<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\usepackage{epsfig}
\topmargin=-0.5cm
\oddsidemargin=-0.5cm
\evensidemargin=-0.5cm
\textheight=23cm
\textwidth=17cm
\begin{document}
\newcommand{\bold}[1]{\mbox{\boldmath $#1$}}
\newcommand{\ems}{{\cal E}}
\begin{bf}
\noindent
Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003: Elektrisitet og magnetisme

\noindent
Spring 2008

\end{bf}

\begin{center}
\begin{large}
{\O}ving 11
\end{large}
\end{center}

\noindent
Veiledning: Fredag 28. og mandag 31. mars

\noindent
Innleveringsfrist: Fredag 4. april


\vspace{0.7cm}
\noindent
{\em Exercise 1}

\vspace{0.2cm}
\noindent
A voltage source $V = 1.5$ V is connected to a resistor with resistance
$R = 20 \; \Omega$ via two 60 cm long copper wires with
cross section 2 mm$^2$.

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=3.0in \epsfbox{ov10_fig1.eps}}
\vspace{0.5cm}
%%END FIGURE

\vspace{0.3cm}
\noindent
$a)$ How big is the voltage drop over the Cu wires and
the resistor, respectively?
[Answers: 0.75 mV and 1.5 V]

\vspace{0.2cm}
\noindent
$b)$ Determine the current and the power dissipated ("lost")
in the resistor.
[Answers: ca 0.075 A and 0.11 W]

\vspace{0.2cm}
\noindent
$c)$ What is the average drift velocity of the free electrons
through the Cu wires? Assume one free electron from each Cu atom.
Compare with the average thermal velocity for an electron at
room temperature. (Average kinetic energy pr electron at temperature
$T$ is $3 k_B T / 2$, where $k_B$ is Boltzmann's constant.)

[Answers: 2.76 $\mu$m/s and ca $10^5$ m/s.]

\vspace{0.2cm}
\noindent
Given information: Mass density of Cu: 8960 kg/m$^3$. Molar mass of Cu: 63.54 g/mol.
Electric conductivity for Cu at room temperature: $5.8\cdot 10^7$ $\Omega^{-1}$m$^{-1}$.
Boltzmann's constant: $k_B = 1.38 \cdot 10^{-23}$ J/K.



\newpage
%\vspace{0.7cm}
\noindent
{\em Exercise 2}

\vspace{0.2cm}
\noindent
The figure below shows an electric circuit with 5 resistors $R_j$, $j = 1,..,5$.

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=3.0in \epsfbox{ov11_fig1.eps}}
\vspace{0.5cm}
%%END FIGURE

\vspace{0.3cm}
\noindent
$a)$ Determine the total resistance $R$ between the points A and B, i.e.:
Determine the resistance $R$ in the equivalent circuit in the following figure:

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=3.0in \epsfbox{ov11_fig2.eps}}
\vspace{0.5cm}
%%END FIGURE


\vspace{0.2cm}
\noindent
$b)$ An ideal voltage source with electromotive force $\cal E$ is connected
to the circuit so that $\Delta V = V_A - V_B = {\cal E}$. Determine the
resulting currents $I_j$ through each of the resistors $R_j$.

\noindent
(Unless otherwise specified, we always assume, in exercises like this one, that
the connecting wires between the various resistors are {\em perfect
conductors}, i.e., with zero resistance.)



\vspace{1.5cm}
\noindent
{\em Exercise 3}

\vspace{0.3cm}
\noindent
A capacitor with capacitance $C$ has charge $\pm Q_0$. The capacitor is at time
$t=0$ connected to a resistor $R$, so that we have a closed circuit, and current
may run in the circuit. Determine the capacitor charge $Q(t)$ and the current $I(t)$
for $t \ge 0$.


\vspace{1.5cm}
\noindent
{\em Exercise 4}

\vspace{0.3cm}
\noindent
For a DC circuit with voltage source $V_0$ delivering a current $I_0$, the
total resistance $R$ of the circuit will be given by Ohm's law, $R = V_0/I_0$.

\noindent
For an AC circuit with voltage source $V_0 \, \cos\omega t$  delivering a current
$I_0 \, \cos(\omega t - \alpha)$, we may analogously define a ''generalized resistance'',
or {\rm impedance} $Z$, for the circuit: $Z = V_0/I_0$.
We allow for a {\em phase difference} $\alpha$ between applied voltage and resulting current.
We will also see, in the examples below, that the current amplitude $I_0$ may become dependent
upon the (angular) frequency $\omega$, which implies that the impedance $Z$ may also be
frequency dependent, i.e., $Z = Z(\omega)$.

\noindent
In the lectures, I did the simplest example, with a voltage source $V_0 \, \cos\omega t$
connected to a single resistor $R$, and found that the current becomes
$I_0 \, \cos\omega t$, with amplitude $I_0 = V_0/R$. So, the impedance for a resistor
is simply $Z_R = R$.

\vspace{1cm}
\noindent
$a)$ A voltage source $\ems(t) = V_0 \, \cos\omega t$ is connected to a capacitor $C$:

%%FIGURE
\vspace{0.8cm}
\centerline{\epsfxsize=2.4in \epsfbox{EC-krets.eps}}
\vspace{0.5cm}
%%END FIGURE

\vspace{0.3cm}
\noindent
Find the charge $Q(t)$ on the capacitor, and also the current $I(t)$ in the circuit.
Sketch $\ems(t)$ and $I(t)$ between $t=0$ and $t=T=2\pi/\omega$.

\noindent
Show that the current may be written in the form
$$I(t) = I_0 \, \cos(\omega t - \alpha)$$
and determine (the frequency dependent) current amplitude $I_0$ and the phase constant $\alpha$.
What is then the impedance $Z_C(\omega)$ of a capacitor $C$?


\vspace{0.4cm}
\noindent
$b)$ A voltage source $\ems(t) = V_0 \, \cos\omega t$ is connected to a
capacitor $C$ and a resistor $R$ connected in parallel:

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=2.8in \epsfbox{ov12_fig4.eps}}
\vspace{0.5cm}
%%END FIGURE

\vspace{0.3cm}
\noindent
Use Kirchhoff's rules to find the currents $I(t)$, $I_C(t)$ and $I_R(t)$.
Show that the total current delivered by the voltage source may be written as
$$I(t) = I_0 \, \cos(\omega t - \alpha)$$
with amplitude
$$I_0 = \frac{V_0}{R} \, \sqrt{1 + (\omega RC)^2}$$
and phase constant
$$\alpha = -\arctan(\omega RC)$$
Hint: Use $\cos(a \pm b) = \cos a \, \cos b \mp \sin a \, \sin b$.

\vspace{0.2cm}
\noindent
Assume the voltage source has amplitude $V_0 = 1.0$ V and frequency
$f = 1.0$ MHz, and also assume $R = 10 \, \Omega$ and $C = 16$ nF.
Determine numerical values for $I_0$ and $\alpha$. Sketch $\ems(t)$, $I(t)$,
$I_R(t)$ and $I_C(t)$ for $t$ between 0 and $T = 1/f$.
(Answers: $I_0 = 0.14$ A, $\alpha = -45^\circ$)

\vspace{0.2cm}
\noindent
With the given numerical values of $V_0$, $R$ and $C$, sketch the functions
$I_0(\omega)$, $\alpha(\omega)$, and the impedance of the circuit, $Z(\omega) = V_0/I_0(\omega)$
for angular frequencies between $10^5$ and $10^9$ s$^{-1}$.
Hint: Use a {\em logarithmic} scale along the $\omega$ axis, i.e., sketch $I_0$
and $\alpha$ for angular frequencies so that $\log\omega$ varies between the values
5 and 9.



\vspace{1.2cm}
\noindent
{\em Exercise 5}

\vspace{0.2cm}
\noindent
The figure shows two spherical conductors with radius $a$ (the inner sphere)
and $b$ (the outer sphere), respectively. The region between the conducting
spheres is filled with a material with resistivity $\rho$.

\noindent
(Note: Here, the symbol $\rho$ denotes resistivity, or inverse
conductivity, since $\rho = 1/\sigma$. So here, $\rho$ does not
mean charge pr unit volumd...!)

\noindent
A thin, isolated conducting wire passes through a little hole in the
outer spherical conductor and in to the inner sphere. A stationary
(i.e., time independent) electric
current runs "through the system" as shown in the figure. Then, the potential
difference between the inner and outer sphere is $\Delta V = V_a - V_b$,
with the largest value of the potential on the inner sphere.
Assume that the connecting wires has a negligible resistance in comparison
with the material between the two spheres. Show that the resistance of
this system is
$R = \rho (a^{-1} - b^{-1})/4\pi$.
You may do this in one of two ways (or both, if you like!):

\begin{enumerate}
\item Start with the resistance of a thin spherical shell with radius $r$
and thickness $dr$, which is $dR = \rho dr/4\pi r^2$.
\item Start by assuming that the inner sphere has charge $Q$, and determine
the two quantities $\Delta V$ and the current $I = \int \bold{j} \cdot d\bold{A} =
\rho^{-1} \int \bold{E} \cdot d\bold{A}$. (So you may use Gauss' law...!)
\end{enumerate}

%%FIGURE
\vspace{0.5cm}
\centerline{\epsfxsize=3.0in \epsfbox{ov10_fig2.eps}}
\vspace{0.5cm}
%%END FIGURE





\end{document}
</pre></body></html>