Chapter 2 - Circuit Elements

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Problems 
Section 2-2 Engineering and Linear Models 
 
P2.2-1 
The element is not linear. For example, doubling the current from 2 A to 4 A does not double the 
voltage. Hence, the property of homogeneity is not satisfied. 
 
P2.2-2 
(a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and 
the line passes through the origin so the equation of the line is 0.12v i= . The element is indeed 
linear. 
 (b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV 
 (c) When v = 4 V, 4 33.3 A
0.12
i = = 
 
P2.2-3 
(a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and 
the line passes through the origin so the equation of the line is 256.5v 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 
 (c) When v = 12 V, 12 0.04678
256.5
i = = A = 46.78 mA. 
 
P2.2-4 
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. 
 
 
P2.2-5 
(a) 0.4 3.2 V
10 40 8
v v v v= + = ⇒ = 
0.08 A
40
vi = = 
(b) 
2
20.4 0.8 0
10 2 5
v v vv= + ⇒ + − =
 
 
Using the quadratic formula 0.2 1.8 0.8, 1.0 V
2
v − ±= = − 
When v = 0.8 V then 
20.8 0.32 A
2
i = = . When v = -1.0 V then ( )21 0.5 A
2
i
−= = . 
 
 
(c) 
2
20.4 0.8 0.8 0
10 2 5
v v vv= + + ⇒ + + = 
 
Using the quadratic formula 0.2 0.04 3.2
2
v − ± −= 
So there is no real solution to the equation. 
 
 
 
Section 2-4 Resistors 
 
P2.4-1 
 
 
3 A and = 7 3 = 21 V
and adhere to the passive convention
 21 3 = 63 W 
is the power absorbed by the resistor.
si i v Ri
v i
P v i
= = = ×
∴ = = × 
 
 
P2.4-2 
 
 
3 3
3 mA and 24 V
24 8000 8 k
.003
= (3 10 ) 24 = 72 10 72 mW
si i v
vR
i
P − −
= = =
= = = = Ω
× × × =
 
 
 
P2.4-3 
 
 
=10 V and 5 
10 2 A
5
and adhere to the passive convention
 2 10 20 W
is the power absorbed by the resistor
sv v R
vi
R
v i
p v i
= = Ω
= = =
∴ = = ⋅ =
 
 
P2.4-4 
 
 
24 V and 2 A
24 12
2
 24 2 = 48 W
sv v i
vR
i
p vi
= = =
= = = Ω
= = ⋅
 
 
 
P2.4-5 
 1 2
1 2
1 1
1
1
1
 150 V; 
50 ; 25 
and adhere to the passive convention so 
150 3 A
50
sv v v
R R
v i
vi
R
= = =
= Ω = Ω
= = =
 
2
2 2 2
2
150 and do not adhere to the passive convention so 6 A
25
vv i i
R
= − = − = − 
 
 1 1 1 1The power absorbed by is 150 3 450 WR P v i= = ⋅ = 
 
2 2 2 2The power absorbed by is 150( 6) 900 WR P v i= − = − − = 
 
 
P2.4-6 
 
 
 1 2 
1 2
 1 1
 1 1 1
 1
 1 1 1
2 A ; 
=4 and 8 
 and do not adhere to the passive convention so 
 4 2 8 V.
The power absorbed by is 
( 8)(2) 16 W.
si i i
R R
v i
v R i
R
P v i
= = =
Ω = Ω
=− =− ⋅ =−
=− =− − =
 
 
2 2 2 2 2
 2 2 2 2
 and do adhere to the passive convention so 8 2 16 V .
The power absorbed by 16 2 32 W.
v i v R i
R is P v i
= = ⋅ =
= = ⋅ = 
 
 
P2.4-7 
 
2 2 2
2 2
Model the heater as a resistor, then
(250)with a 250 V source: 62.5
1000
(210)with a 210 V source: 705.6 W
62.5
v vP R
R P
vP
R
= ⇒ = = =
= = =
Ω 
 
P2.4-8 
 
2
2 2
5000 125The current required by the mine lights is: A
120 3
Power loss in the wire is : 
Thus the maximum resistance of the copper wire allowed is 
0.05 0.05 5000 0.144 
(125/3)
now 
Pi
v
i R
PR
i
= = =
×= = = Ω
6
6
2
since the length of the wire is 2 100 200 m 20,000 cm
thus / with = 1.7 10 cm from Table 2.5 1
1.7 10 20,000 0.236 cm
0.144
L
R L A
LA
R
ρ ρ
ρ
−
−
= × = =
= × Ω⋅
× ×= = =
−
 
 
 
 
 
*P2.4-9 
380 4200.7884 0.8108
102 380 98 420
gain= ≤ ≤ =+ + 
 
0.7884 0.8108 0.7996
2
nominal gain += = 
 
0.7996 0.7884 0.8108 0.7996 100 100 1.40%
0.7996 0.7996
gain tolerance − −= × = × = 
So 
0.7996 1.40%gain = ± 
 
 
P2.4-10 
Label the current i as shown. That current is 
the element current in both resistors. First 
 
a
40
v
i = 
Next 
a b
b
a
40
40
v v
v R i R R
v
= = ⇒ = 
For example, 
7.0540 24 
11.75
R = = Ω 
 
 
 
Section 2-5 Independent Sources 
 
P2.5-1 
 (a) ( )2215 3 A and 5 3 45 W
5
svi P R i
R
= = = = = = 
(b) and do not depend on .si P i 
 The values of and are 3 A and 45 W, both when 3 A and when 5 A.s si P i i= = 
 
P2.5-2 
 
(a) 
2 2
 
10 5 2 10 V and 20 W
5s
vv R i P
R
= = ⋅ = = = = 
(b) and do not depend on . sv P v 
 The values of and are 10V and 20 W both when 10 V and when 5 Vs sv P v v= = 
 
 
P2.5-3 
 
 
 
Consider the current source: 
 and do not adhere to the passive convention, 
so 3 12 36 W 
is the power supplied by the current source.
s s
cs s s
i v
P i v= = ⋅ = 
 
 
 
 
 
Consider the voltage source: 
 and do adhere to the passive convention,
so 3 12 36 W 
is the power absorbed by the voltage source.
 The voltage source supplies 36 W.
s s
vs s s
i v
P i v= = ⋅ =
∴ −
 
 
 
P2.5-4 
 
 
 
Consider the current source: 
 and adhere to the passive convention
so 3 12 36 W 
is the power absorbed by the current source.
Current source supplies 36 W.
s s
cs s s
i v
P i v= = ⋅ =
−
 
 
 
 
 
 
 
Consider the voltage source:
 and do not adhere to the passive convention 
so 3 12 36 W 
is the power supplied by the voltage source.
s s
vs s s
i v
P i v= = ⋅ = 
 
 
 
 
P2.5-5 
 (a) 2(2 cos ) (10 cos ) 20 cos mWP v i t t t= = = 
(b) 
1
1 1 2
0 0
0
1 1 20 cos = 20 sin 2 10 5 sin 2 mJ 
2 4
 w P dt t dt t t⎛ ⎞= = + = +⎜ ⎟⎝ ⎠∫ ∫ 
 
 
 
P2.5-6 
(a) capacity 800 mAhtime to discharge 16 hours
current 50 mA
= = = 
 
Section 2-6 Voltmeters and Ammeters 
 
P2.6-1 
 
 
(a) 5 10
0.5
vR
i
= = = Ω 
 
(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. 
 
P2.6-2 
 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 20 Vv v= ⇒ = 
 
 
 
 
 
P2.6-3 
(a) 
m
900 12 10.8 V
900 100
v ⎛ ⎞= =⎜ ⎟+⎝ ⎠ 
 
12 10.8 0.1 10%
12
− = = 
(b) We require 
m
m m
m
m
12 12
100
0.02 0.98 4900 
12 100
R
R R
R
R
⎛ ⎞− ⎜ ⎟⎜ ⎟+⎝ ⎠≥ ⇒ ≥ ⇒+ ≥ Ω 
 
(checked: LNAP 6/16/04) 
 
 
P2.6-4 
(a) 
m
1000 2 1.98 A
1000 10
i ⎛ ⎞= =⎜ ⎟+⎝ ⎠ 
 
2 1.98% error 100 0.99%
2
−= × = 
(b) 
m
m
m
10002 2
1000 10000.05 0.95 52.63 
2 1000
R
R
R
⎛ ⎞− ⎜ ⎟⎜ ⎟+⎝ ⎠≥ ⇒ ≥ ⇒+ ≤ Ω 
 
(checked: LNAP 6/17/04) 
 
P2.6-5 
a.) 
( )R R25 25 2 50 Vv i= = − = − 
 
( )m R12 12 50 62 Vv v= − = − − = 
 
b.) 
Element Power supplied 
voltage source ( ) ( )s12 12 2 24 Wi− = − = − 
current source ( )62 2 124 W= 
 
resistor ( ) ( )R R 50 2 100 Wv i− × = − − − = − 
total 0 
 
 
P2.6-6 
a.) 
R
R
12 0.48 A
25 25
v
i = = = 
 
m R 2 0.482 1.52 Ai i= − = − = − 
 
b.) 
Element Power supplied 
voltage source ( ) ( )m12 12 1.52 18.24 Wi = − = − 
current source ( ) ( )s 2 12 2 =24 Wv = 
resistor ( )( )R R 12 0.48 5.76 Wv i− × = − = − 
total 0 
 
 
Section 2-7 Dependent Sources 
 
P2.7-1 
 
 
 
 
8 4
2
b
a
vr
i
= = = Ω 
 
P2.7-2 
 
 
2 A8 V ; 2 A ; 0.25 
8 V
a
b b a
b
iv g v i g
v
= = = = = = 
 
P2.7-3 
 
 
32 A 8 A ; 32A ; 4 
8 A
a
b b a
b
ii d i i d
i
= = = = = = 
 
P2.7-4 
 8 V2 V ; 8 V ; 4 
2 V
b
a a b
a
vv b v v b
v
= = = = = = 
 
 
P2.7-5 
4 2 
2
R = − = Ω− and 
2 V4 
0.5 A
A = = −− 
 
(checked: LNAP 6/6/04) 
 
 
P2.7-6 
 
 
 
2 V, 4 8 A and 2.2 Vc d c dv i v v= − = = − = 
id and vd adhere to the passive convention so 
(2.2) ( 8) 17.6 Wd dP v i= = − = − 
is the power received by the dependent source. The power supplied by the 
dependent source is 17.6 W. 
 
P2.7-7 
 
 
1.25 A, 2 2.5 V and 1.75 Ac d c di v i i= = = = 
id and vd adhere to the passive convention so 
(2.5) (1.75) 4.375 Wd dP v i= = = 
is the power received by the dependent source. 
 
 
Section 2-8 Transducers 
 
P2.8-1 
 
 
 
360 = , =
360 
(360)(23V)= = 75.27
(100 k )(1.1 mA)
m
p
va
R I
 
θ θ
θ °Ω
 
 
P2.8-2 
 
 
AAD590 : =1 , 
K
=20 V (voltage condition satisfied)
k
v
μ
° 
 
4 A < < 13 A
4 K< <13 K
 
i
TiT
k
μ μ
° °
⎫⎪ ⇒⎬= ⎪⎭
 
 
 
Section 2-9 Switches 
 
P2.9-1 
 
 
At t = 1 s the left switch is open and the 
right switch is closed so the voltage 
across the resistor is 10 V. 
 
3
10= = 2 mA
5 10
vi
R
=× 
 
 
At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is 
15 V. 
3
15= = 3 mA
5 10
vi
R
=× 
 
P2.9-2 
 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. 
 
 
 
 
P2.9-3 
(a) v = 12 V 
(b) 100 12 11.43 V
105
v ⎛ ⎞= =⎜ ⎟⎝ ⎠ 
(c) v = 0 V 
(d) 100 12 0.1188 0.12 V
10100
v ⎛ ⎞= =⎜ ⎟⎝ ⎠ � 
 
 
Section 2-10 How Can We Check…? 
 
P2.10-1 
 
 
 
=40 V and = ( 2) 2 A. (Notice that the ammeter measures rather than .)
40 V So 20
2 A
Your lab partner is wrong.
o s s s
o
s
v i i
v
i
− − = −
= =
i
 
 
P2.10-2 
 
12We expect the resistor current to be = 0.48 A. The power absorbed by
25
this resistor will be = (0.48) (12) = 5.76 W. 
A half watt resistor can't absorb this much power. You should n
s
s
vi
R
P i v
= =
=
ot try another resistor.
 
 
 
 
	CH2sec2
	Problems 
	Section 2-2 Engineering and Linear Models 
	P2.2-1 
	P2.2-2 
	P2.2-3 
	P2.2-4 
	P2.2-5 
	CH2sec4
	Section 2-4 Resistors 
	P2.4-1 
	P2.4-2 
	P2.4-3 
	P2.4-4 
	P2.4-5 
	P2.4-6 
	P2.4-7 
	P2.4-8 
	*P2.4-9 
	P2.4-10 
	CH2sec5
	Section 2-5 Independent Sources 
	P2.5-1 
	P2.5-2 
	P2.5-3 
	P2.5-4 
	P2.5-5 
	P2.5-6 
	CH2sec6
	Section 2-6 Voltmeters and Ammeters 
	P2.6-1 
	P2.6-2 
	P2.6-3 
	P2.6-4 
	P2.6-5 
	P2.6-6 
	CH2sec7
	Section 2-7 Dependent Sources 
	P2.7-1 
	P2.7-2 
	P2.7-3 
	P2.7-4 
	P2.7-5 
	P2.7-6 
	P2.7-7 
	CH2sec8
	Section 2-8 Transducers 
	P2.8-1 
	P2.8-2 
	CH2sec9
	Section 2-9 Switches 
	P2.9-1 
	P2.9-2 
	P2.9-3 
	CH2sec10
	Section 2-10 How Can We Check…? 
	P2.10-1 
	P2.10-2