C HAPTER 1
Sections 1-2 and 1-3: Dimensions, Charge, and Current
Problem 1.1 Use appropriate multiple and submultiple prefixes to express the
following quantities:
(a) 3,620 watts (W)
(b) 0.000004 amps (A)
(c) 5.2 × 10−6 ohms (!)
(d) 3.9 × 1011 volts (V)
(e) 0.02 meters (m)
(f) 32 × 105 volts (V)
Solution:
(a) 3,620 W = 3.62 kW.
(b) 0.000004 A = 4 µ A.
(c) 5.2 × 10−6 ! = 5.2 µ !.
(d) 3.9 × 1011 V = 390 GV.
(e) 0.02 m = 20 mm.
(f) 32 × 105 V = 3.2 MV.
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.2 Use appropriate multiple and submultiple prefixes to express the
following quantities:
(a) 4.71 × 10−8 seconds (s)
(b) 10.3 × 108 watts (W)
(c) 0.00000000321 amps (A)
(d) 0.1 meters (m)
(e) 8,760,000 volts (V)
(f) 3.16 × 10−16 hertz (Hz)
Solution:
(a) 4.71 × 10−8 s = 47.1 ns.
(b) 10.3 × 108 W = 1.03 GW.
(c) 0.00000000321 A = 3.21 nA.
(d) 0.1 = 10 cm.
(e) 8,760,000 V = 8.76 MV.
(f) 3.16 × 10−16 Hz = 316 aHz.
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.3 Convert:
(a) 16.3 m to mm
(b) 16.3 m to km
(c) 4 × 10−6 µ F (microfarad) to pF (picofarad)
(d) 2.3 ns to µ s
(e) 3.6 × 107 V to MV
(f) 0.03 mA (milliamp) to µ A
Solution:
(a) 16.3 m = 16, 300 mm.
(b) 16.3 m = 0.0163 km.
(c) 4 × 10−6 µ F = 4 pF.
(d) 2.3 ns = 2.3 × 10−3 µ s.
(e) 3.6 × 107 V = 36 MV.
(f) 0.03 mA = 30 µ A.
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.4 Convert:
(a) 4.2 m to µ m
(b) 3 hours to µ seconds
(c) 4.2 m to km
(d) 173 nm to m
(e) 173 nm to µ m
(f) 12 pF (picofarad) to F (farad)
Solution:
(a) 4.2 m = 4.2 × 106 µ m.
(b) 3 hours = 1.08 × 1010 µ s.
(c) 4.2 m = 4.2 × 10−3 km.
(d) 173 nm = 1.73 × 10−7 m.
(e) 173 nm = 0.173 µ m.
(f) 12 pF = 1.2 × 10−11 F.
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.5 The total charge contained in a certain region of space is −1 C. If that
region contains only electrons, how many does it contain?
Solution:
ne =
Q
−1
=
= 6.25 × 1018 electrons.
qe −1.6 × 10−19
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.6 A certain cross section lies in the x–y plane. If 3 × 1020 electrons
go through the cross section in the z-direction in 4 seconds, and simultaneously,
1.5 × 1020 protons go through the same cross section in the negative z-direction, what
is the magnitude and direction of the current flowing through the cross section?
Solution: Negatively charged electrons moving along +z-direction constitute a
current in the −z-direction:
Ie =
!Q 3 × 1020 × 1.6 × 10−19
=
= 12 A,
!t
4
along −z-direction.
Positively charged protons moving along −z-direction constitute a current in the
−z-direction:
Ip =
!Q 1.5 × 1020 × 1.6 × 10−19
=
= 6 A,
!t
4
along −z-direction.
Total net current is:
I = Ie + Ip = 12 + 6 = 18 A,
along −z-direction.
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.7 Determine the current i(t) flowing through a resistor if the cumulative
charge that has flowed through it up to time t is given by
(a) q(t) = 3.6t mC
(b) q(t) = 5 sin(377t) µ C
(c) q(t) = 0.3[1 − e−0.4t ] pC
(d) q(t) = 0.2t sin(120! t) nC
Solution:
d
−3
−3 = 3.6 (mA).
(a) i(t) = dq
dt = dt (3.6t × 10 ) = 3.6 × 10
d
−6
−6
(b) i(t) = dq
dt = dt [(5 sin 377t) × 10 ] = 5 × 377 × 10 cos 377t = 1.885 cos 377t
(mA).
d
−0.4t ) × 10−12 ] = 0.3 × 10−12 × (−0.4) × (−e−0.4t )
(c) i(t) = dq
dt = dt [0.3(1 − e
= 0.12e−0.4t (pA).
d
−9
(d) i(t) = dq
dt = dt [(0.2t sin 120! t) × 10 ]
= (0.2 sin 120! t + 0.2t × 120! cos 120! t) × 10−9
= 0.2 sin 120! t + 75.4t cos 120! t
(nA).
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.8 Determine the current i(t) flowing through a certain device if the
cumulative charge that has flowed through it up to time t is given by
(a) q(t) = −0.45t 3 µ C
(b) q(t) = 12 sin2 (800! t) mC
(c) q(t) = −3.2 sin(377t) cos(377t) pC
(d) q(t) = 1.7t[1 − e−1.2t ] nC
Solution:
(a)
i(t) =
dq
d
= [−0.45t 3 × 10−6 ] = −0.45 × 3t 2 × 10−6 = −1.35t 2
dt
dt
(µ A).
(b)
i(t) =
dq
d
= [(12 sin2 800! t) × 10−3 ] = 2 × 12 × 800! × 10−3 sin 800! t cos 800! t
dt
dt
= 60.32 sin 800! t cos 800! t
(A).
(c)
i(t) =
dq
d
= [(−3.2 sin 377t cos 377t) × 10−12 ]
dt
dt
= [(−3.2 × 377 cos2 377t + 3.2 × 377 sin2 377t) × 10−12 ]
= 1.21(sin2 377t − cos2 377t)
(nA)
(d)
i(t) =
d
dq
= [1.7t(1 − e−1.2t ) × 10−9 ]
dt
dt
= [1.7(1 − e−1.2t ) + 1.7t(−1.2)(−e−1.2t )] × 10−9
= 1.7(1 − e−1.2t + 1.2te−1.2t )
(nA).
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.9 Determine the net charge !Q that flowed through a resistor over the
specified time interval for each of the following currents:
(a) i(t) = 0.36 A, from t = 0 to t = 3 s
(b) i(t) = [40t + 8] mA, from t = 1 s to t = 12 s
(c) i(t) = 5 sin(4" t) nA, from t = 0 to t = 0.05 s
(d) i(t) = 12e−0.3t mA, from t = 0 to t = #
Solution:
(a)
!Q(0, 3) =
! 3
0
i dt =
! 3
0
0.36 dt = 0.36t|30 = 1.08
(C).
(b)
!Q(1, 12) =
! 12
1
i dt =
"!
=
$
1
(c)
!Q(0, 0.05) =
! 0.05
0
12
#
(40t + 8) dt × 10−3
%&12
&
40t 2
+ 8t && × 10−3 = 2.948
2
1
i dt =
"!
0.05
0
(C).
#
5 sin 4" t dt × 10−9
&
−5 cos 4" t &&0.05
−9
=
& × 10
4"
0
= (−0.32 + 0.40) × 10−9 = 80
(pC).
(d)
!Q(0, #) =
! #
0
i dt =
"!
0
#
−0.3t
12e
#
−3
dt × 10
&#
−12e−0.3t &&
=
× 10−3 = 40
0.3 &0
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
(mC).
Problem 1.10 Determine the net charge !Q that flowed through a certain device
over the specified time intervals for each of the following currents:
(a) i(t) = [3t + 6t 3 ] mA, from t = 0 to t = 4 s
(b) i(t) = 4 sin(40" t) cos(40" t) µ A, from t = 0 to t = 0.05 s
(c) i(t) = [4e−t − 3e−2t ] A, from t = 0 to t = #
(d) i(t) = 12e−3t cos(40" t) nA, from t = 0 to t = 0.05 s
Solution:
(a)
!Q(0, 4) =
! 4
0
i dt =
"!
=
$
(b)
!Q(0, 0.05) =
! 0.05
0
4
0
3
−3
(3t + 6t ) dt × 10
#
%&4
3t 2 6t 4 &&
+
× 10−3 = 408
2
4 &0
i dt =
"!
0.05
0
(mC).
#
4 sin 40" t cos 40" t dt × 10−6
4
sin2 40" t|0.05
× 10−6 = 0.
0
2 × 40"
=
(c)
!Q(0, #) =
! #
0
i dt =
! #
0
(d)
!Q(0, 0.05) =
−t
−2t
(4e − 3e
! 0.05
0
i dt =
) dt =
"!
0.05
0
$
%&
3 −2t &&#
−4e + e
& = 2.5
2
0
−t
−3t
12e
(C).
#
cos 40" t dt × 10−9 .
From Tables of Integrals,
!
eax cos bx dx = eax
(a cos bx + b sin bx)
.
a2 + b2
Hence,
'
!Q(0, 0.05) = 12e−3t
& (
(−3 cos 40" t + 40" sin 40" t) &&0.05
× 10−9 = 0.32
&
9 + (40" )2
0
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
(pC).
Problem 1.11 If the current flowing through a wire is given by i(t) = 3e−0.1t mA,
how many electrons pass through the wire’s cross section over the time interval from
t = 0 to t = 0.3 ms?
Solution:
!Q(0, 0.3 ms) =
! 0.3 ms
0
i dt =
"!
0
0.3 ms
−0.1t
3e
#
dt × 10−3
$0.3×10−3
$
−3e−0.1t
−3 $
=
× 10 $
0.1
0
−5
= −30(e−3×10 − 1) × 10−3 = 9 × 10−7
ne =
!Q
9 × 10−7
=
= 5.62 × 1012 electrons.
e
1.6 × 10−19
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
(C).
Problem 1.12
by
The cumulative charge in mC that entered a certain device is given


for t < 0,
0,
q(t) = 5t,
for 0 ≤ t ≤ 10 s,


60 − t, for 10 s ≤ t ≤ 60 s
(a) Plot q(t) versus t from t = 0 to t = 60 s.
(b) Plot the corresponding current i(t) entering the device.
Solution:
(a)
q(t)
50 mC
t (s)
10 20 30 40 50 60
(b)
i(t)
5 mA
10 20 30 40 50 60
t (s)
−1 mA
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.13 A steady flow resulted in 3 × 1015 electrons entering a device in
0.1 ms. What is the current?
Solution:
i=
!Q ne e 3 × 1015 × 1.6 × 10−19
=
=
= 4.8 A.
!t
!t
0.1 × 10−3
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.14
Given that the current in mA flowing through a wire is given by


for t < 0,
0,
i(t) = 6t,
for 0 ≤ t ≤ 5 s,

30e−0.6(t−5) , for t ≥ 5 s,
(a) Sketch i(t) versus t.
(b) Sketch q(t) versus t.
Solution:
(a)
i(t)
30 mA
20 mA
10 mA
t (s)
2
4
6
8
10 12 14
t
(b) q(t) = −!
i(t) dt.
For 0 ≤ t ≤ 5 s,
)t
&' t
(
6t 2 ))
−3
× 10−3 = 3t 2
q(t) =
6t dt × 10 =
2 )0
0
%
(mC).
For t ≥ 5 s,
q(t) =
=
*'
,
5
0
6t dt +
' t
5
−0.6(t−5)
30e
+
dt × 10−3
)5
' t
6t 2 ))
+3
−0.6t
+ 30e
e
dt × 10−3
2 )0
5
= [75 + 50(1 − e−0.6(t−5) )]
(mC).
q (mC)
125
100
75
50
25
t (s)
2
4
6
8
10 12 14
All rights reserved. Do not reproduce or distribute. ©2009 National Technology and Science Press
Problem 1.15 The plot in Fig. P1.15 displays the cumulative amount of charge q(t)
that has entered a certain device up to time t. What is the current at
(a) t = 1 s
(b) t = 3 s
(c) t = 6 s
q(t)
4C
0
2s
4s
t
6s
8s
−4 C
Figure P1.15: q(t) for Problem 1.15.
Solution:
(a) a = 42 = 2 A @ t = 1 s (slope of first segment).
(b) i = 0 @ t = 3 s (slope of q(t) = 0 at t = 3 s).
(c) i = −8
4 = −2 A @ t = 6 s (negative slope of third segment).
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.16 The plot in Fig. P1.16 displays the cumulative amount of charge q(t)
that has exited a certain device up to time t. What is the current at:
(a) t = 2 s
(b) t = 6 s
(c) t = 12 s
q(t)
4C
4e−0.2(t−8)
2C
0
4s
8s
t
Figure P1.16: q(t) for Problem 1.16.
Solution:
(a) i = 0 @ t = 2 s (slope = 0 of first segment).
2
(b) i = 4−2
8−4 = 4 = 0.5 A (slope of second segment).
(c)
i=
d
d
dq
= (4e−0.2(t−8) ) = 4e1.6 e−0.2t = −4 × 0.2e1.6 e−0.2t
dt
dt
dt
= −0.36 A @ t = 12 s.
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Sections 1-4 and 1-5: Voltage, Power, and Circuit Elements
Problem 1.17 For each of the eight devices in the circuit of Fig. P1.17, determine
whether the device is a supplier or a recipient of power and how much power it is
supplying or receiving.
+6 V_
4A
1A
2
+
1 16 V
_
+8 V_
+4 V_
4
5
2A
1A
+
10 V 3
_
6
+ 12 V _
3A
_ 7V+
9V 7
_
+
8
Figure P1.17: Circuit for Problem 1.17.
Solution:
Device 1:
Device 2:
Device 3:
Device 4:
Device 5:
Device 6:
Device 7:
Device 8:
p = vi = 16 × (−4) = −64 W (supplier)
p = vi = 6 × 4 = 24 W (recipient)
p = vi = 10 × 1 = 10 W (recipient)
p = vi = 8 × 1 = 8 W (recipient)
p = vi = 4 × 1 = 4 W (recipient)
p = vi = 12 × 4 = 24 W (recipient)
p = vi = 9 × (−3) = −27 W (supplier)
p = vi = 7 × 3 = 21 W (recipient)
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.18 For each of the seven devices in the circuit of Fig. P1.18, determine
whether the device is a supplier or a recipient of power and how much power it is
supplying or receiving.
+6 V_
V_
24 V 1
V
5
2A
3A
7
10
_V
+ V
6_
6
+
_
4A
+4 V_
1A
+
4
3
_ 12
8
+
5A
+
2
2A
Figure P1.18: Circuit for Problem 1.18.
Solution:
Device 1:
Device 2:
Device 3:
Device 4:
Device 5:
Device 6:
Device 7:
p = vi = 24 × (−5) = −120 W (supplier)
p = vi = 6 × 5 = 30 W (recipient)
p = vi = 8 × 1 = 8 W (recipient)
p = vi = 12 × 4 = 48 W (recipient)
p = vi = 4 × (−2) = −8 W (supplier)
p = vi = 10 × 3 = 30 W (recipient)
p = vi = 6 × 2 = 12 W (recipient)
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.19 An electric oven operates at 120 V. If its power rating is 0.6 kW,
what amount of current does it draw and how much energy does it consume in 12
minutes of operation?
Solution:
p 0.6 × 103
=
=5
(A).
v
120
w = p !t = 0.6 × 103 × 12 × 60 = 432
i=
All rights reserved. Do not reproduce or distribute.
(kJ).
©2009 National Technology and Science Press
Problem 1.20 A 9-V flashlight battery has a rating of 1.8 kWh. If the bulb draws a
current of 100 mA when lit, determine the following:
(a) For how long will the flashlight provide illumination?
(b) How much energy in joules is contained in the battery?
(c) What is the battery’s rating in ampere-hours?
Solution:
(a)
!t =
1.8 × 103
w W
=
=
hours
p
vi
9 × 100 × 10−3
= 2, 000 hours.
(b) W = 1.8 × 103 × 3600 = 6.48 (MJ).
(c) Ampere-hours = 1.89kWh
V = 0.2 (kAh).
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.21
The voltage across and current through a certain device are given by
v(t) = 5 cos(4! t) V,
i(t) = 0.1 cos(4! t) A.
Determine:
(a) The instantaneous power p(t) at t = 0 and t = 0.25 s.
(b) The average power pav , defined as the average value of p(t) over a full time
period of the cosine function (0 to 0.5 s).
Solution:
(a)
p(t) = vi = (5 cos 4! t)(0.1 cos 4! t) = 0.5 cos2 4! t
(W).
p(0) = 0.5 W @ t = 0
p(0.25 s) = 0.5 cos2 (4! × 0.25) = 0.5 W @ t = 0.25 s.
(b)
1
pav =
T
! T
0
0.5
1
p(t) dt =
0.5 cos2 4! t dt
0.5 0
1
1
[sin 4! t cos 4! t + 4! t]|0.5
=
0 =
8!
4
!
All rights reserved. Do not reproduce or distribute.
(W).
©2009 National Technology and Science Press
Problem 1.22
The voltage across and current through a certain device are given by
v(t) = 100(1 − e−0.2t ) V
i(t) = 30e−0.2t mA.
Determine:
(a) The instantaneous power p(t) at t = 0 and t = 3 s.
(b) The cumulative energy delivered to the device from t = 0 to t = !.
Solution:
(a)
p(t) = vi = 100(1 − e−0.2t ) × 30e−0.2t × 10−3
= 3(e−0.2t − e−0.4t )
(W).
(b)
W=
! !
0
p(t) dt = 3(e−0.2t − e−0.4t ) dt
"
#$!
−3e−0.2t 3e−0.4t $$
=
+
$
0.2
0.4
0
=
3
3
−
= 7.5
0.2 0.4
All rights reserved. Do not reproduce or distribute.
(J).
©2009 National Technology and Science Press
Problem 1.23 The voltage across a device and the current through it are shown
graphically in Fig. P1.23. Sketch the corresponding power delivered to the device
and calculate the energy absorbed by it.
i(t)
10 A
5A
0
t
v(t)
1s
2s
1s
2s
5V
t
0
Figure P1.23: i(t) and v(t) of the device in Problem 1.23.
Solution: For 0 ≤ t ≤ 1 s,
p = vi = 5t × 10 = 50t
For 1 s ≤ t ≤ 2 s,
p = vi = (10 − 5t) × 5 = 50 − 25t.
W=
! 2
0
p dt =
! 1
0
50t dt +
! 2
1
(50 − 25t) dt
"1 #
$"2
50t 2 ""
25t 2 ""
=
+ 50t −
2 "0
2 "1
= 37.5
(J).
p(t)
50 W
25 W
t (s)
1s
All rights reserved. Do not reproduce or distribute.
2s
©2009 National Technology and Science Press
Problem 1.24 The voltage across a device and the current through it are shown
graphically in Fig. P1.24. Sketch the corresponding power delivered to the device
and calculate the energy absorbed by it.
i(t)
10 A
t
0
v(t)
1s
2s
1s
2s
5V
t
0
Figure P1.24: i(t) and v(t) of the device in Problem 1.24.
Solution: For 0 ≤ t ≤ 1 s,
p(t) = vi = (5t)(10t) = 50t 2
For 1 s ≤ t ≤ 2 s,
v = 5(2 − t)
i = 10(2 − t)
p(t) = 50(2 − t)2
p(t)
50 W
t
1s
w=
=
! 2
0
! 1
0
2s
p(t) dt
50t 2 dt +
! 2
1
50(2 − t)2 dt
= 33.3 J.
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
Problem 1.25 For the circuit in Fig. P1.25, generate circuit diagrams that include
only those elements that have current flowing through them for
(a) t < 0
(b) 0 < t < 2 s
(c) t > 2 s
t=0
R1
R2
V0
+
_
R3
R4
t=2s
R5
R6
Figure P1.25: Circuit for Problem 1.25.
Solution:
(a) t < 0
R1
V0
R2
R3
R5
R6
+
_
(b) 0 < t < 2 s
R1
R2
V0
+
_
R5
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press
(c) t > 2 s
R1
R2
V0
+
_
R4
R5
All rights reserved. Do not reproduce or distribute.
R6
©2009 National Technology and Science Press
Problem 1.26 For the circuit in Fig. P1.26, generate circuit diagrams that include
only those elements that have current flowing through them for
(a) t < 0
(b) 0 < t < 2 s
(c) t > 2 s
R1
V1
SPST
t=0
+
_
R3
SPDT
R2
t=2s
R5
R4
+
V2 _
t=0
R6
SPST
Figure P1.26: Circuit for Problem 1.26.
Solution:
(a) t < 0
R3
R2
R4
+
V2 _
R6
(b) 0 < t < 2 s
R1
V1
+
_
R3
R2
(c) t > 2 s
R1
V1
+
_
R3
R4
R5
All rights reserved. Do not reproduce or distribute.
©2009 National Technology and Science Press