CHAPTER 1
Basic Concepts
Section 1.2 Solutions
1.2.1 If the current in an electric conductor is 1.2 A, how many coulombs of charge pass any point in a
30-s interval?
Solution:
I = 1.2 A, ∆t = 30 s
Q = I ⋅ ∆t
Q = 36 C
1.2.2 Determine the time interval required for a 15-A battery charger to deliver 6000 C.
Solution:
I = 15 A, Q = 6000 C
Q
∆t = _
I
∆t = 400 s
1.2.3 If a 12-V battery delivers 100 J in 5 s, find (a) the amount of charge delivered and (b) the current
produced.
Solution:
V = 12 V, ∆W = 100 J in 5 s
∆W
a. ∆Q = _
V
∆Q = 8.33 C
∆Q
b. I = _, ∆t = 5 s
∆t
I = 1.67 A
1.2.4 The current in a conductor is 2.5 A. How many coulombs of charge pass any point in a time interval
of 2.5 min?
Solution:
I = 2.5 A, ∆t = 2.5 min = 150 s
Q = I ⋅ ∆t
Q = 375 C
1
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1.2.5 If 90 C of charge pass through an electric conductor in 45 s, determine the current in the conductor.
Solution:
Q = 90 C, ∆t = 45 s
Q
I=_
∆t
I=2A
1.2.6 Calculate the power absorbed by element A in Fig. P1.2.6.
2A
−
10 V
A
+
FIGURE P1.2.6
Solution:
−
A
10 V
+
−2 A
PA = (10)(−2)
PA = −20 W absorbed
1.2.7 Calculate the power supplied by element A in Fig. P1.2.7.
5A
+
25 V
A
−
FIGURE P1.2.7
Solution:
−5 A
+
25 V
A
−
PA = (25)(−5)
PA = −125 W supplied
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Solutions to Problems
3
1.2.8 Determine the number of coulombs of charge produced by a 12-A battery charger in 1 h.
Solution:
I = 12 A, ∆t = 1 hr = 60 min = 3600 s
Q = I ⋅ ∆t
Q = 43.2 kC
1.2.9 The current at a given point in a certain circuit may be written as i(t) = –3 + t A. Find the total
charge passing the point between t = 99 and t = 102 s.
Solution: The charge passing through the circuit is
q=
102
102
∫99 i(t)dt = ∫99
(−3 + t) dt
99 = 292.5 C
_ − −3 × 99 + _
= (−3 × 102 + 102
2) (
2)
2
2
1.2.10 The charge entering the positive terminal of an element is given by the expression q(t) = –10 e–t
mC. The power delivered to the element is p(t) = 2 e–2t mW. Calculate the current in the element, voltage
across the element and the energy delivered to the element in the interval 0 < t < 100 ms.
Solution:
dq = 10 e −t mA
I =_
dt
Now P = VI, thus V = 0.2 e −t V, is the voltage across the element.
∫
E = p(t)dt =
2
∫0 2 e −tdt = 0.362 mJ
1.2.11 The voltage across an element is 12 e −2t V. The current entering the positive terminal of the
element is 2 e −2t A. Find the energy absorbed by the element in 1.5 seconds starting from t = 0.
Solution:
V(t) = 12 e −2t V
i(t) = 2 e −2t A
W=
1.5
t2
∫t1 V ⋅ idt = ∫0
|
(12 e −2t) ⋅ (2 e −2t)dt
−4t) 1.5
(
24 e
W=_
−4
W = 5.99 J
0
1.2.12 The power absorbed by the BOX in Fig. P1.2.12 is 2 e −2t W. Calculate the amount of charge that
enters the BOX between 0.1 and 0.4 seconds.
5e−t V
+
−
BOX
FIGURE P1.2.12
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C HA PTER 1
Basic Concepts
Solution:
P(t) = 2 e −2t W
V(t) = 5 e −t V
P(t)
= 0.4 e −t A
i(t) = _
V(t)
∆q(t) =
0.4
∫0.1 i(t) dt
= (−0.4 e −t)| 0.4
0.1
q(t) = 93.8 mC,
0.1 s < t < 0.4 s
1.2.13 The power absorbed by the BOX in Fig. P1.2.13 is 0.1 e −4t W for t ≥ 0s. Calculate the energy
absorbed by the BOX during this same time interval.
10e−2t V
+
−
BOX
FIGURE P1.2.13
Solution:
P(t) = 0.1 e −4t W
∫
W = P(t)dt =
|
(0.1 e −4t)
W=_
−4
W = 25 mJ
∫0
∞
0.1 e −4t dt
∞
0
1.2.14 Five coulombs of charge pass through element E in Fig. P1.2.14 from point A to point B.
If the energy absorbed by the element is 150 J, determine the voltage across the element.
B
+
V1
−
A
FIGURE P1.2.14
Solution:
W = 150 J, Q = 5 C
W = −V1 ⋅ Q
W
V1 = − _
Q
V1 = −30 V
1.2.15 The current that enters an element is shown in Fig. P1.2.15. Find the charge that enters the
element in the time interval 0 < t < 20 s.
i(t) mA
10
0
10
20
t (s)
FIGURE P1.2.15
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Solutions to Problems
5
Solution:
i(t) = m ⋅ t + b
10m − 0 = −1m
m = ________
10 − 20
i(t) = −1m ⋅ t + b
10m = −1m ⋅ (10 s) + b
b = 20 m
i(t) = (−t + 20) mA
20
q(t) =
∫0
q(t) =
∫0
10
i(t)dt
10 × 10 −3 dt +
20
∫10
20 − t dt
_
1000
|
t2
1
_
_
q(t) = 10 × 10 −3 ⋅ t| 10
0 + 1000 (20t − 2 )
20
10
q(t) = 0.15 C, 0 < t < 20 s
1.2.16 Element A in the diagram in Fig. P1.2.16 absorbs 30 W of power. Calculate Vx.
3A
+
Vx
A
−
FIGURE P1.2.16
Solution:
−3 A
+
Vx
A
−
30 = Vx ⋅ (−3)
Vx = −10 V
1.2.17 Element B in the diagram in Fig. P1.2.17 supplies 84 W of power. Calculate Ix.
−
24 V
B
+
Ix
FIGURE P1.2.17
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Basic Concepts
Solution:
Ix
+
−24 V
B
−
84 = (−24) ⋅ Ix
Ix = −3.5 A
1.2.18 Element B in the diagram in Fig. P1.2.18 supplies 72 W of power. Calculate VA.
3A
+
VA B
–
FIGURE P1.2.18
Solution:
–3 A
72 = VA ⋅ (−3)
+
VA B
–
VA = −24 V
1.2.19 Element B in the diagram in Fig. P1.2.19 supplies 90 W of power. Calculate Ix.
+
18 V B
−
Ix
FIGURE P1.2.19
Solution:
90 = (18) ⋅ Ix
Ix = 5 A
1.2.20 Two elements are connected in series, as shown in Fig. P1.2.20. Element 1 supplies 27 W of power.
Is element 2 absorbing or supplying power, and how much?
1
2
+
3V
−
−
5V
+
FIGURE P1.2.20
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Solutions to Problems
7
Solution:
+
3V
−
−
5V
+
1
2
I
P1 = 27 = V1 ⋅ I
27 = 9 A
I=_
3
P2 = V2 ⋅ I = (5)(9)
P2 = 45 W absorbed
1.2.21 The energy absorbed by the BOX in Fig. P1.2.21 is given below. Calculate and sketch the current
flowing into the BOX. Also calculate the charge that enters the BOX between 0 and 12 s.
i (t)
+
−
12 V
BOX
w(t) (J)
5
6
1
2
3
4
7
8
12
10
5
t (s)
11
9
−2.5
FIGURE P1.2.21
Solution:
i(t) (A)
5
24
5
48
4
1
2
3
6
5
7
8
9
10
11
12
t (s)
−5
24
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Basic Concepts
dw
P=_
dt
P = V ⋅ i = (12) ⋅ i
0s≤t≤2s
‾
2.5 = _
5 A
P =_
i=_
V 12 24
5 − 0 = 2.5 W,
P=_
2−0
2s≤t≤4s
‾
5 − 5 = 0 W,
P=_
4−2
4s≤t≤6s
‾
i=0A
0 − 5 = −2.5 W,
P=_
6−4
6s≤t≤7s
‾
0 − 0 = 0 W,
P=_
7−6
7s≤t≤8s
‾
2.5 = − _
5 A
P =−_
i=_
12
24
V
i=0A
−2.5 − 0 = −2.5 W,
P=_
8−7
8 s ≤ t ≤ 10 s
‾
10 s ≤ t ≤ 12 s
‾
2.5 = − _
5 A
P =−_
i=_
12
24
V
−2.5 − (−2.5) = 0 W,
P = ___________
10 − 8
i=0A
0 − −2.5 = 1.25 W,
P=_
12 − 10
1.25 = _
5 A
P =_
i=_
12
48
V
(
q=
)
∫ i dt
5 ⋅ (2) + − _
5
5
5
_
_
q = (_
( 24 ) ⋅ (2) + (− 24 ) ⋅ (1) + ( 48 ) ⋅ (2)
24 )
q=0C
1.2.22 The energy absorbed by the BOX in Fig. P1.2.22 is shown below. How much charge enters the
BOX between 0 and 10 milliseconds?
i (t)
+
−
15 V
BOX
w(t) (mJ)
15
10
5
1
2
3
4
5
6
7
8
9
10
t (ms)
−5
−10
−15
FIGURE P1.2.22
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Solutions to Problems
9
Solution:
dw
P=_
dt
P = V ⋅ i = (15) ⋅ i
0 s ≤ t ≤ 1 ms
‾
5m − 0 = 5 W,
P=_
1m − 0
5 =_
1A
P =_
i=_
V 15 3
1 ms ≤ t ≤ 3 ms
‾
5m − 5m = 0 W,
P=_
3m − 1m
i=0A
3 ms ≤ t ≤ 4 ms
‾
15m − 5m = 10 W,
P=_
4m − 3m
4 ms ≤ t ≤ 6 ms
‾
15m − 15m = 0 W,
P = ___________
6m − 4m
6 ms ≤ t ≤ 7 ms
‾
10m − 15m = −5 W,
P = ___________
7m − 6m
7 ms ≤ t ≤ 8 ms
‾
10m − 10m = 0 W,
P = ___________
8m − 7m
8 ms ≤ t ≤ 10 ms
‾
0 − 10m = −5 W,
P=_
10m − 8m
10 = _
P =_
2A
i=_
V 15 3
i=0A
5 = −_
P = −_
1A
i=_
15
3
V
i=0A
5 = −_
P = −_
1A
i=_
15
3
V
∫
∆q = i dt
1 (1m) + _
2
−1
−1
_
_
∆q = (_
( 3 )(1m) + ( 3 )(1m) + ( 3 )(2m)
3)
∆q = 0 C
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C H A PTER 1
Basic Concepts
1.2.23 The energy absorbed by the BOX in Fig. P1.2.23 is shown in the graph below. Calculate and sketch
the current flowing into the BOX between 0 and 10 milliseconds.
i (t)
12 V
+
−
BOX
w(t) (mJ)
30
20
10
5
1
2
3
6
7
4
8
9
10
t (ms)
−10
−20
−30
FIGURE P1.2.23
Solution:
i(t) (A)
5
6
t (ms)
−5
6
−5
3
dw
P=_
dt
P = V ⋅ i = (12) ⋅ i
0 s ≤ t ≤ 1 ms
‾
1 ms ≤ t ≤ 3 ms
‾
3 ms ≤ t ≤ 4 ms
‾
4 ms ≤ t ≤ 5 ms
‾
c01SolutionsToProblems_V2.indd 10
10m − 0 = 10 W,
P=_
1m − 0
5A
10 = _
P =_
i=_
V 12 6
10m − 10m = 0 W,
P=_
3m − 1m
i=0A
0 − 10m = −10 W,
P=_
4m − 3m
5A
10 = −_
P = −_
i=_
12
6
V
0 − 0 = 0 W,
P=_
5m − 4m
i=0A
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Solutions to Problems
5 ms ≤ t ≤ 6 ms
‾
6 ms ≤ t ≤ 7 ms
‾
7 ms ≤ t ≤ 9 ms
‾
t > 9 ms
‾
−20m − 0 = −20 W,
P=_
6m − 5m
5A
20 = −_
P = −_
i=_
3
12
V
−20m − −20m = 0 W,
P = ____________
7m − 6m
(
)
0 − −20m = 10 W,
P=_
9m − 7m
(
P = 0 W,
)
11
i=0A
5A
10 = _
P =_
i=_
V 12 6
i=0A
1.2.24 (a) In Fig. P1.2.24(a), P1 = 42 W. Is element 2 absorbing or supplying power, and how much?
(b) In Fig. P1.2.24(b), P2 = −72 W. Is element 1 absorbing or supplying power, and how much?
1
2
+
14 V
−
+
6V
−
1
2
(a)
−
5V
+
+
18 V
−
(b)
FIGURE P1.2.24
Solution:
I
+
14 V
−
+
6V
−
1
2
a) P1 = 42 = V1 ⋅ I
42 = 3 A
I=_
14
P2 = V2 ⋅ I = (6)(3)
P2 = 18 W absorbed
I
1
2
−
5V
+
+
18 V
−
b) P2 = −72 = −V2 ⋅ I
−72 = 4 A
I=_
−18
P1 = V1 ⋅ I = (5)(4)
P1 = 20 W absorbed
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C H A PTER 1
Basic Concepts
1.2.25 The charge that enters the BOX in Fig. P1.2.25 is shown in the graph below. Calculate and sketch
the current flowing into and the power absorbed by the BOX between 0 and 10 milliseconds.
i (t)
12 V
+
−
BOX
q(t) (mC)
3
2
1
5
1
2
3
4
6
7
8
9
10
t (ms)
−1
−2
−3
FIGURE P1.2.25
Solution:
dq
i(t) = _
dt
P = V ⋅ i = (12) ⋅ i
0 s ≤ t ≤ 1 ms
‾
1 ms ≤ t ≤ 2 ms
‾
2 ms ≤ t ≤ 3 ms
‾
3 ms ≤ t ≤ 5 ms
‾
5 ms ≤ t ≤ 6 ms
‾
6 ms ≤ t ≤ 8 ms
‾
c01SolutionsToProblems_V2.indd 12
1m − 1m = 0 A,
i=_
1m − 0
P=0W
0 − 1m = −1 A,
i=_
2m − 1m
0 − 0 = 0 A,
i=_
3m − 2m
P = (12)(−1) = −12 W
P=0W
−2m − 0 = −1 A,
i=_
5m − 3m
3m − −2m = 5 A,
i = ___________
6m − 5m
(
)
1m − 3m = −1 A,
i=_
8m − 6m
P = (12)(−1) = −12 W
P = (12)(5) = 60 W
P = (12)(−1) = −12 W
29-Jan-22 12:39:08 PM
Solutions to Problems
8 ms ≤ t ≤ 9 ms
‾
1m − 1m = 0 A,
i=_
9m − 8m
9 ms ≤ t ≤ 10 ms
‾
13
P=0W
0 − 1m = −1 A, P = (12)(−1) = −12 W
i=_
10m − 9m
i(t) (A)
5
1
0
2
3
4
5
6
7
8
9
10
t (ms)
−1
P(t) (W)
60
0
1
2
3
4
5
6
7
8
9
10
t (ms)
−12
Section 1.3 Solutions
1.3.1 Find Vx in the network in Fig. P1.3.1 using Tellegen’s theorem.
1A
+
10 V
+
1
16 V
−
−+
2
36 V
+ V −
x
20 V
−
3
−
+
−
+
15 V
FIGURE P1.3.1
Solution:
P10 V = (10)(1) = 10 W → 10 W absorbed
P1 = (16)(1) = 16 W → 16 W absorbed
P36 V = (−36)(1) = −36 W → 36 W supplied
P2 =Vx(1) = Vx W → Vx W absorbed
P3 = (20)(1) = 20 W → 20 W absorbed
P15 V = (−15)(1) = −15 W → 15 W supplied
Psup = Pabs
36 + 15 = 10 + 16 + Vx + 20
Vx = 5 V
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C H A PTER 1
Basic Concepts
1.3.2 Find the power that is absorbed or supplied by the circuit elements in Fig. P1.3.2.
+
9V
− 3A
1
+
30 V
−
8V
1
−
+
+
−
3A
+
Ix = 4 A
21 V
16 V
+
−
4A
−
2Ix
4A
3A
(a)
(b)
FIGURE P1.3.2
Solution:
a. P3 A = (−30) ⋅ (3) = −90 W
P3 A = 90 W supplied
P1 = (9) ⋅ (3) = 27 W absorbed
P21 V = (21) ⋅ (3) = 63 W absorbed
b. P4 A = (−16)(4) = −64 W
P4 A = 64 W supplied
P1 = (8)(4) = 32 W absorbed
P2IX = [2(4)] ⋅ (4) = 32 W absorbed
1.3.3 Compute the power that is absorbed or supplied by the elements in the network in Fig. P1.3.3.
Ix = 4 A +
12 V
−+
1
2A
36 V
1Ix
−
+
−
2A
+
2 24 V
−
+
3 28 V
−
FIGURE P1.3.3
Solution:
P36 V = (−36) ⋅ Ix = (−36)(4) = −144 W
P36 V = 144 W supplied
P1 = (12) ⋅ Ix = (12) ⋅ (4) = 48 W absorbed
P2 = (24) ⋅ (2) = 48 W absorbed
P1Ix = [−1(4)] ⋅ (2) = −8 W
P1Ix = 8 W supplied
P3 = (28) ⋅ (2) = 56 W absorbed
1.3.4 Find Ix in the circuit in Fig. P1.3.4 using Tellegen’s theorem.
4V
−
+
8V
−
+
18 V
2A
+
−
24 V
+
12 V
−
12 V
−
+−
2A +
Ix
2A
Ix
+
6V
−
FIGURE P1.3.4
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Solutions to Problems
15
Solution:
P24 V = (−24)(2) = −48 W → 48 W supplied
P4 V = (4)(2) = 8 W → 8 W absorbed
P8 V = (8)(2) = 16 W → 16 W absorbed
P2 A = (−12)(2) = −24 W → 24 W supplied
P18 V = (18)(Ix) = 18Ix W → 18Ix W absorbed
P12 V = (−12)(Ix) = −12Ix W → 12Ix W supplied
P6 V = (6)(Ix) = 6Ix W → 6Ix W absorbed
P sup = P abs
48 + 24 + 12Ix = 8 + 16 + 18Ix + 6Ix
Ix = 4 A
1.3.5 Is the source, Vs, in the network in Fig. P1.3.5 absorbing or supplying power, and how much?
4V
+
VS
−
2A
−+
6A
4A
−
−
−
8V
6A
+
12 V
6V
+
2A
+
FIGURE P1.3.5
Solution:
P8 V = (8)(2) = 16 W → 16 W absorbed
P4 V = (4)(2) = 8 W → 8 W absorbed
P6 A = (−12)(6) = −72 W → 72 W supplied
PVs = Vs(4) = 4 Vs W → 4 Vs W absorbed
P6 V = (6)(4) = 24 W → 24 W absorbed
Psup = Pabs
72 = 16 + 8 + 4 Vs + 24
Vs = 6 V
PVs = (6)(4) = 24 W absorbed
1.3.6 Calculate the power absorbed by each element in the circuit in Fig. P1.3.6. Also, verify that
Tellegen’s theorem is satisfied by this circuit.
+
4A
+
1A
40 V
2
−
+
5V
4
−
3A
4A
30 V
+
−
15 V
+
5A
1
3
−
+
−
−
5V
5V
10 V
−
1A
+
5
10 V
−
+
FIGURE P1.3.6
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C H A PTER 1
Basic Concepts
Solution:
P40 V = (−40)(5) = −200 W → 200 W supplied
P4 A = (30)(4) = 120 W → 120 W absorbed
P15 V = (15)(1) = 15 W → 15 W absorbed
P1 = (5)(5) = 25 W → 25 W absorbed
P2 = (5)(1) = 5 W → 5 W absorbed
P3 = (10)(4) = 40 W → 40 W absorbed
P4 = (−5)(3) = −15 W → 15 W supplied
P5 = (10)(1) = 10 W → 10 W absorbed
Psup − Pabs = 0
(200 + 15) − (120 + 15 + 25 + 5 + 40 + 10) = 0
(215) − (215) = 0
1.3.7 Find the power that is absorbed or supplied by the network elements in Fig. P1.3.7.
Ix = 4 A +
16 V
1
− 4A
+
−
+
−
24 V
2Ix
4A
(a)
24 V
−+
2A
+
−
+
20 V
2A
1
−
Ix = 2 A
+
4Ix
2
12 V
−
2A
(b)
FIGURE P1.3.7
Solution:
a. P24 V = (−24)(4) = −96 W
P24 V = 96 W supplied
P1 = (16)(4) = 64 W absorbed
P2Ix = [2 ⋅ (4)] ⋅ (4) = 32 W absorbed
b. P4Ix = [−4(2)] ⋅ (2) = −16 W
P4Ix = 16 W supplied
P24 V = (−24)(2) = −48 W
P24 V = 48 W supplied
P1 = (20) ⋅ (2) = 40 W absorbed
P2 = (12) ⋅ (2) = 24 W absorbed
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Solutions to Problems
17
1.3.8 Find the power absorbed or supplied by element 1 in Fig. P1.3.8.
−
6V
1
Ix
+
+
4V
−
2
2A
18 V +
−
24 V
+
−
+
20 V
2Ix
−
Ix
FIGURE P1.3.8
Solution:
P18 V = (−18) ⋅ Ix = −18Ix W → 18 ⋅ Ix W supplied
P1 = (−6) ⋅ Ix = −6Ix W → 6 ⋅ Ix W supplied
P24 V = (−24) ⋅ (2) = −48 W → 48 W supplied
P2 = (4) ⋅ (2Ix) = 8Ix W → 8 ⋅ Ix W absorbed
P2Ix = (20)(2Ix) = 40Ix W → 40Ix W absorbed
Psup = Pabs
18Ix + 6Ix + 48 = 8Ix + 40Ix
Ix = 2 A
P1 = (−6) ⋅ (2) = −12 W
P1 = 12 W supplied
1.3.9 Find Ix in the network in Fig. P1.3.9.
1Ix
Ix + 8 V −
1
2A
24 V
+
−
−+
+
2
16 V
−
2A
+
3
20 V
−
FIGURE P1.3.9
Solution:
P24 V = (−24) ⋅ Ix = −24Ix W → 24 ⋅ Ix W supplied
P1 = (8) ⋅ Ix = 8Ix W → 8 ⋅ Ix W absorbed
P2 = (16) ⋅ (2) = 32 W → 32 W absorbed
P1Ix = [−1(Ix)] ⋅ 2 = −2 ⋅ Ix W → 2 · Ix W supplied
P3 = (20) ⋅ (2) = 40 W → 40 W absorbed
Psup = Pabs
24x + 2 ⋅ Ix = 8Ix + 32 + 40
Ix = 4 A
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18
C H A PTER 1
Basic Concepts
1.3.10 Find the power that is absorbed or supplied by the network elements in Fig. P1.3.10.
FIGURE P1.3.10
Solution:
a. P1 = (2)(2) = 4 W absorbed
P2Ix = (2.2) (2) = 8 W absorbed
P6v = (−6)(2) = −12 W
P6v = 12 W supplied
b. P1 = (24)(1) = 24 V absorbed
P2v = (14)(1) = 14V absorbed
P4Ix = 4 W supplied
P34v = 34 W supplied
1.3.11 Determine the power absorbed by element 1 in Fig. P1.3.11.
Ix +
16 V
1
−
+
12 V
2
−
2A
+
48 V
+
−
32 V
−
2Ix
+
3 20 V
−
FIGURE P1.3.11
Solution:
P48 V = (−48) ⋅ Ix = −48 ⋅ Ix W → 48 ⋅ Ix W supplied
P1 = (16) ⋅ Ix = 16Ix W → 16 ⋅ Ix W absorbed
P2Ix = (32) ⋅ (2Ix) = 64 ⋅ Ix W → 64 Ix W absorbed
P2 = (−12)(2) = −24 W → 24 W supplied
P3 = (−20)(2) = −40 W → 40 W supplied
Psup = Pabs
48Ix + 24 + 40 = 16Ix + 64Ix
Ix = 2 A
P1 = (16)(2) = 32 W absorbed
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Solutions to Problems
19
1.3.12 Find the power absorbed or supplied by element 1 in Fig. P1.3.12.
4V
+
1
12 V
−
−+
+
4Ix
+
12 V
2
−
4A
+
3
8V
+
4
20 V
−
2A
−
4A
Ix
20 V
−
FIGURE P1.3.12
Solution:
P4Ix = (−12) ⋅ (4Ix) = −48Ix W → 48Ix W supplied
P1 = (4)(4Ix) = 16Ix W → 16Ix W absorbed
P2 = (8)(4) = 32 W → 32 W absorbed
P12 V = (−12)(4) = −48 W → 48 W supplied
P3 = (20)(2) = 40 W → 40 W absorbed
P4 = (20) ⋅ Ix = 20Ix W → 20Ix W absorbed
Psup = Pabs
48Ix + 48 = 16Ix + 32 + 40 + 20Ix
Ix = 2 A
P1 = 16(2) = 32 W absorbed
1.3.13 Find Io in the network in Fig. P1.3.13 using Tellegen’s theorem.
6A+
8V
1
−
4A
+
+
−
24 V
2
−
6V
3
3A
Io
8V
6
4 16 V
−
+
6V
5
−
+
+
+
−
4Ix
Ix=2 A
10 V
−
−
+
1A
3A
FIGURE P1.3.13
Solution:
P24 V = (−24)(6) = −144 W → 144 W supplied
P4Ix = [−4(2)](3) = −24 W → 24 W supplied
P1 = (8)(6) = 48 W → 48 W absorbed
P2 = (10)(4) = 40 W → 40 W absorbed
P3 = (6) ⋅ Io = 6Io W → 6Io W absorbed
P4 = (16)(2) = 32 W → 32 W absorbed
P5 = (6)(1) = 6 W → 6 W absorbed
P6 = (8)(3) = 24 W → 24 W absorbed
Psup = Pabs
144 + 24 = 48 + 40 + 6Io + 32 + 6 + 24
Io = 3 A
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20
C H A PTER 1
Basic Concepts
1.3.14 Calculate the power absorbed by each element in the circuit in Fig. P1.3.14. Also, verify that
Tellegen’s theorem is satisfied by this circuit.
3Ix
+
−+
24 V
5
−
2A
2A
+
24 V
+
−
12 V
1
−
+
2A
+
12 V
4A
4A
6A
−
6V
2
−
−
+
6V 3
−
9V
4
+
4A
15 V
+
−
Ix = 2 A
FIGURE P1.3.14
Solution:
P3Ix = [−3(2)] ⋅ (2) = −12 W → 12 W supplied
P24 V = (−24)(4) = −96 W → 96 W supplied
P6 A = (12)(6) = 72 W → 72 W absorbed
P15 V = (−15)(4) = −60 W → 60 W supplied
P1 = (12)(2) = 24 W → 24 W absorbed
P2 = (−6)(4) = −24 W → 24 W supplied
P3 = (6)(2) = 12 W → 12 W absorbed
P4 = (9)(4) = 36 W → 36 W absorbed
P5 = (24)(2) = 48 W → 48 W absorbed
Psup − Pabs = 0
(12 + 96 + 60 + 24) − (72 + 24 + 12 + 36 + 48) = 0
(192) − (192) = 0
1.3.15 In the circuit in Fig. P1.3.15, element 1 absorbs 40 W, element 2 supplies 50 W, element 3 supplies
25 W, and element 4 absorbs 15 W. How much power is supplied by element 5?
FIGURE P1.3.15
Solution:
Psupplied = Pabsorbed
P2 + P3 + P5 = P1 + P4
50 + 25 P5 = 40 + 15
P5 = −20 W supplied or P5 = 20 W absorbed
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