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Section 2-2 Engineering and Linear Models
P 2.2-
Solution:
The element is not linear. For example, changing the current from 4 A to 6 A does not change
the voltage from 10 V to 15 V. Instead the voltage changes from 10 V to 34 V. Hence, the
property of homogeneity is not satisfied.
P 2.2-
Solution:
(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 = 50 mA, v = (0.12 V/A)×(50 mA) = (0.12 V/A)×(0.05 A) = 6 mV
(c) When v = 6 V,
50 A
i = =
P 2.2-
Solution:
(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 = 5 mA, v = (256.5 V/A)×(5 mA) = (256.5 V/A)×(.005 A) = 1.282 V
(c) When v = 15 V,
i = = A = 58.47 mA.
P 2.2-
Solution:
Let i = 1 A , then v = 6 i + 10 = 16 V. Next 2 i = 2A but 16 = 2 v ≠ 6(2 i ) + 10 = 22. Hence,
the property of homogeneity is not satisfied. The element is not linear.
P 2.2-
Solution:
(a) 0.4 3.2 V 10 40 8
v v v = + = ⇒ v =
0.08 A
v i = =
(b)
Using the quadratic formula
0.8, 1.0 V
v
When v = 0.8 V then
2
0.32 A 2
i = =. When v = -1.0 V then
( )
2 1 0.5 A 2
i
(c)
2 2 0.4 0.8 0.8 0 10 2 5
v v v = + + ⇒ v + + =
Using the quadratic formula
v
So there is no real solution to the equation.
2 2 0.4 0.8 0 10 2 5
v v v = + ⇒ v + − =
P 2.4-
Solution
to the pa
The power
The power
P 2.4-
Solution
2 and
The pow
v i
n:
assive conven
absorbed by R 1
absorbed by R 2
n:
2 do adhere
wer absorbed
ntion so i 2 =
is P = v i 1 1 1 =
=
is P = 2 v i2 2 =
=
−
2 2
to the passiv
by R is P
2
2
v
R
0 4
0W
⋅
= 200( 8)
= 1600W
− −
1 2 s
1 1
i i = i =
v and i do n
The power ab
−
2 2 2
ve conventio
= v i =(
1
1
1
v =v
v an
v i = R
−8A
1
1
50 mA; R 8
not adhere to the
bsorbed by R is
= Ω
on so 2
v = R
2 s
1
1 1
v =v =200V; R
nd i adhere to th
v 200 4A R 50
= =
2
1 1 1
and R 16
e passive conven
s P v i (
Ω = Ω
= − = −
2 2 16 0.(
40 mW.
R i =
1 2
2 2
R =50 ; R =
he passive conve
v and i do no
Ω Ω
ntion so v 1 = R
( 0.4) (0.050)
−
− =
.050 (^) )=0.8 V
ntion so
ot adhere
Ω
R i 1 1 = 8(.050) =
20W
−
=
V.
−0.4V.
P 2.4-
Solution:
2
2 2
2 2
v Model the heater as a resistor, then from P= R
v (200) R = 40 P 1000
with a 210 V source
v (210) P= 1210 W R 40
⇒ = = Ω
= =
P 2.4-
Solution:
2
2 2
The 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
(125/3)
now
P
i v
i R
P
R
i
×
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
0.236 cm
L
R L A
L
A
R
ρ ρ
ρ
−
−
= × = =
= × Ω⋅ −
× ×
P 2.4-
Solution:
380 420 0.7884 0. 102 380 98 420
= ≤ gain ≤ =
nominal gain
gain tolerance
= × = × =
So
gain = 0.7996 ±1.40%
Section
P 2.5-
Solution
P 2.5-
Solution
P 2.5-
Solution
n 2-5 Indep
n:
(a) 2
i v^ s R
P R i
=
=
(b) i and P
The values of
n:
(a) From O
voltage sou
(b) Since v
both when
n:
cs s s
Consider the c
do not adh
so
P i v =
is the power su
v s
=
vs
Consider the
do adhere
so
P
is the power a
The voltag
v s
∴
pendent So
2 2
1 0 2 a 5
5 ( 2 ) 2
= = A
= =
P do not dep
f i and P are 2A
Ohm’s law v
urce voltage
v and P do n
n v =15V an s
current source.
here to the passive
5 10 = 50 W
upplied by the cu
i s
⋅
s s s
voltage source.
to the passive c
= i v =5 10 =
absorbed by the
ge source supplie
⋅
ources
an d
0 W
pend on is.
A and 20W both w
v=R i =5(5)=25Vs
e.) Next P =
not depend o
nd when v =s
and
e convention,
urrent source.
and
onvention,
50 W
voltage source.
es 50 W.
i s
−
when i s = 2A an
V. (The resi
2 2 25
5
v
R
on v (^) s the val
10V.
nd when i (^) s=4A
istor voltage
= 125 W.
lues of v and
A.
e does not de
d P are 25 V
epend on the
V and 125 W
e
P 2.5-
Solution
P 2.5-
Solution
P 2.5-
Solution
(a) time
(b) energ
n: Consider t
passive co
is the recei
supplies −
Consider t
the passive
is the supp
n:
(a) P = vi =
(b)
1
0
W = (^) ∫ P
n:
to discharge
gy = (12 V) (
the current so
onvention so
ived by the c
30 W.
the voltages
e convention
plied by the v
=(5 cos t ) (
1
0
100
1 100 2
P dt
t
=
⎛ = (^) ⎜ + ⎝
∫
capacity e current
(0.025 A) (
ource. i S and
o PCS = i vS S
current sourc
source. i S an
n so PCS = i
voltage sour
cos t ) =100 co
2
1
0
cos
1 sin 2 5 4
t dt
t
⎞
y 800 mA
t 25 mA
2 * 60*60 se
d v S adhere to
S =^ 3 10^ (^ )=^3
ce. The curre
nd v S do not
i vS S =3 10 ( )
rce.
s 2 t mW
50 +25 sin 2 mJ
h 32 hours A
econds) =
o the
30 W
ent source
adhere to
) =30 W
J
s
4.56 kJ
Solution:
(a)
1000 2 1.96 A 1000 20
i (^) m
% error 100 2% 2
= × =
(b)
m m m
R
R
R
(checked: LNAP 6/17/04)
P 2.6-
Solution:
a.)
vR = 30 i R= 30( 5)− = −150V
v m (^) = 15 − v R= 15 − ( −150) = 165V
b.)
Element Power supplied
voltage source
− 15 ( i s) = −15 5 ( ) = −75 W
current source 165 5( ) =825 W
resistor ( ) ( )
− v (^) R × i R = − − 150 − 5 = −750 W
total 0
P 2.6-
Solution:
a.)
R R
0.5 A
v i = = =
i (^) m = i R − 2 = 0.5 − 2 = −1.5 A
b.)
Element Power supplied
voltage source
16 ( i m) = 16 ( −1.5 ) = −24 W
current source ( ) ( )
v (^) s 2 =15 2 =30 W
resistor ( ) ( )
− v (^) R × i R = − 15 0.5 = −6.0 W
total 0
P 2.7-
Solution:
v (^) c = − 2 V, id = 4 v (^) c = −8 A and vd =5 V
i d and v d adhere to the passive convention so
P = v (^) d id = (5) ( 8)− = −40 W
is the power received by the dependent source. The power supplied by the
dependent source is 40 W.
P2.7-
Solution:
a
0.09 A 90 mA 220
i = − = − = −
k i (^) a = −450 mA
a
a
450 A
90 A
k i k i
P2.7-
Solution:
v (^) a = 100 0.05 ( ) =5 V
mA A 100 0. V V
k = =
i (^) b = − (^) ( 0.1) ( ) 5 = −0.5 A = −500 mA
Section 2-9 Switches
P 2.9-
Solution:
At t = 1 s the left switch is open and the
right switch is closed so the voltage
across the resistor is 20 V.
20 3 2 mA 10 10
v i R
= = = ×
At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is
20 V.
20 3 2 mA 10 10
v i R
= = = ×
P 2.9-
Solution:
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.
P 2.9-
Solution:
(a) v = 12 V
(b)
12 11.43 V
v
(c) v = 0 V
(d)
v
≃ 0.12 V
Section 2-10 How Can We Check…?
P 2.10-
Solution:
=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 i
v
i
P 2.10-
Solution:
12 We 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
v i R
P i v
ot try another resistor.
DP 2-3 Resistors are given a power rating.
For example, resistors are available with
ratings of 1/8 W, 1/4 W, 1/2 W, and 1 W. A
1/2-W resistor is able to safely dissipate 1/2 W
of power, indefinitely. Resistors with larger
power ratings are more expensive and bulkier
than resistors with lower power ratings. Good
engineering practice requires that resistor
power ratings be specified to be as large as, but
not larger than, necessary.
Figure DP 2-
Consider the circuit shown in Figure DP 2-3. The values of the resistances are
R 1 (^) = 1000 Ω, R 2 = 2000 Ω, and R 3 (^) = 4000 Ω
The value of the current source current is
is =30 mA
Specify the power rating for each resistor.
Solution::
( ) ( ) ( ) ( )
2 2 P 1 (^) = 30 mA ⋅ 1000 Ω = .03 1000 = 0.9 W <1 W
( ) ( ) ( ) ( )
2 2 P 2 (^) = 30 mA ⋅ 2000 Ω = .03 2000 = 1.8 W <2 W
( ) ( ) ( ) ( )
2 2 P 3 (^) = 30 mA ⋅ 4000 Ω = .03 4000 = 3.6 W <4 W
R 1 R 2 R 3
i r = i s
i s
Chapter 2 Circuit Elements
Exercises
Exercise 2.4-1 Find the power absorbed by a 100-ohm resistor when it is connected directly
across a constant 10-V source.
Answer: 1-W
Solution:
( )
2 2 10 1 W 100
v P R
Exercise 2.4-2 A voltage source v = 10 cos t V is connected across a resistor of 10 ohms. Find
the power delivered to the resistor.
Answer: 10 cos
2 t W
Solution:
2 2 (10 cos ) (^2) 10 cos W 10
v t P t R
