SOLUTIONS MANUAL
FUNDAMENTALS OF ELECTRICAL ENGINEERING 2ND
EDITION
CHAPTER NO. 01: FUNDAMENTALS OF ELECTRICAL
CIRCUITS
Sections 1.2-1.3: Charge, Current, and Kirchhoff’s Current Law;
Voltage and Kirchhoff’s Voltage Law
PROBLEM 1.1
A free electron has an initial potential energy per unit charge (voltage) of 17 kJ/C and a
velocity of 93 Mm/s. Later, its potential energy per unit charge is 6 kJ/C. Determine the
change in velocity of the electron.
Solution:
Known quantities:
m
Initial Coulombic potential energy, Vi = 17kJ /C ; initial velocity, U i = 93M ; final Coulombic
s
potential energy, V f = 6kJ /C .
Find:
The change in velocity of the electron.
Assumptions:
∆PEg << ∆PEc
Analysis:
Using the first law of thermodynamics, we obtain the final velocity of the electron:
Qheat − W = ∆KE + ∆PEc + ∆PEg + ...
Heat is not applicable to a single particle. W=0 since no external forces are applied.
∆KE = −∆PEc
1
me (U 2f −U i2 ) = −Qe (V f − Vi )
2
, 2Qe
U 2f = U i2 − (V f − Vi )
me
m
= 93 M −
(
2 2 −1.6 × 10−19 C
)
(6kV − 17kV )
s 9.11 × 10−37 g
m2 m2
= 8.649 × 1015 2
− 3.864 × 1015
s s2
m
U f = 6.917 ×10 7
s
m m m
U f −U i = 93 M − 69.17 M = 23.83 M .
s s s
PROBLEM 1.2
The units for voltage, current, and resistance are the volt (V), the ampere (A), and the
ohm (Ω), respectively. Express each unit in fundamental MKS units.
Solution:
Known quantities:
MKSQ units.
Find:
Equivalent units of volt, ampere and ohm.
Analysis:
Joule J
Voltage = Volt = V=
Coulomb C
Coulomb C
Current = Ampere = a=
second s
Volt Joule × second J ⋅s
Resistance = Ohm = = 2
Ω= 2
Ampere Coulomb C
Ampere C 2
Conductance = Siemens or Mho = =
Volt J ⋅s
PROBLEM 1.3
A particular fully charged battery can deliver 2.7 x 106 coulombs of charge.
a. What is the capacity of the battery in ampere-hours?
b. How many electrons can be delivered?
,Solution:
Known quantities:
qBattery = 2.7 · 106 C.
Find:
The current capacity of the battery in ampere-hours
The number of electrons that can be delivered.
Analysis:
There are 3600 seconds in one hour. Amperage is defined as 1 Coulomb per second
and is directly proportional to ampere-hours.
1ℎ𝑟𝑟
2.7 ∙ 106 𝐶𝐶 ∙ = 750 𝐴𝐴𝐴𝐴
3600𝑠𝑠
a) The charge of a single electron is -1.602·10-19 C. The negative sign is negligible. Simple
division gives the solution:
2.7 ∙ 106 𝐶𝐶 25
�1.602 ∙ 10−19 𝐶𝐶 = 1.685 ∙ 10 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
1 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
PROBLEM 1.4
The charge cycle shown in Figure P1.4 is an example of a three-rate charge. The
current is held constant at 30 mA for 6 h. Then it is switched to 20 mA for the next 3 h.
Find:
a. The total charge transferred to the battery.
b. The energy transferred to the battery.
Hint: Recall that energy w is the integral of power, or P = dw/dt.
Solution:
Known quantities:
See Figure P1.4
Find:
a) The total charge transferred to the battery.
b) The energy transferred to the battery.
, Analysis:
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
Current is equal to , therefore given the current and a duration of that current,
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆
the transferred charge can be calculated by the following equation:
𝐴𝐴 ∙ 𝑡𝑡 = 𝐶𝐶
The two durations should be calculated independently and then added together.
0.030𝐴𝐴 ∙ 21600𝑠𝑠 = 648𝐶𝐶
0.020𝐴𝐴 ∙ 10800𝑠𝑠 = 216𝐶𝐶
648𝐶𝐶 + 216𝐶𝐶 = 𝟖𝟖𝟖𝟖𝟖𝟖𝑪𝑪
P=V·I, therefore, an equation for power can be found by multiplying the two graphs
together.
First separate the voltage graph into three equations:
−6
0 h → 3 h :𝑉𝑉 = 9.26 · 10 𝑡𝑡 + 0.5
−5
3 h → 6 h :𝑉𝑉 = 5.55 ∙ 10 𝑡𝑡
−4
6 h → 9 h :𝑉𝑉 = 1.11 ∙ 10 𝑡𝑡 − 1.6
Next, multiply the first two equations by 0.03A and the third by 0.02A.
−7
0 h → 3 h :𝑃𝑃 = 2.77 ∙ 10 𝑡𝑡 + 0.015
−6
3 h → 6 h :𝑃𝑃 = 1.66 ∙ 10 𝑡𝑡
−6
6 h → 9 h :𝑃𝑃 = 2.22 ∙ 10 𝑡𝑡 − 0.032
Finally, since Energy is equal to the integral of power, take the integral of each of the
equations for their specified times and add them together.
2.77∙10−7 𝑡𝑡 2 10800
0 h → 3 h :𝐸𝐸 = � + 0.015𝑡𝑡� | = 178.2 J
2 0
1.66∙10−6 𝑡𝑡 2 21600
3 h → 6 h :𝐸𝐸 = � 2
�|
10800= 290.43 J
2.22∙10−6 𝑡𝑡 2 32400
6 h → 9 h :𝐸𝐸 = � + 0.032𝑡𝑡� | = 992.95 J
2 21600
𝑬𝑬𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑱𝑱
FUNDAMENTALS OF ELECTRICAL ENGINEERING 2ND
EDITION
CHAPTER NO. 01: FUNDAMENTALS OF ELECTRICAL
CIRCUITS
Sections 1.2-1.3: Charge, Current, and Kirchhoff’s Current Law;
Voltage and Kirchhoff’s Voltage Law
PROBLEM 1.1
A free electron has an initial potential energy per unit charge (voltage) of 17 kJ/C and a
velocity of 93 Mm/s. Later, its potential energy per unit charge is 6 kJ/C. Determine the
change in velocity of the electron.
Solution:
Known quantities:
m
Initial Coulombic potential energy, Vi = 17kJ /C ; initial velocity, U i = 93M ; final Coulombic
s
potential energy, V f = 6kJ /C .
Find:
The change in velocity of the electron.
Assumptions:
∆PEg << ∆PEc
Analysis:
Using the first law of thermodynamics, we obtain the final velocity of the electron:
Qheat − W = ∆KE + ∆PEc + ∆PEg + ...
Heat is not applicable to a single particle. W=0 since no external forces are applied.
∆KE = −∆PEc
1
me (U 2f −U i2 ) = −Qe (V f − Vi )
2
, 2Qe
U 2f = U i2 − (V f − Vi )
me
m
= 93 M −
(
2 2 −1.6 × 10−19 C
)
(6kV − 17kV )
s 9.11 × 10−37 g
m2 m2
= 8.649 × 1015 2
− 3.864 × 1015
s s2
m
U f = 6.917 ×10 7
s
m m m
U f −U i = 93 M − 69.17 M = 23.83 M .
s s s
PROBLEM 1.2
The units for voltage, current, and resistance are the volt (V), the ampere (A), and the
ohm (Ω), respectively. Express each unit in fundamental MKS units.
Solution:
Known quantities:
MKSQ units.
Find:
Equivalent units of volt, ampere and ohm.
Analysis:
Joule J
Voltage = Volt = V=
Coulomb C
Coulomb C
Current = Ampere = a=
second s
Volt Joule × second J ⋅s
Resistance = Ohm = = 2
Ω= 2
Ampere Coulomb C
Ampere C 2
Conductance = Siemens or Mho = =
Volt J ⋅s
PROBLEM 1.3
A particular fully charged battery can deliver 2.7 x 106 coulombs of charge.
a. What is the capacity of the battery in ampere-hours?
b. How many electrons can be delivered?
,Solution:
Known quantities:
qBattery = 2.7 · 106 C.
Find:
The current capacity of the battery in ampere-hours
The number of electrons that can be delivered.
Analysis:
There are 3600 seconds in one hour. Amperage is defined as 1 Coulomb per second
and is directly proportional to ampere-hours.
1ℎ𝑟𝑟
2.7 ∙ 106 𝐶𝐶 ∙ = 750 𝐴𝐴𝐴𝐴
3600𝑠𝑠
a) The charge of a single electron is -1.602·10-19 C. The negative sign is negligible. Simple
division gives the solution:
2.7 ∙ 106 𝐶𝐶 25
�1.602 ∙ 10−19 𝐶𝐶 = 1.685 ∙ 10 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
1 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
PROBLEM 1.4
The charge cycle shown in Figure P1.4 is an example of a three-rate charge. The
current is held constant at 30 mA for 6 h. Then it is switched to 20 mA for the next 3 h.
Find:
a. The total charge transferred to the battery.
b. The energy transferred to the battery.
Hint: Recall that energy w is the integral of power, or P = dw/dt.
Solution:
Known quantities:
See Figure P1.4
Find:
a) The total charge transferred to the battery.
b) The energy transferred to the battery.
, Analysis:
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
Current is equal to , therefore given the current and a duration of that current,
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆
the transferred charge can be calculated by the following equation:
𝐴𝐴 ∙ 𝑡𝑡 = 𝐶𝐶
The two durations should be calculated independently and then added together.
0.030𝐴𝐴 ∙ 21600𝑠𝑠 = 648𝐶𝐶
0.020𝐴𝐴 ∙ 10800𝑠𝑠 = 216𝐶𝐶
648𝐶𝐶 + 216𝐶𝐶 = 𝟖𝟖𝟖𝟖𝟖𝟖𝑪𝑪
P=V·I, therefore, an equation for power can be found by multiplying the two graphs
together.
First separate the voltage graph into three equations:
−6
0 h → 3 h :𝑉𝑉 = 9.26 · 10 𝑡𝑡 + 0.5
−5
3 h → 6 h :𝑉𝑉 = 5.55 ∙ 10 𝑡𝑡
−4
6 h → 9 h :𝑉𝑉 = 1.11 ∙ 10 𝑡𝑡 − 1.6
Next, multiply the first two equations by 0.03A and the third by 0.02A.
−7
0 h → 3 h :𝑃𝑃 = 2.77 ∙ 10 𝑡𝑡 + 0.015
−6
3 h → 6 h :𝑃𝑃 = 1.66 ∙ 10 𝑡𝑡
−6
6 h → 9 h :𝑃𝑃 = 2.22 ∙ 10 𝑡𝑡 − 0.032
Finally, since Energy is equal to the integral of power, take the integral of each of the
equations for their specified times and add them together.
2.77∙10−7 𝑡𝑡 2 10800
0 h → 3 h :𝐸𝐸 = � + 0.015𝑡𝑡� | = 178.2 J
2 0
1.66∙10−6 𝑡𝑡 2 21600
3 h → 6 h :𝐸𝐸 = � 2
�|
10800= 290.43 J
2.22∙10−6 𝑡𝑡 2 32400
6 h → 9 h :𝐸𝐸 = � + 0.032𝑡𝑡� | = 992.95 J
2 21600
𝑬𝑬𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑱𝑱
