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\begin{document}
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\begin{bf}
\noindent
Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003 Elektrisitet og magnetisme

\noindent
V{\aa}r 2007

\end{bf}

\begin{center}
\begin{large}
L{\o}sningsforslag til {\o}ving 12
\end{large}
\end{center}

\noindent
Veiledning uke 13


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

\vspace{0.2cm}
\noindent
Kirchhoff's voltage rule (K2) yields
$$\ems = \frac{Q}{C} = R I_R$$
whereas Kirchhoff's current rule (K1) yields
$$I = I_C + I_R$$
Furthermore, we have
$$I_C = \frac{dQ}{dt}$$
Hence:
\begin{eqnarray}
I_R(t) &=& \frac{V_0}{R} \, \cos\omega t \nonumber \\
Q(t) &=& V_0 C \, \cos\omega t \nonumber \\
I_C(t) &=& -\omega C V_0 \, \sin\omega t = \omega C V_0 \, \cos(\omega t +\pi/2) \nonumber
\end{eqnarray}
Total current delivered by the voltage source is then
$$I(t) = \frac{V_0}{R} \, \cos\omega t - \omega C V_0 \, \sin\omega t$$
We want $I(t)$ on the form
$$I(t) = I_0 \, \cos(\omega t - \alpha)$$
with amplitude $I_0 = V_0/Z$, where $Z$ is the impedance of $R$ and $C$ coupled in
parallel, whereas $\alpha$ becomes the phase angle, i.e., the phase difference between
$\ems(t)$ and $I(t)$. We have
$$\cos(\omega t - \alpha) = \cos\omega t \, \cos\alpha + \sin\omega t \, \sin\alpha$$
Hence, by direct comparison:
\begin{eqnarray}
\frac{\cos\alpha}{Z} &=& \frac{1}{R} \nonumber \\
\frac{\sin\alpha}{Z} &=& -\omega C \nonumber
\end{eqnarray}
These two equations, with two unknowns $Z$ and $\alpha$, are easily solved,
and we find
\begin{eqnarray}
Z &=& \frac{R}{\sqrt{1 + (\omega R C)^2}} \nonumber \\
I_0 &=& \frac{V_0}{R} \, \sqrt{1 + (\omega R C)^2} \nonumber \\
\alpha &=& -\arctan(\omega RC) \nonumber
\end{eqnarray}
In the limit $\omega \rightarrow 0$ we should recover well known results from the DC examples
given in the lectures, and so we do:
$Z \rightarrow R$ and $\alpha \rightarrow 0$
so that $I_0 \rightarrow V_0/R$. All the current runs through the resistor $R$, while
the capacitor $C$ now represents an open circuit, where no direct current can run.

\vspace{0.2cm}
\noindent
With given numerical values, we have
$$\omega RC = 2\pi \cdot 10^6 \cdot 10 \cdot 16 \cdot 10^{-9} = 1.0$$
so that
\begin{eqnarray}
I_0 &=& \frac{1.0}{10} \cdot \sqrt{2} = 0.14 \; {\rm A} \nonumber \\
\alpha &=& - \arctan 1.0 = -45^\circ \nonumber
\end{eqnarray}

\vspace{0.2cm}
\noindent
Sketch of $\ems(t)$, $I(t)$, $I_R(t)$ and $I_C(t)$:

%%FIGURE
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\centerline{\epsfxsize=3.0in \epsfbox{los12_skisse_1.eps}}
\vspace{1.0cm}
%%END FIGURE


\vspace{0.5cm}
\noindent
Sketch of $\alpha$, $I_0$ and $Z$ (with $\alpha$ in radians and
$\omega RC$ between 0.016 and 160 along the horizontal axis):

%%FIGURE
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\centerline{\epsfxsize=3.0in \epsfbox{los12_skisse_2.eps}}
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%%END FIGURE





\end{document}
