Charge and Current
\[\newcommand{\evalAt}[3]{\left. #1 \right\rvert_{#2}^{#3}}
\newcommand{\kilo}{\mathrm{k}}
\newcommand{\milli}{\mathrm{m}}
\newcommand{\micro}{\mu}
\newcommand{\nano}{\mathrm{n}}
\newcommand{\pico}{\mathrm{p}}
\newcommand{\ampere}{\mathrm{A}}
\newcommand{\coulomb}{\mathrm{C}}
\newcommand{\second}{\mathrm{s}}\]
Exercise 1.1
There are \(\frac{1}{1.602 \times 10^{-19}} = 6.24 \times 10^{18}\) electrons
per \(1 \coulomb\) of charge.
(a)
\[\left( 6.482 \times 10^{17} \right) \left( -1.602 \times 10^{-19} \right) =
-103.84 \milli\coulomb\]
(b)
\[\left( 1.24 \times 10^{18} \right) \left( -1.602 \times 10^{-19} \right) =
-198.65 \milli\coulomb\]
(c)
\[\left( 2.46 \times 10^{19} \right) \left( -1.602 \times 10^{-19} \right) =
-3.941 \coulomb\]
(d)
\[\left( 1.628 \times 10^{20} \right) \left( -1.602 \times 10^{-19} \right) =
-26.08 \coulomb\]
Exercise 1.2
(a)
\[i = \frac{dq}{dt}
= \frac{d}{dt} \left( 3t + 8 \right)
= 3 \milli\ampere\]
(b)
\[i = \frac{dq}{dt}
= \frac{d}{dt} \left( 8t^2 + 4t - 2 \right)
= \left( 16t + 4 \right) \ampere\]
(c)
\[i = \frac{dq}{dt}
= \frac{d}{dt} \left( 3e^{-t} - 5e^{-2t} \right)
= \left( -3e^{-t} + 10e^{-2t} \right) \nano\ampere\]
(d)
\[i = \frac{dq}{dt}
= \frac{d}{dt} \left( 10 \sin 120 \pi t \right)
= \left( 1200 \pi \cos 120 \pi t \right) \pico\ampere\]
(e)
\[\begin{split}i = \frac{dq}{dt}
= \frac{d}{dt} \left( 20 e^{-4t} \cos 50t \right)
&= -80 e^{-4t} \cos 50t - 1000 e^{-4t} \sin 50t\\
&= -40 e^{-4t} \left( 2 \cos 50t +50 \sin 50t \right) \micro\ampere\end{split}\]
Exercise 1.3
(a)
Given \(i(t) = 3 \ampere\) and \(q(0) = 1 \coulomb\),
\[q(t) = \int i dt
= 3t + C\]
where
\[\begin{split}q(0) &= 3(0) + C\\
1 &= C.\end{split}\]
(b)
Given \(i(t) = (2t + 5) \milli\ampere\) and \(q(0) = 0\),
\[q(t) = \int (2t + 5) dt
= t^2 + 5t + C\]
where
\[\begin{split}q(0) &= (0)^2 + 5(0) + C\\
0 &= C.\end{split}\]
(c)
Given \(i(t) = 20 \cos\left( 10t + \pi / 6 \right) \micro\ampere\) and
\(q(0) = 2 \micro\coulomb\),
\[q(t) = \int i dt
= 2 \sin\left( 10t + \pi / 6 \right) + C\]
where
\[\begin{split}q(0) &= 2 \sin\left( 10(0) + \pi / 6 \right) + C\\
2 &= 2 \sin \frac{\pi}{6} + C\\
1 &= C.\end{split}\]
(d)
Given \(i(t) = 10e^{-30t} \sin\left( 40t \right) \ampere\) and
\(q(0) = 0\),
\[q(t) = \int i dt
= -\frac{1}{25} e^{-30t} \left( 3 \sin 40t + 4 \cos 40t \right) + C\]
where
\[\begin{split}q(0) &= -\frac{1}{25} e^{-30(0)} \left(
3 \sin 40 (0) + 4 \cos 40 (0)
\right) + C\\
0 &= -\frac{1}{25} 4 + C\\
\frac{4}{25} &= C.\end{split}\]
Exercise 1.4
Suppose a current of \(7.4 \ampere\) flows through a conductor. The
amount of charge passing through any cross-section of the conductor in
\(20 \second\) is
\[Q = \int_{t_0}^t i dt = \evalAt{it}{0}{20} = 148 \coulomb.\]
Exercise 1.5
The total charge transferred over the time interval of
\(0 \leq t \leq 10 \second\) when
\(i(t) = \frac{1}{2} t \ampere\) is
\[Q = \int_{t_0}^t i dt
= \left. \frac{t^2}{4} \right\rvert_0^{10}
= 25 \coulomb.\]
Exercise 1.6
The charge (\(\milli\coulomb\)) entering a certain element is defined as
\[\begin{split}q(t) = \begin{cases}
15t & \text{if } 0 \leq t \leq 2\\
30 & \text{if } 2 \leq t \leq 8\\
30 - 7.5 (t - 8) & \text{if } 8 \leq t \leq 12\\
0 & \text{otherwise.}
\end{cases}\end{split}\]
(a)
At \(t = 1 \milli\second\), the current is
\[i(t) = \frac{dq}{dt} = 15 \ampere.\]
(b)
At \(t = 6 \milli\second\), the current is
\[i(t) = \frac{dq}{dt} = 0 \ampere.\]
(b)
At \(t = 10 \milli\second\), the current is
\[i(t) = \frac{dq}{dt} = -7.5 \ampere.\]
Exercise 1.7
The charge (\(\coulomb\)) flowing in a wire is defined as
\[\begin{split}q(t) = \begin{cases}
25t & \text{if } 0 \leq t \leq 2\\
50 - 25 (t - 2) & \text{if } 2 \leq t \leq 4\\
-25 (t - 4) & \text{if } 4 \leq t \leq 6\\
-50 + 25 (t - 6) & \text{if } 6 \leq t \leq 8\\
0 & \text{otherwise.}
\end{cases}\end{split}\]
The corresponding current (\(\ampere\)) is
\[\begin{split}i(t) = \frac{dq}{dt}
= \begin{cases}
25 & \text{if } 0 \leq t \leq 2\\
-25 & \text{if } 2 \leq t \leq 6\\
25 & \text{if } 6 \leq t \leq 8\\
0 & \text{otherwise.}
\end{cases}\end{split}\]
Exercise 1.8
The current (\(\milli\ampere\)) flowing past a point in a device is
characterized as
\[\begin{split}i(t) = \begin{cases}
10t & \text{if } 0 \leq t \leq 1 \milli\second\\
10 & \text{if } 1 \leq t \leq 2 \milli\second\\
0 & \text{otherwise.}
\end{cases}\end{split}\]
The total charge through the point is
\[Q(t) = \int_{t_0 = 0}^{t = 2} i dt
= \int_0^1 10t dt + \int_1^2 10 dt
= \evalAt{5t^2}{0}{1} + \evalAt{10t}{1}{2}
= 15 \micro\coulomb.\]
Exercise 1.9
The current (\(\ampere\)) through an element is characterized by
\[\begin{split}i(t) = \begin{cases}
i_1 = 10 & \text{if } 0 \leq t \leq 1\\
i_2 = 10 - 5 (t - 1) & \text{if } 1 \leq t \leq 2\\
i_3 = 5 & \text{if } 2 \leq t \leq 4\\
i_4 = 5 - 5 (t - 4) & \text{if } 4 \leq t \leq 5\\
0 & \text{otherwise.}
\end{cases}\end{split}\]
(a)
The total charge that passed through the element at \(t = 1 \second\) is
\[Q(t) = \int_{t_0 = 0}^{t = 1} i dt
= \int_0^1 i_1 dt
= \evalAt{10t}{0}{1}
= 10 \coulomb.\]
(b)
The total charge that passed through the element at \(t = 3 \second\) is
\[Q(t) = \int_{t_0 = 0}^{t = 3} i dt
= \int_0^1 i_1 dt + \int_1^2 i_2 dt + \int_2^3 i_3 dt
= Q(1) +
\evalAt{10t - \frac{5}{2} t^2 + 5t}{1}{2} +
\evalAt{5t}{2}{3}
= 22.5 \coulomb.\]
(c)
The total charge that passed through the element at \(t = 5 \second\) is
\[Q(t) = \int_{t_0 = 0}^{t = 5} i dt
= \int_0^1 i_1 dt + \int_1^2 i_2 dt + \int_2^4 i_3 dt + \int_4^5 i_4 dt
= Q(3) + \evalAt{5t}{3}{4} +
\evalAt{5t - \frac{5}{2} t^2 + 20t}{4}{5}
= 30 \coulomb.\]
Voltage, Power, and Energy
\[\newcommand{\joule}{\mathrm{J}}
\newcommand{\volt}{\mathrm{V}}
\newcommand{\watt}{\mathrm{W}}
\newcommand{\hour}{\mathrm{h}}\]
Exercise 1.10
A lightning bolt with \(10 \kilo\ampere\) strikes an object for
\(15 \micro\second\). The amount of charge deposited on the object is
\[Q = \int i dt
= \left( 10 \times 10^3 \ampere \right)
\left( 15 \times 10^{-6} \second \right)
= 0.15 \coulomb.\]
Exercise 1.11
A rechargeable flashlight battery can deliver \(90 \milli\ampere\) for about
\(12 \hour\). The amount of charge it can release at that rate is
\[Q = \int i dt
= \evalAt{\left( 90 \times 10^{-3} \right) t}{0}{12 \times 3600}
= 3.888 \kilo\coulomb.\]
If the terminal voltage is \(1.5 \volt\), the battery can delivery
\[\begin{split}w &= \int v dq
& \quad \text{(1.3)}\\
&= \int v i dt
& \quad \text{(1.1), (1.9)}\\
&= \evalAt{\left( 1.5 \times 0.09 \right) t}{0}{12 \times 3600}\\
&= 5.832 \kilo\joule\\
&= 1.62 \watt\hour.\end{split}\]
Exercise 1.12
The current (\(\ampere\)) through an element is given by
\[\begin{split}i(t) = \begin{cases}
3t & \text{if } 0 \leq t < 6\\
18 & \text{if } 6 \leq t < 10\\
-12 & \text{if } 10 \leq t < 15\\
0 & \text{if } t \geq 15.
\end{cases}\end{split}\]
The charge (\(\coulomb\)) stored in the element over
\(0 < t < 20 \second\) is given by
\[\begin{split}Q(t) = \begin{cases}
\frac{3}{2} t^2 & \text{if } 0 \leq t < 6\\
54 + 18 (t - 6) & \text{if } 6 \leq t < 10\\
126 - 12 (t - 10) & \text{if } 10 \leq t < 15\\
66 & \text{if } t \geq 15.
\end{cases}\end{split}\]
Exercise 1.13
Given the charge (\(\milli\coulomb\)) entering the positive terminal of an
element is
\[q = 5 \sin 4 \pi t,\]
the current (\(\milli\ampere\)) is
\[i = \frac{dq}{dt}
= 20 \pi \cos 4 \pi t.\]
Suppose the voltage (\(\volt\)) across the element (passive sign convention)
is
\[v = 3 \cos 4 \pi t.\]
(a)
The power delivered to the element at \(t = 0.3 \second\) is
\[\begin{split}p &= vi & \quad \text{(1.7)}\\
&= 60 \pi \cos^2 4 \pi t\\
&= 30 \pi \left( 1 + \cos 8 \pi t \right)
& \quad \text{cosine identity} \cos 2 \alpha = 2 \cos^2 \alpha - 1\\
&= 123.37 \milli\watt.\end{split}\]
(b)
The energy delivered to the element between \(0\) and \(0.6 \second\) is
\[\begin{split}w &= \int_{t_0}^t p dt
& \quad \text{(1.9)}\\
&= \evalAt{30 \pi t + \frac{15}{4} \sin 8 \pi t}{0}{0.6}\\
&= 58.76 \milli\joule.\end{split}\]
Exercise 1.14
The voltage (\(\volt\)) across a device and the current
(\(\milli\ampere\)) through it are
\[v(t) = 10 \cos 2 t
\quad \text{and} \quad
i(t) = 20 \left( 1 - \exp -\frac{t}{2} \right).\]
(a)
The total charge in the device at \(t = 1 \second\) is
\[\begin{split}Q &= \int_{t_0 = 0}^t i dt
& \quad \text{(1.2)}\\
&= \evalAt{20t + 40 \exp -\frac{t}{2}}{0}{1}\\
&= 4.26 \milli\coulomb.\end{split}\]
(b)
The power consumed by the device at \(t = 1 \second\) is
\[\begin{split}p &= vi
& \quad \text{(1.7)}\\
&= 200 \left( 1 - \exp -\frac{t}{2} \right) \cos 2t\\
&= -32.75 \milli\watt.\end{split}\]
Exercise 1.15
The current (\(\milli\ampere\)) entering the positive terminal of a device
is \(i(t) = 6 e^{-2t}\) and the voltage (\(\milli\volt\)) across the
device is
\[v(t) = 10 \frac{di}{dt} = -120 \exp -2t.\]
(a)
The total charge delivered to the device between \(t = 0\) and
\(t = 2 \second\) is
\[\begin{split}Q &= \int_{t_0 = 0}^t i dt
& \quad \text{(1.2)}\\
&= \evalAt{-3 \exp -2t}{0}{2}\\
&= 2.95 \milli\coulomb.\end{split}\]
(b)
The power (\(\micro\watt\)) absorbed is
\[\begin{split}p &= vi
& \quad \text{(1.7)}\\
&= -720 \exp -4t.\end{split}\]
(c)
The energy absorbed in \(3 \second\) is
\[\begin{split}w &= \int_{t_0 = 0}^t p dt
& \quad \text{(1.9)}\\
&= \evalAt{180 \exp -4t}{0}{3}\\
&= -180 \micro\joule.\end{split}\]
Circuit Elements
Exercise 1.16
The current (\(\milli\ampere\)) through and the voltage (\(\volt\))
across an element are given by
\[\begin{split}i(t) = \begin{cases}
30t & \text{if } 0 \leq t < 2\\
60 - 30 (t - 2) & \text{if } 2 \leq t < 4\\
0 & \text{otherwise.}
\end{cases}
\qquad \text{and} \qquad
v(t) = \begin{cases}
5 & \text{if } 0 \leq t < 2\\
-5 & \text{if } 2 \leq t < 4\\
0 & \text{otherwise.}
\end{cases}\end{split}\]
(a)
The power (\(\milli\watt\)) delivered to the element for \(t > 0\) is
given by
\[\begin{split}p(t) = \begin{cases}
150t & \text{if } 0 \leq t < 2\\
-300 + 150 (t - 2) & \text{if } 2 \leq t < 4\\
0 & \text{otherwise.}
\end{cases}\end{split}\]
(b)
The total energy absorbed by the element for the period of
\(0 < t < 4 \second\) is
\[\begin{split}w &= \int_{t_0 = 0}^t p dt
& \quad \text{(1.9)}\\
&= \int_0^2 p dt + \int_2^4 p dt\\
&= \evalAt{75t^2}{0}{2} + \evalAt{-600t + 75t^2}{2}{4}\\
&= 0 \milli\joule.\end{split}\]
Exercise 1.17
The amount of power absorbed by element 3 is
\[\begin{split}p_1 + p_2 + p_3 + p_4 + p_5 &= 0 & \quad \text{(1.8)}\\
p_3 &= 205 - 60 - 45 - 30\\
&= 70 \watt.\end{split}\]
Exercise 1.18
Given \(I = 10 \ampere\),
\[\begin{split}p_1 &= -\left( 30 \times 10 \right) = -300 \watt\\
\\
p_2 &= I \times 10 = 100 \watt\\
\\
p_3 &= 20 \times 14 = 280 \watt\\
\\
p_4 &= -\left( 8 \times 4 \right) = -32 \watt\\
\\
p_5 &= -\left( 12 \times 0.4I \right) = -48 \watt.\end{split}\]
Exercise 1.19
The power (\(\watt\)) absorbed by each element is given by
\[\begin{split}p_1 &= - \left( 9 \times 8 \right) = -72\\
\\
p_2 &= 9 \times 2 = 18\\
\\
p_3 &= 3I\\
\\
p_4 &= 6I\end{split}\]
where
\[\begin{split}0 &= \sum_i p_i
& \quad \text{(1.8)}\\
&= -54 + 9I\\
I &= 6 \ampere.\end{split}\]
Exercise 1.20
Given \(I_o = 2 \ampere\), the power (\(\watt\)) absorbed by each
element is given by
\[\begin{split}p_1 &= - \left( 30 \times 6 \right) = -180\\
\\
p_2 &= 12 \times 6 = 72\\
\\
p_3 &= 3 V_o\\
\\
p_4 &= 28 \times I_o = 56\\
p_5 &= 28 \times 1 = 28\\
p_6 &= -\left( 5 I_o \times 3 \right) = -30\\\end{split}\]
where
\[\begin{split}0 &= \sum_i p_i
& \quad \text{(1.8)}\\
&= -54 + 3 V_o\\
V_o &= 18 \volt.\end{split}\]
Applications
\[\newcommand{\minute}{\mathrm{min}}
\newcommand{\per}{\mathrm{/}}
\newcommand{\cent}{\mathrm{cents}}
\newcommand{\day}{\mathrm{day}}
\newcommand{\watthour}{\mathrm{Wh}}
\newcommand{\amperehour}{\mathrm{Ah}}
\newcommand{\mega}{\mathrm{M}}\]
Exercise 1.21
A \(60 \watt\) incandescent bulb operates at \(120 \volt\). The amount
of coulombs flowing through the bulb in one day is
\[\begin{split}Q &= \int \frac{p}{v} dt
& \quad \text{(1.6)}\\
&= \frac{60}{120} \times 3600\\
&= 43.2 \kilo\coulomb,\end{split}\]
which is equivalent to saying the number of electrons is
\[N_e = Q \left( 6.24 \times 10^{18} \right) = 2.696 \times 10^{23}.\]
Exercise 1.22
A lightning bolt strikes an airplane with \(40 \kilo\ampere\) for
\(1.2 \milli\second\). The charge deposited on the plane is
\[\begin{split}Q &= \int_{t_0 = 0}^t i dt
& \quad \text{(1.2)}\\
&= \evalAt{40000t}{0}{0.0012}\\
&= 48 \coulomb.\end{split}\]
Exercise 1.23
A \(1.8 \kilo\watt\) electric heater takes \(15 \minute\) to boil a
quantity of water i.e.
\[\begin{split}w &= \int_{t_0 = 0}^{t} p dt
& \quad \text{(1.9)}\\
&= \evalAt{1800t}{0}{15 \times 60}\\
&= 1620000 \joule\\
&= 450 \watthour
& \quad 1 \watthour = 3600 \joule\\
&= 0.45 \kilo\watthour.\end{split}\]
Suppose this is done once a day and power costs
\(10 \cent\per\kilo\watthour\). The cost of its
operation for \(30\) days is
\[c = 30 \times 10 \times w = 135 \cent.\]
Exercise 1.24
A utility company charges \(8.2 \cent\per\kilo\watthour\). Suppose a
consumer operates a \(60 \watt\) light bulb continuously for one day i.e.
\[\begin{split}w &= \int_{t_0 = 0}^{t} p dt
& \quad \text{(1.9)}\\
&= \evalAt{60t}{0}{24}\\
&= 1.44 \kilo\watthour.\end{split}\]
That consumer would pay
\[c = 8.2 \times w = 11.81 \cent.\]
Exercise 1.25
A \(1.5 \kilo\watt\) toaster takes roughly \(3.5\) minutes to heat four
slices of bread i.e.
\[\begin{split}w &= \int_{t_0 = 0}^{t} p dt
& \quad \text{(1.9)}\\
&= \evalAt{1500t}{0}{3.5 \times 60}\\
&= 315000 \joule\\
&= 0.0875 \kilo\watthour.\end{split}\]
Assuming energy costs \(8.2 \cent\per\kilo\watthour\), the cost of
operating the toaster once per day for one month is
\[c = 8.2 \times 30 \times w = 21.53 \cent.\]
Exercise 1.26
A flashlight battery has a rating of \(0.8 \amperehour\) and a lifetime of
\(10 \hour\).
(a)
The battery can deliver
\[i = \frac{0.8}{10} = 80 \milli\ampere.\]
(b)
If its terminal voltage is \(6 \volt\), the battery can give
\[\begin{split}p &= vi
& \quad \text{(1.7)}\\
&= 6 \times 0.08\\
&= 0.48 \watt.\end{split}\]
(c)
The amount of energy stored in the battery is
\[\begin{split}w &= \int_{t_0 = 0}^{t = 10} p
& \quad \text{(1.9)}\\
&= \evalAt{pt}{0}{10}\\
&= 4.8 \watthour.\end{split}\]
Exercise 1.27
A typical automotive battery requires a constant current of \(3 \ampere\)
for 4 hours to become fully charged. Suppose the terminal voltage
(\(\volt\)) is \(10 + \frac{t}{2}\) where \(t\) is in hours.
(a)
The amount of charge transported is
\[\begin{split}Q &= \int_{t_0}^t i dt
& \quad \text{(1.2)}\\
&= \evalAt{3t}{0}{4 \times 3600}\\
&= 43.2 \kilo\coulomb.\end{split}\]
(b)
The terminal voltage reformulated in seconds is \(10 + \frac{0.5t}{3600}\).
The energy expended is
\[\begin{split}w &= \int_{t_0}^t vi dt
& \quad \text{(1.9)}\\
&= i \left(
\evalAt{10t + \frac{0.25 t^2}{3600}}{0}{4 \times 3600}
\right)\\
&= 475.2 \kilo\joule\\
&= 0.132 \kilo\watthour
& \quad 1 \kilo\watthour = 3600 \kilo\joule.\end{split}\]
(c)
Assuming electricity costs \(9 \cent\per\kilo\watthour\), the charging cost
is
\[c = 9 \times w = 1.188 \cent.\]
Exercise 1.28
A \(60 \watt\) incandescent lamp is connected to a \(120 \volt\) source.
(a)
The current through the lamp is
\[\begin{split}i &= \frac{p}{v}
& \quad \text{(1.7)}\\
&= 0.5 \ampere.\end{split}\]
(b)
The energy consumed in a non-leap year is
\[\begin{split}w &= \int_{t_0}^t p dt
& \quad \text{(1.9)}\\
&= \evalAt{pt}{0}{365 \times 24}\\
&= 525.6 \kilo\joule.\end{split}\]
If electricity costs \(9 \cent\per\kilo\watthour\), the cost of operating
the light is
\[c = 9 \times w = \$47.304.\]
Exercise 1.29
An electric stove with four burners and an oven is used as follows:
Burner 1: 20 minutes.
Burner 2: 40 minutes.
Burner 3: 15 minutes.
Burner 4: 45 minutes.
Oven: 30 minutes.
Suppose each burner is rated at \(1.2 \kilo\watt\) and the oven at
\(1.8 \kilo\watt\).
The total amount of energy expended by the burners is
\[\begin{split}w_b &= \int_{t_0}^t p_b dt
& \quad \text{(1.9)}\\
&= \evalAt{p_b t}{0}{(20 + 40 + 15 + 45) \times 60}\\
&= 8640 \kilo\joule\\
&= 2.4 \kilo\watthour\end{split}\]
and the oven is
\[\begin{split}w_o &= \int_{t_0}^t p_o dt
& \quad \text{(1.9)}\\
&= \evalAt{p_o t}{0}{30 \times 60}\\
&= 3240 \kilo\joule\\
&= 0.9 \kilo\watthour.\end{split}\]
If electricity costs \(12 \cent\per\kilo\watthour\), the cost of electricity
used in preparing the meal is
\[c = 12 \times (w_b + w_o) = 39.6 \cent.\]
Exercise 1.30
Reliant Energy charges customers as follows:
If a customer uses \(2.436 \mega\watthour\) in one month, Reliant Energy
will charge
\[c = 6 + 0.02 \times 250 + 0.07 \times (2436 - 250) = \$164.02.\]
Exercise 1.31
Suppose a \(120 \watt\) PC runs for \(4 \hour\per\day\) while a
\(60 \watt\) bulb runs for \(8 \hour\per\day\). The energy consumed by
the PC is
\[\begin{split}w_{pc} &= \int_{t_0}^t p_{pc} dt
& \quad \text{(1.9)}\\
&= \evalAt{p_{pc} t}{0}{4 \times 365}\\
&= 175200 \watthour\\
&= 175.2 \kilo\watthour\end{split}\]
and the bulb is
\[\begin{split}w_b &= \int_{t_0}^t p_b dt
& \quad \text{(1.9)}\\
&= \evalAt{p_b t}{0}{8 \times 365}\\
&= 175200 \watthour\\
&= 175.2 \kilo\watthour.\end{split}\]
If electricity costs \(12 \cent\per\kilo\watthour\), this household pays per
year
\[c = 12 \times (w_{pc} + w_b) = \$42.048 \cent.\]
Comprehensive Problems
\[\newcommand{\horsepower}{\mathrm{hp}}\]
Exercise 1.32
A telephone wire has a current of \(20 \micro\ampere\) flowing through it.
The time it takes for a charge of \(15 \coulomb\) to pass through the wire
is
\[\begin{split}Q &= \int_{t_0}^t i dt
& \quad \text{(1.2)}\\
15 &= \evalAt{it}{0}{t}\\
t &= \frac{15}{i}\\
&= 208.\bar{3} \hour.\end{split}\]
Exercise 1.33
A lightning bolt carried a current of \(2 \kilo\ampere\) and lasted for
\(3 \milli\second\). The amount of charge contained in the lightning bolt
is
\[\begin{split}Q &= \int_{t_0}^t i dt
& \quad \text{(1.2)}\\
&= \evalAt{it}{0}{0.003}\\
&= 6 \coulomb.\end{split}\]
Exercise 1.34
The power (\(\watt\)) consumption of a certain household in one day is given
by
\[\begin{split}p(t) = \begin{cases}
200 & \text{if } 0 \leq t < 6\\
800 & \text{if } 6 \leq t < 8\\
200 & \text{if } 8 \leq t < 18\\
1200 & \text{if } 18 \leq t < 22\\
200 & \text{if } 18 \leq t \leq 24.
\end{cases}\end{split}\]
(a)
The total energy consumed is
\[\begin{split}w &= \int_{t_0 = 0}^t p dt
& \quad \text{(1.9)}\\
&= \int_0^6 200 dt +
\int_6^8 800 dt +
\int_8^{18} 200 dt +
\int_{18}^{22} 1200 dt +
\int_{22}^{24} 200 dt\\
&= 200 (6 - 0) + 800 (8 - 6) + 200 (18 - 8) +
1200 (22 - 18) + 200 (24 - 22)\\
&= 10 \kilo\watthour.\end{split}\]
(b)
The average power per hour over the entire day is
\[p_{avg} = \frac{w}{24} = 416.\bar{6} \watt.\]
Exercise 1.35
The power (\(\mega\watt\)) drawn by an industrial plant between 8:00 and
8:30 A.M. is given by
\[\begin{split}p(t) = \begin{cases}
5 & \text{if } 0 \leq t < 5\\
4 & \text{if } 5 \leq t < 10\\
3 & \text{if } 10 \leq t < 15\\
8 & \text{if } 15 \leq t < 20\\
4 & \text{if } 20 \leq t \leq 30.
\end{cases}\end{split}\]
The total energy consumed by the plant is
\[\begin{split}w &= \int_{t_0 = 0}^t p dt
& \quad \text{(1.9)}\\
&= \int_0^5 5 dt +
\int_5^{10} 4 dt +
\int_{10}^{15} 3 dt +
\int_{15}^{20} 8 dt +
\int_{20}^{30} 4 dt\\
&= \frac{
5 (5 - 0) + 4 (10 - 5) + 3 (15 - 10) +
8 (20 - 15) + 4 (30 - 20)
}{60}
& \quad \text{convert } \minute \text{ to } \hour\\
&= 2.\bar{3} \mega\watthour.\end{split}\]
Exercise 1.36
A lead-acid battery is rated at \(160 \amperehour\).
(a)
The maximum current it can supply for \(40 \hour\) is
\[\begin{split}i &= \frac{\Delta q}{\Delta t}
& \quad \text{(1.1)}\\
&= \frac{160}{40}\\
&= 4 \ampere.\end{split}\]
(a)
If it is discharged at \(1 \milli\ampere\), the battery can last
\[\begin{split}\Delta t &= \frac{\Delta q}{i}
& \quad \text{(1.1)}\\
&= \frac{160 \times 3600}{0.001}\\
&= 6666.\bar{6} \day.\end{split}\]
Exercise 1.37
A \(12 \volt\) battery requires a total charge of \(40 \amperehour\)
during recharging. The energy supplied to the battery is
\[\begin{split}v &= \frac{\Delta w}{\Delta q}
& \quad \text{(1.3)}\\
\Delta w &= v \Delta q\\
&= 12 (40 \times 3600)\\
&= 1.728 \mega\joule.\end{split}\]
Exercise 1.38
Assuming \(1 \horsepower = 746 \watt\), a \(10 \horsepower\) motor
delivers in 30 minutes
\[\begin{split}w &= \int_{t_0 = 0}^{t} p dt
& \quad \text{(1.9)}\\
&= \evalAt{10 \times 746 t}{0}{30 / 60}\\
&= 12 (40 \times 3600)\\
&= 3.73 \kilo\watthour.\end{split}\]
Exercise 1.39
A \(600 \watt\) TV receiver is turned on for \(4 \hour\) with nobody
watching it. The total energy expended is
\[\begin{split}w &= \int_{t_0 = 0}^{t} p dt
& \quad \text{(1.9)}\\
&= \evalAt{600t}{0}{4}\\
&= 2.4 \kilo\watthour.\end{split}\]
If electricity costs \(10 \cent\per\kilo\watthour\), the consumer wastes
\[c = 10 \times 2.4 = 24 \cent.\]