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Unformatted text preview: Chapter 2  Circuit Elements
Exercises
Ex. 2.31 m ( i1 + i 2 ) = mi1 + mi 2 ⇒ superposition is satisfied m ( a i1 ) = a ( mi1 ) ⇒ homogeneity is satisfied
Therefore the element is linear.
Ex. 2.32 m ( i1 + i 2 ) + b = mi1 + mi 2 + b ≠ ( mi1 + b ) + ( mi 2 + b ) ⇒ superposition is not satisfied
Therefore the element is not linear. Ex. 2.51
v 2 (10 )
P= =
=1 W
R 10 0
2 Ex. 2.52
P= v 2 (10 cos t ) 2
=
= 10 cos 2 t W
R
10 Ex. 2.81 ic = − 1 .2 A , vd = 2 4 id = 4 ( − 1 .2) = − 4 .8 V
A id and vd adhere to the passive convention so
P = vd id = (24) (−4.8) = −115.2 W is the power received by the dependent source 21 Ex. 2.82 vc = −2 V, id = 4 vc = −8 A and vd = 2.2 V
id and vd adhere to the passive convention so
P = vd id = (2.2) (−8) = −17.6 W
is the power received by the dependent source. The power supplied by the
dependent source is 17.6 W.
Ex. 2.83 ic = 1.25 A, vd = 2 ic = 2.5 V and id = 1.75 A
id and vd adhere to the passive convention so
P = vd id = (2.5) (1.75) = 4.375 W
is the power received by the dependent source. 22 Ex. 2.91 θ = 4 5 ° , I = 2 m A, R p = 2 0 k Ω
a= θ 45
(20 kΩ) = 2.5 kΩ
⇒ aR =
p 360
360 vm = (2 ×10−3 )(2.5 ×103 ) = 5 V Ex. 2.92 v = 10 V, i = 280 µA, k = 1 µA
°K for AD590 °K i = 280° K
i = kT ⇒ T = = (280µA)1 µA k Ex. 2.101 At t = 4 s both switches are open, so i = 0 A.
Ex. 2.10.2 At t = 4 s the switch is in the up position, so v = i R = (2 mA)(3 kΩ) = 6V .
At t = 6 s the switch is in the down position, so v = 0 V. Problems
Section 23 Engineering and Linear Models
P2.31
The element is not linear. For example, doubling the current from 2 A to 4 A does not double the
voltage. Hence, the property of homogeneity is not satisfied.
P2.32 (a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and
the line passes through the origin so the equation of the line is v = 0.12 i . The element is indeed
linear.
(b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV
4
(c) When v = 4 V, i =
= 33 A = 33 A.
0.12
23 P2.33 (a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and
the line passes through the origin so the equation of the line is v = 256.5 i . The element is indeed
linear.
(b) When i = 4 mA, v = (256.5 V/A)×(4 mA) = (256.5 V/A)×(0.004 A) = 1.026 V
12
(c) When v = 12 V, i =
= 0.04678 A = 46.78 mA.
256.5
P2.34 Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence,
the property of homogeneity is not satisfied. The element is not linear. Section 25 Resistors
P2.51 i = is = 3 A and v = R i = 7 × 3 = 21 V
v and i adhere to the passive convention
∴ P = v i = 21 × 3 = 63 W
is the power absorbed by the resistor. P2.52 i = is = 3 m A a n d v = 2 4 V
v
24
=
= 8 0 00 = 8 k Ω
i
.003
P = (3×1 0 −3 )× 2 4 = 7 2×10 −3 = 7 2 m W R= P2.53 v = vs =10 V an d R = 5 Ω
v
10
=
=2 A
R
5
v and i adhere to the passive convention
∴ p = v i = 2⋅10 = 2 0 W
is the pow er absorbed by the resistor
i= 24 P2.54 v = vs = 24 V and i = 2 A
v 24
=
= 12 Ω
i
2
p = vi = 24⋅2 = 48 W R= P2.55 v1 = v 2 = vs = 150 V;
R1 = 50 Ω; R2 = 25 Ω
v 1 and i1 adhere to the passive convention so v 1 150
=
=3 A
R 1 50
v
150
v2 and i 2 do not adhere to the passive convention so i 2 = − 2 = −
= −6 A
R2
25
i1 = The power absorbed by R1 is P = v1 i1 = 150 ⋅ 3 = 450 W
1
The power absorbed by R 2 is P 2 = − v 2i 2 = −150(−6) = 900 W P2.56 i1 = i 2 = is = 2 A ;
R1 =4 Ω and R2 = 8 Ω
v 1 and i 1 do not adhere to the passive convention so
v 1 =− R 1 i 1 =−4⋅2= −8 V.
The power absorbed by R 1 is
P1 =−v 1i 1 =−(−8)(2) = 16 W. v2 and i 2 do adhere to the passive convention so v2 = R 2 i 2 = 8 ⋅ 2 = 16 V .
The power absorbed by R 2 is P 2 = v 2i 2 = 16 ⋅ 2 = 32 W.
P2.57
Model the heater as a resistor, then
(250) 2
v2
v2
⇒R=
=
= 62.5 Ω
1000
R
P
v 2 (210) 2
= 705.6 W
with a 210 V source: P = =
R
62.5
with a 250 V source: P = 25 P2.58
The current required by the mine lights is: i = P 5000 125
A
=
=
3
v 120 Power loss in the wire is : i 2 R
Thus the maximum resistance of the copper wire allowed is
0.05P 0.05×5000
=
= 0.144 Ω
(125/3) 2
i2
now since the length of the wire is L = 2×100 = 200 m = 20,000 cm
R= thus R = ρ L / A with ρ = 1.7×10−6 Ω ⋅ cm from Table 2.5−1
A= ρL
R = 1.7×10−6 ×20,000
= 0.236 cm 2
0.144 Section 26 Independent Sources
P2.61
v s 15
2
=
= 3 A a nd P = R i 2 = 5 ( 3 ) = 4 5 W
R
5
(b) i and P do not depend on is . (a) i = The values of i and P are 3 A and 45 W, both when i s = 3 A and when i s = 5 A.
P2.62 v 2 102
= 20 W
(a) v = R i s = 5 ⋅ 2 = 10 V and P = =
R
5
(b) v and P do not depend on v s .
The values of v and P are 10V and 20 W both when v s = 10 V and when v s = 5 V 26 P2.63 Consider the current source:
i s and v s do not adhere to the passive convention,
so Pcs =i s v s =3⋅12 = 36 W
is the power supplied by the current source. Consider the voltage source:
i s and v s do adhere to the passive convention,
so Pvs = i s vs =3 ⋅12 = 36 W
is the power absorbed by the voltage source.
∴ The voltage source supplies −36 W.
P2.64 Consider the current source:
i s and vs adhere to the passive convention
so Pcs = i s vs =3 ⋅12 = 36 W
is the power absorbed by the current source.
Current source supplies − 36 W.
Consider the voltage source:
i s and vs do not adhere to the passive convention
so Pvs = i s vs = 3 ⋅12 =36 W
is the power supplied by the voltage source. P2.65 (a) P = v i = (2 cos t ) (10 cos t ) = 20 cos 2 t mW
1 1 1 (b) w = ∫0 P dt = ∫0 20 cos t dt = 20 t + sin 2t = 10 + 5 sin 2 mJ
2 4
0
1 1 2 27 Section 27 Voltmeters and Ammeters
P2.71 (a) R = v
5
=
= 10 Ω
i 0.5 (b) The voltage, 12 V, and the
current, 0.5 A, of the voltage
source adhere to the passive
convention so the power
P = 12 (0.5) = 6 W
is the power received by the
source. The voltage source
delivers 6 W.
P2.72 The voltmeter current is zero
so the ammeter current is
equal to the current source
current except for the
reference direction:
i = 2 A
The voltage v is the voltage of
the current source. The power
supplied by the current source
is 40 W so
40 = 2 v ⇒ v = 20 V 28 Section 28 Dependent Sources
P2.81
r= vb 8
= =4 Ω
ia 2 P2.82
vb = 8 V ; g v b = i a = 2 A ; g = ia 2
A
= = 0.25
vb 8
V i b = 8 A ; d i b = i a = 32 A ; d = i a 32
A
=
=4
ib
8
A va = 2 V ; b va = vb = 8 V ; b = vb 8
V
= =4
va 2
V P2.83 P2.84 Section 29 Transducers
P2.91
a= θ= θ
360 , θ= 360 vm
Rp I (360)(23V)
= 75.27°
(100 kΩ)(1.1 mA ) P2.92
AD590 : k =1 µA ,
K
v =20 V (voltage condition satisfied)
° 4 µ A < i < 13 µ A i
T= k ⇒ 4 ° K < T < 1 3° K 29 Section 210 Switches
P2.101 At t = 1 s the left switch is open and the
right switch is closed so the voltage
across the resistor is 10 V.
i= v
10
=
= 2 mA
R
5×103 At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is
15 V.
i= v
15
=
= 3 mA
R
5×103 P2.102 At t = 1 s the current in the resistor
is 3 mA so v = 15 V.
At t = 4 s the current in the resistor
is 0 A so v = 0 V. Verification Problems
VP21 vo =40 V and i s = − (−2) = 2 A. ( Notice that the ammeter measures − i s rather than i s .)
So vo
40
V
=
= 20
2
A
is Your lab partner is wrong.
VP22 vs 12
= 0.48 A. The power absorbed by
=
R 25
this resistor will be P = i vs = (0.48) (12) = 5.76 W.
We expect the resistor current to be i = A half watt resistor can't absorb this much power. You should not try another resistor. 210 Design Problems
DP21 1.) 10
10
> 0.04 ⇒ R <
= 250 Ω
R
0.04 2.) 102 1
<
⇒ R > 200 Ω
R
2 Therefore 200 < R < 250 Ω. For example, R = 225 Ω.
DP22 1.) 2 R > 40 ⇒ R > 20 Ω
15
2.) 2 2 R < 15 ⇒ R < = 3.75 Ω
4
Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously. DP23 P = ( 30 mA ) ⋅ (1000 Ω ) = (.03) (1000 ) = 0.9 W < 1 W
1
2 2 P2 = ( 30 mA ) ⋅ ( 2000 Ω ) = (.03) ( 2000 ) = 1.8 W < 2 W
2 2 P3 = ( 30 mA ) ⋅ ( 4000 Ω ) = (.03) ( 4000 ) = 3.6 W < 4 W
2 2 211 ...
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