = −
Rf
R1
V1 +
Rf
R2
V2 +
Rf
R3
V3





.

SOLUTION 4.16. Referring figure P4.15, the value of the resistance must satisfy the following
constraints:
R1 = R2 = R3 = 3R
Rf = R
These will yield the inverted average. If polarity is a concern, a second inverting stage should be added
with a unity gain, i.e. both R’s equal.
SOLUTION 4.17. Using the topology of 4.12 the following parameters are chosen,
Ga1 = 3, Ga2 = 5, Gb1 = 2, Gb2 = 4
For the time being assume Gf = 1. Now we calculate δ = (1+ 3 + 5) −(2 + 4) = 3, this sets Gg = 3.
(a) The requirement for Gf = 10µS sets the scaling factor K = 10µ / 1 =10µ . This then yields the
following set of parameters,
Ga1 = 30µS, Ga2 = 50 µS, Gb1 = 20µS, Gb2 = 40µS, Gf = 10µS, Gg = 30µS
(b) The requirement for Gf = 2µS , sets the scaling constant to 2uS. So the following parameters are
obtained:
Ga1 = 6µS, Ga2 =10 µS, Gb1 = 4 µS, Gb2 = 8µS
Furthermore for Gg = 12µS , ∆G = 6µS in order to make the incident conductance equal at both terminal.
(c) Using the starting values from (a), one could choose a scaling constant of 5 µS. This will yield the
following resistances:
Ra1 = 66.67kΩ, Ra2 = 40kΩ, Rb1 = 100kΩ, Rb2 = 50kΩ, Rf = 200kΩ, Rg = 66.67kΩ
These are all reasonable values for circuit implementation.
SOLUTION 4.18. (a) Choosing the following initial values:
Ga1 = 3S, Ga2 = 5S, Gb1 = 11S, Gb2 = 4S, Gf =1S

then calculate δ = (1+ 3 + 5) −(11+ 4) =−6 . Thus Gg = 1S, and ∆G =1 + 6 = 7S . Scaling everything by
1µS, yield this final set of parameters, which meet the requirements.
Ga1 = 3µS, Ga2 = 5µS, Gb1 =11µS, Gb2 = 4µS, Gf = 1µS, Gg =1µS, ∆G = 7uS
(b) The set of parameters remains unchanged, except for ∆G which now becomes 6uS in order to
maintain the equal termination conductance requirement due to Gg = 0S.
(c) Scale the initial parameters of (a) by 5uS, and get the following set of resistances:
Ra1 = 66.67kΩ, Ra2 = 40kΩ, Rb1 = 18.18kΩ, Rb2 = 50kΩ, Rf = 200kΩ, Rg = 200kΩ, ∆R = 28.57kΩ
SOLUTION 4.19. (a) Choosing the following initial set of parameters:
Ra1 = 1/ (4S) = 0.25Ω, Ra2 = 1/ (2S) = 0.5Ω, Rb1 = 1/ (5S) = 1/ 5Ω, Rb2 =1/ (4S) = 0.25Ω, Rf = 1Ω
and δ = (1+ 4 + 2) − (5 + 4) = −2 , thus choose Rg =1/ (1S) = 1Ω and ∆R = 1/ (1+ 2) = 1/ 3Ω. To meet
the Rf = 50kΩ requirement, all the parameters must be scaled by 50k, which gives
Ra1 = 12.5kΩ, Ra2 = 25kΩ, Rb1 = 10kΩ, Rb2 =12.5kΩ, Rf = 50kΩ, Rg = 50kΩ, ∆R = 16.67kΩ
(b) Same as (a) with a 100k scaling constant:
Ra1 = 25kΩ, Ra2 = 50kΩ, Rb1 = 20kΩ, Rb2 = 25kΩ, Rf = 100kΩ, Rg = 100kΩ, ∆R = 33.33kΩ
SOLUTION 4.20. (a) When the op-amp is in its active region vout / vs = −5 . Thus it will operate in its
active region when −3 ≤ vs ≤ 3, and will saturate at 15V when vs ≤ −3 , and at –15V when vs ≥ 3. SPICE
yield the following plot:

(b) Using SPICE the following plot is obtained:

SOLUTION 4.21. The first stage is in a summing configuration, thus its output is, assuming it's in the
active region of operation, -15 V which means it is just about to saturate. The second stage is in the
inverting configuration with a gain of -1.5, which means that the overall output will be saturated at 15V.
SOLUTION 4.22. When vin −80k
vin +1.5
100k




> 0, or vin > 6 the output of the comparator saturates at
–15 V, when it is vin < 6 , it will saturate at 15 V. The following plot is obtained from SPICE.
SOLUTION 4.23. When vin −10k
vin + 20
110k




> 0 , or vin > 2 the output of the comparator will be
saturated at -15V. Otherwise when it is < 2V the output saturates at 15V. In SPICE:

SOLUTION 4.24. Based on the same reasoning as the previous questions,
The output will be +Vsat, when vin < vref 1 −
R1 + R2
R2





 = −
R1
R2
vref , and –Vsat for
vin > vref 1 −
R1 + R2
R2





 = −
R1
R2
vref .
SOLUTION 4.25. Using the previously derived relationship, and the topology of figure P4.24, set
vref = −1.5V , and R1 = 2kΩ and R2 = 3kΩ. Set the power supplies to the Op-amp to +/– 10V to satisfy
the Vsat requirement. Also the input to the inverting and non-inverting terminal are reversed for fig. P4.24.
Verifying in SPICE we obtain the following,

SOLUTION 4.26. The design that fulfills the requirement is the same as for P4.25, with the input to the op
amp reversed. The following is obtained from SPICE,

SOLUTION 4.27. First, for the comparator to give +Vsat for the lower voltages, the inputs to the op amp
in the topology of P4.24 must be interchanged. Then the components are chosen to satisfy the following
relationship, vswitch = vref 1−
R1 + R2
R2





 =−
R1
R2
vref . Choose vref = −1.5V , and R1 = 2k and R2 = 1k.
Verifying in SPICE,

SOLUTION 4.28. Write KCL at the inverting terminal, noting that the no current flows into it:
(V− − vin)/ R = (vout − V− )/ R. Use the following relationship vout = A(V+ − V− ) = −AV− . Solving
using the previous two equations yields vout / vin = −
A
A + 2
.
SOLUTION 4.29. (a) By inspection the voltage gain for the ideal case is –1. When A=1000, the gain
becomes –0.998, thus 0.2%.
(b) Repeating the method of P4.28, and setting vout / vin = −
ARf
Rf + R1 + AR1
to –1 and solving for
Rf =10.417kΩ.
(c) Solving the previous equation when the gain is –1, Rf / R1 = (A+ 1)/ ( A−1).
SOLUTION 4.30. (a)The first part was obtained in P4.29. Rearranging the equation yields
vout / vin = −
Rf
R1
1
1 + 1+ Rf / R1
( )/ A








.

(b) The error is caused by (1+ Rf / R1) / A in the denominator, and may be defined, in percent, as
100 −
1
1+ 1+ Rf / R1
( )/ A








⋅100. Thus for the conditions listed in the problem, it will always be less than
2.05%. With A = 10000 it will be less than 0.21%.
SOLUTION 4.31. (a) Substituting the non-ideal model, and writing KCL at the inverting terminal,
(V− − vin)/ R1 + V− / Rin = (vout − V− ) / Rf is obtained. Now observe the following dependencies,
iout = vout / RL, and vout =− AV− − (iout + (vout − V− )/ Rf )Rout . Using these three equations, substitute
the second into the third and then solve for vout / vinusing the last two. This yields
V− = vout + vin
Rf
R1
1/
Rf
R1
+
Rf
Rin
+1





 = vout
−1 −
Rout
RL
−
Rout
Rf
A −
Rout
Rf












and
vout / vin = −
Rf
R1
1
1+
1 +
Rout
Rf
+
Rout
RL





 1+
Rf
R1
+
Rf
Rin






A −
Rout
Rf










































A gain of –9.988
(b) For an ideal op-amp the gain is −Rf / R1=-10.
(c) The error is about 0.1175%.
SOLUTION 4.32. The gain is –9.883, and the error 1.16%
SOLUTION 4.33. This derivation was performed in P4.31.

SOLUTION 4.34. Assume that the sliding contact is at the bottom of Rp. Then, writing KCL at the
inverting terminal yields vin / Ro = (vout − vin)/ Rp . This implies vout / vin = 1 + Rp / Ro . When the
slider is at the top, it is evident that vout = vin . Therefore 1 ≤ vout / vin ≤1 +
Rp
Ro
.
SOLUTION 4.35. Writing KCL at inverting input, and making use of voltage division,
−vin / R1 = βvout / Rf
[ ] where β is the fraction of vout that appears across Rf. Hence,
vout
vin
=−
Rf
βR1
.
When the slider is at the top β= 1 and
vout
vin
=−
Rf
R1
. When the slider is at the bottom, the fraction of vout
appearing across Rf is β=
Rf / /R0
Rf / /R0 + Rp
=
Rf R0
Rf + R0
×
1
Rf R0
Rf + R0
+ Rp
=
Rf R0
Rf R0 + Rp(Rf + R0)
. Hence
1
β
=
Rf R0 + Rp(Rf + R0 )
Rf R0
=1 +
Rp
R0
+
Rp
Rf
. It follows that
vout
vin
=−
Rf
βR1
=−
Rf
R1
1 +
Rp
R0
+
Rp
Rf





 .
Therefore the range of achievable voltage gain is
−
Rf
R1
≥
vout
vin
≥ −
Rf
R1
1 +
Rp
R0
+
Rp
Rf






SOLUTION 4.36. Using the basic non-inverting configuration of figure 4.10 characterized by
vout / vin = 1+
Rf
R1





 , i.e., µ = 1+
Rf
R1





 .
SOLUTION 4.37. At first glance, one might use two inverting configurations, figure 4.5, in cascade.
However, such would not have infinite input resistance. To circumvent this problem we add a buffer
amplifier as per figure 4.7 at the front end of a cascade of two inverting configurations. The resulting
overall gain is µ =
Rf1
R11






Rf 2
R12





 . Indeed, such a configuration can achieve theoretically any gain greater
than zero.

SOLUTION 4.38. Using a single inverting amplifier configuration, figure 4.5, preceded by a buffer stage
of figure 4.7. The gain is µ = −
Rf
R1
.
SOLUTION 4.39. By KVL for figure P4.39a, Vo = −i1Rf . Thus to achieve Vo = −i1rm in figure P4.39b,
we set Rf = rm .
SOLUTION 4.40. Writing KCL at the inverting node of the ideal op amp yields IL = Vi / Ra , which is
indeed independent of the load resistor which has no effect on the load current.
SOLUTION 4.41. The current through the LED is IL = 10
R1
10k




/ 3.8k, so for (a) it is 1.32mA and for
(b) 2.11mA.
SOLUTION 4.42. Applying KCL at the inverting terminal, IL = vin / R1. Again, ideally, RL does not
affect IL.
SOLUTION 4.43. (a) Defining a temporary voltage Vo at the output of the op-amp, we can write KCL at
the inverting and non-inverting terminal:
(V− − 2)/ 1k = (Vo − V− ) / 2k
V− /100 + (V− − Vo) / 200 = Iout
Substituting the first equation into the second and simplifying causes Vo to drop out and Iout = 20mA .
(b) The answer remains the same as the value of the load resistance was not used for finding the load
current.
SOLUTION 4.44. Using the same approach as for the previous question, but with resistor labels instead,
the following equations are obtained from KCL:
V− =
R2Vs + R1Vo
R2 + R1
Iout = V−
R2 + R1
R1R2





 −
Vo
R2
Substituting the first into the second yields Iout = Vs / R1.

SOLUTION 4.45. (a) Vs = 5 V, (b) Iout = 10mA sets R1 = Vs / Iout = 500 Ω. (c) From KVL and Ohm's
law, Is = (Vs + RLIout) / αR1. We require Is < 0.5 mA. This means that in the worst case, RL = 500 Ω,
(Vs + RLIout )
0.5 ×10
−3
R1
=
5 + 500 × 0.01
0.25
= 40 <α .
(d) From KVL and Ohm's,
Vo = −RLIout − (Iout +(RLIout )/ R1)R2 ≥ −20 V
Hence
R2 ≤
20 − RLIout
(Iout + (RLIout) / R1)
=
20 − 5
0.01 + 5/ 500
= 750
Hence one design is to pick R2 = 750 Ω and α = 40 which impliesαR2 = 30 kΩ.
SOLUTION 4.46. The exact same design as P4.45 can be used with the isolation buffer of figure 4.7
placed at the input of it in order to provide the infinite input resistance needed by P4.46b.
SOLUTION 4.47. The general expression for this summing circuit is
Vout = −
Rf
Ro
Vo −
Rf
R1
V1 −
Rf
R2
V2 −
Rf
R3
.
(a) Using the expression above |Vout |=| −1 − 0 − 0 − 8| E = 9E.
(b) |Vout |=| −0 − 2 − 4 − 0| E = 6E.
(c) It has to be a linear combination of 8, 4, 2, 1, thus [1 1 0 1] would yield 13E.
(d) With the same approach, [0 1 1 1].
SOLUTION 4.48. For this implementation we add an extra R-2R branch along with an extra summing
input to the op amp. From the theory developed in Example 4.9 the total resistance seen by the source is
2R. For the total power supplied by the source to be less than 0.02 W, we require R ≥
E2
2 0.02
( )
= 2.5 kΩ.

SOLUTION 4.49. The same steps as in the previous questions are repeated. Because the resistance seen
by the source is unchanged no matter how many branches are added to the R-2R network,
R ≥
E2
2 0.01
( )
= 5 kΩ.
SOLUTION 4.50. (a) If the input is 3vmax / 8, then the first comparator will give –Vsat, keeping S2 down.
The next comparator will output +Vsat, causing S1 to go up. After subtraction, the input to the last
comparator is vmax / 8 yielding +Vsat at its output since its input is slightly above the reference input
level. Thus the logic output values are [0 1 1].
(b) Putting in 6vmax / 8, will cause +Vsat and S2 to go up. The input to the second comparator will be
2vmax / 8, which will cause +Vsat and S1 to go up. The input to the last comparator will be 0, thus it will
output –Vsat. The corresponding logic output is [1 1 0].
SOLUTION 4.51. Simply add a subtractor and switch to the last comparator, followed by an additional
comparator. The reference level to the new (additional) comparator will be vmax
−
/16, and its output will be
the new least significant bit.
SOLUTION 4.52. (a) Writing the node equation for figure P4.52c,
Vout
RL
' +
(Vout − V1)
10k
=
A(0 − V1) − Vout
Rout
which implies that
Vout / V1 =
1
10k
−
A
Rout






1
RL
'
+
1
10k
+
1
Rout






For figure P4.52a, the corresponding node equation is
Vout
RL
=
A(0 − Vin) − Vout
Rout
which leads to

Vout / Vin =
−
A
Rout






1
RL
+
1
Rout






Note that
1
RL
=
1
RL
1 +
1
10k
, which when substituted into the later equation make both of them
approximately the same since the 1/10k term in the numerator of Vout / V1 has a negligible contribution.
(b) Writing the node equation for figure P4.52d, yields
Vout
RL
' +
(Vout − V2)
10k
=
A(0 − V1) − Vout
Rout
Hence
V1 = V2
100|| Rin
(100 ||Rin) +10k





 ≈
V2
101
Solving produces Vout / V2 =
1
10k
−
A
101Rout






1
RL
1
+
1
10k
+
1
Rout






. Note that as in (a) the 1/10k term in the numerator is
negligible; after eliminating this negligible term, one sees that
Vout
V2
is 101 time smaller than
Vout
Vin
.
SOLUTION 4.53. (a) Using the equation just derived, after substituting in the values, the gain is –980.392
(b) From the previous equation, Vout / V1 = −980.382 ; write KCL at the non-inverting terminal to obtain,
Vin − V1
10k
=
V1
Rin
+
V1 − Vout
10k
; substitute Vout = −980.382V1; solve for V1 / Vin , and then multiply both gains
to obtain (Vout / V1)(V1 / Vin ) = Vout / Vin = −0.9979 .
(c) They only differ by about 0.01%, thus they are very similar.
SOLUTION 4.54. Writing out the transfer equation, Vout =
R2
R1
Vs2 −
R2
R1
Vs1, thus R2 / R1 = 4 . Using
R2 = 100kΩ, R1 = 25kΩ. As expected SPICE shows to noticeable difference in outputs when the source
resistances are varied.
SOLUTION 4.55. Due to the ideal nature of the op-amp, the voltage VRb
= Vs2 − Vs1. By KVL

V2 = Vs2 + Ra(Vs2 − Vs1)/ Rb
V1 = Vs1 − Ra(Vs2 − Vs1) / Rb
Next, V1 − V2 = (Vs1 − Vs2)(1+ 2Ra / Rb) .
SOLUTION 4.56. (a) Noticing that the final stage is a summing op-amp in which Vout =
R2
R1
V1 −
R2
R1
V2 .
From the previous question, Vout =
R2
R1
(V1 −V2) =
R2
R1
(1+ 2Ra / Rb)(Vs1 − Vs2) . Thus
α =
R2
R1
(1+ 2Ra / Rb). The gain α can be varied by adjusting the single resistance Rb.
(b) Picking the set of values below will satisfy the requirement:
R2 = 100kΩ, R1 =100kΩ, Ra = 20kΩ, Rb = 10kΩ.
(c) Doing the SPICE simulation using the parameters from (b) yield 5 V at the output for
Vs1 −Vs2 = 2 −1 V. Setting Rb arbitrarily to 20 kΩ, the output now becomes 3 V, which agrees with the
relationships developed earlier.

Solution 5.1. (a) Vs = 10 V, P = 20 W and P = Vs×Is implies Is = 2 A.
(b) Rin = Vs/Is = 10/2 = 5 Ω
(c) By the linearity/proportionality property:
Vs
new
Vs
old
=
Is
new
Is
old
which implies
2
10
=
Is
new
2
implies
Is
new
= 0.4 A.
(d) P
new
= Vs
new
× Is
new
= 2 × 0.4 = 0.8 watts. Observe that
P
new
Pold
=
0.8
20
≠
Vs
new
Vs
old
=
2
10
It follows that the proportionality property does not hold for power calculations.
Solution 5.2 First note that the ratio IR/VS is constant. With the given values of voltage and current, this
ratio is:
IR/VS = 0.25/25 = 0.01
Power dissipated in the resistor is
P = IR
2
R = 2.5 è IR
2
= 2.5/R = 0.25 è IR = 0.5
Since IR is always 0.01×VS, it follows that VS = 50V.
Solution 5.3 Label the resistances R1, R2, and so on in the manner shown in Example 5.11. In this
problem, we have R1 to R10 (the last being the 2 Ohm resistance at the voltage source). First, assume that
V1 (the voltage across R1) is 1V. Then evaluate the rest of the currents and voltages until you deduce the

resulting VS. It should be noted that the equivalent resistance looking into R3, R5, R7, and R9 is always
2Ω.
V1 =1 ⇒ I1 =
V1
4
= 0.25 ⇒ I2 = 0.25 ⇒ V2 = I2 × 2 = 0.5 V
V3 = V1 + V2 = 1.5 è I3 =
V3
3
= 0.5 è I4 = I3 + I2 = 0.75 è V4 = I4 × 4 = 3
V5 = V3 + V4 = 4.5 è I5 =
V5
3
= 1.5 è I6 = I5 + I4 = 2.25 è V6 = I5 × 4 = 9
V7 = V6 + V5 =13.5 è I7 =
V7
3
= 4.5 è I8 = I7 + I6 = 6.75 è V8 = I8 × 4 = 27
V9 = V8 + V7 = 40.5 è I9 =
V9
3
= 13.5 è I10 = I9 + I8 = 20.25 è V10 = I10 × 2 = 40.5
VS = V9 + V10 = 40.5 + 40.5 = 81 V
Thus, an 81 V input produces a 1 V output è Vout = (1/81)×VS = 2 V.
Solution 5.4 Label the resistances R1 to R10 progressively from right to left just like in the previous
problem. Then, assume Iout = 1 and proceed as follows:
Iout = 1 è V1 = Iout × 4 = 4 è I2 =
V1
4
= 1
I3 = I1 + I2 = 2 è V3 = I3 × 4 = 8 è V4 = V3 + V1 = 12 è I4 =
V4
3
= 4
I5 = I4 + I3 = 6 è V5 = I5 × 4 = 24 è V6 = V5 + V4 = 36 è I6 =
V6
3
= 12
I7 = I6 + I5 = 18 è V7 = I7 × 4 = 72 è V8 = V7 + V6 = 108 è I8 =
V8
3
= 36
I9 = I8 + I7 =54 è V9 = I9 × 4 = 216 è V10 = V9 + V8 =324 è I10 =
V10
3
= 108

IS = I10 + I9 = 162
è Iout/IS = 1/162 è Iout = (1/162)×40.5 = 0.25A.
Solution 5.5 (a) MATLAB code given in problem.
(b) Subsitute to obtain V1 = 10V.
(c) Req = VS/IS = 11.6667Ω.
(d) First, define r1 = 1:0.25:10;
then create an outermost loop around the code of part (a) as: for j=1:length(r1)
then, in the statement defining R, do R = [R1(j), R2, R3, R4, R5, R6, R7, R8]’;
Finally, replace the last statement with Vs(j) = V(n) + V(n-1); end;
The following is the resulting plot:
Solution 5.6 (a)
The following code can be used:
n = 9;

v = zeros(n,1);
i = zeros(n,1);
r = [r1 r2 r3 r4 r5 r6 r7 r8 r9]’;
i(1) = 1;
v(1) = i(1)*r(1);
i(2) = i(1);
for k=2:2:n-2
v(k) = r(k)*i(k);
v(k+1) = v(k)+v(k-1);
i(k+1) = v(k+1)/r(k+1);
i(k+2) = i(k+1) + i(k);
end;
v(8) = i(8)*r(8);
v(9) = v(8) + v(7);
i(9) = v(9)/r(9);
Is = i(9) + i(8);
It follows that Is =16.9877A.
(b) By the proportionality property:
I1
new
I1
old
=
Is
new
Is
old
→ I1
new
=
1
16.9877
× 200mA =11.77mA
(c) Req = v(9)/Is = 38.15Ω.
Solution 5.7 Va = 12V, iB = 60m
By inspection:
Vout_a = 300/900×12 = 4V
Vout_b = (300||600)×60m = 12

_ Vout = 4 + 12 = 16V.
*SOLUTION 5.8. Part 1: Set the 3 A current source to zero. This generates an open circuit in place of
the current source eliminating the effect of the series 0.1 S resistor. The equivalent circuit is:
By voltage division,
VL
12V
=
1
0.25 + 0.2 + 0.05
1
0.25 + 0.2 + 0.05
+
1
0.1
×12 =
2
2 +10
×12 = 2 V
Part 2: Set the 12 V source to zero. This generates a short circuit in place of the voltage source which
shorts out the effect of the 0.5 S resistor. The equivalent circuit is:
Note that the 0.1 S resistor in series with the 3 A source is redundant to the calculation of VL. Hence, by
Ohm's law,
VL
3A
=
1
0.25 + 0.2 + 0.05 + 0.1
× 3 =
3
0.6
= 5 V

Therefore by superposition,
VL = VL
12V
+ VL
3A
= 2 + 5 = 7 V
Solution 5.9 Replace the dependent source by an independent voltage source VS:
In the following analysis, we have to always compute Va because It defines the constraint on Vs. So, when
only the 88V source Is active, VA Is the result of voltage division between the 60||30 Ω resistor and the
120||30 Ω resistor. So,
Va_1 = 40V
And, since deactivated VS, Vout_1 = 0.
Now, due to the 55V source, we have
VS

Now, the 120 and 60 Ω resistors are in parallel, and the same can be said about the 30 and 30 Ω resistors.
Thus, another voltage divider gives:
Va_2 = – 15V and Vout_2 = 0V.
Finally, when VS1 is active, the left part of the circuit consists only of resistances, so Va_3 = 0. Vout is given
by another divider formula:
Vout_3 = 90/100×VS
Now add all contributions:
Va = 40 –15 + 0 = 25V
Vout = 0 + 0 + 0.9VS, where VS = 2Va.
Vout = 0.9×2×25 = 45V.
Finally, P = V2
/R = 22.5W.
Solution 5.10
Due to 3A source:
iout = 1A by current division between the two paths. So, vout_1 = 2 V.
Due to 1A source:

iout = 2/3A again by current division. So, vout_2 = 4/3V.
Due to 1vV source:
iout = 14/6 = 7/3A (by Ohm’s Law). So, vout_3 = 14/3V.
Finally, vout = 6/3 + 4/3 + 14/3 = 8V, and the power delivered by the source is 8×1 = 8 W.
Solution 5.11
Due to 22 V source:
Req = 900||225 = 180. Now, by voltage divider:

Linear circuit analysis - solution manuel (R. A. DeCarlo and P. Lin) (z-lib.org).pdf

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