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

\noindent
TFY4155/FY1003: Elektrisitet og magnetisme

\noindent
Spring 2005

\end{bf}

\begin{center}
\begin{large}
Summary, week 13 (March 30)
\end{large}
\end{center}

\vspace{0.5cm}
\noindent
{\bf DC circuits} 

\noindent
[FGT 27; YF 26; TM 25; AF 24.7; LHL 22]

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\vspace{0.2cm}
\noindent
Voltage source (e.g., chemical battery, solar cell etc.):

\noindent
"Delivers {\em electromotive force} (emf), i.e., makes sure that 
we always have a constant potential difference $\ems$ between the
two "poles" + and -.

\vspace{0.2cm}
\noindent
Voltage around the circuit in the figure above ($\Delta V$ is change in
electric potential):

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\vspace{0.2cm}
\noindent
$a\rightarrow b: \Delta V \simeq 0$ (metal wire, good conductor, $R_m \simeq 0$, $\bold{E} \simeq 0$)

\noindent
$b\rightarrow c: \Delta V = -RI$ (resistor, bad conductor, $R \gg R_m$, potential energy is lost
as heat because of collisions, $\bold{E} \neq 0$)

\noindent
$c\rightarrow d: \Delta V \simeq 0$ (as $a\rightarrow b$)

\noindent
$d\rightarrow a: \Delta V = \ems = RI$ ({\em receives} charge carriers with low potential
energy, {\em delivers} charge carriers with high potential energy.)

\newpage
\noindent
A {\em real} source always has a (usually small) inner resistance $R_i$:

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\noindent
When a real voltage source is connected to an electric circuit, the inner resistance $R_i$ comes
in addition to the circuit resistance $R$. We then have a power loss both in the source
($P_i = R_i I^2$) and in the rest of the circuit ($P_R = R I^2$).

\vspace{0.2cm}
\noindent
An {\em ideal} source has $R_i=0$.

\vspace{1.0cm}
\noindent
{\bf Kirchhoff's rules} 

\noindent
[FGT 27.2, 27.3; YF 26.2; TM 25.5; AF 24.8; LHL 22.3]

\vspace{0.3cm}
\noindent
Calculations on electric circuits are done with the help of Kirchhoff's rules.

\vspace{0.3cm}
\noindent
Rule 1 (Current rule): Because of {\em charge conservation},
$$\sum_j I_j = 0$$
in all junctions in a circuit.

\noindent
If not, we would obtain charge accumulation in the junction.

\noindent
Sign convention: {\em Positive} $I$ when it goes {\em out of} the junction.

\vspace{0.5cm}
\noindent
Rule 2 (Voltage rule): Because of {\em energy conservation},
$$\sum ({\rm voltage}\,\,\,{\rm changes}) = 0$$
for all closed loops in a circuit.

\noindent
If not, we would not have a uniquely determined potential energy for the charge
carriers at a given place in the circuit.

\noindent
Sign convention: {\em Positive} contribution means {\em voltage increase}.

%%FIGURE
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\newpage
\noindent
Kirchhoff's rules yields a sufficient number of independent equations to determine
the unknown quantities, e.g., the currents $I_j$ in the various branches of
the circuit.

\vspace{1.0cm}
\noindent
{\bf $RC$ circuits} 

\noindent
[FGT 27.5; YF 26.4; TM 25.6; AF Note 25.1; LHL 22.4]

\vspace{0.3cm}
\noindent
The space between the two conductors in a capacitor is filled with an {\em insulator}, and
through an (ideal) insulator runs {\em zero} electric current.

%%FIGURE
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\noindent
However, we may have a {\em time dependent} current $I(t)$ into and out of the
conductors of the capacitor ({\em the plates}, if we have a parallel plate
capacitor).

\noindent
Hence, we obtain a time dependent charge $Q(t)$ on the capacitor plates.

\vspace{0.3cm}
\noindent
Are we allowed to use Kirchhoff's rules to analyze circuits with time dependent
$I(t), V(t), Q(t)$?

\noindent
Yes: For "slowly" varying currents, where slow means in comparison to how fast a change
at one place in the circuit is "detected" in the rest of the circuit. Since electromagnetic
signals (waves) propagate with the speed of light, $c$, this is in practice usually no problem.

\vspace{0.5cm}
\noindent
Example 1: Charging of capacitor in $RC$ circuit.

%%FIGURE
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\noindent
The voltage source $\ems$ is connected at time $t=0$. At that instant, we have
zero charge on the capacitor, $Q(0) = 0$.

\noindent
Kirchhoff's voltage rule $\Rightarrow$
$$\ems - V_C - V_R = 0$$
Voltage drop across $C$:
$$V_C = Q/C$$
Voltage drop across $R$:
$$V_R = RI = R \frac{dQ}{dt}$$
Yields 1. order differential equation for the charge $Q$:
$$R \frac{dQ}{dt} + \frac{1}{C} Q = \ems$$
which has solution
$$Q(t) = \ems C \left( 1 - e^{-t/RC} \right)$$
Here, we have used the {\em initial condition} $Q(0)=0$.

\noindent
The current is
$$I(t) = \frac{dQ}{dt} = \frac{\ems}{R}e^{-t/RC}$$

\vspace{0.3cm}
\noindent
{\em Time constant} for the charging process: $\tau = RC$

\noindent
The value of $\tau$ provides a {\em time scale} for how long time it takes
to charge the capacitor to its maximum charge
$$Q(t\rightarrow\infty) = \ems C$$

\vspace{0.5cm}
\noindent
Example 2: Discharge of capacitor in $RC$ circuit.

%%FIGURE
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%%END FIGURE

\noindent
We assume that the capacitor has been charged with a voltage source $\ems$, and
that it is fully charged. Hence, the initial condition now is $Q(t=0) = \ems C$.

\noindent
Kirchhoff's voltage rule $\Rightarrow$
$$-V_R - V_C = 0$$
Gives, as above, 1. order differential equation for the charge $Q$:
$$-R \frac{dQ}{dt} - \frac{1}{C} Q = 0$$
which has solution
$$Q(t) = \ems C e^{-t/RC}$$
Here, we have used $Q(0)=\ems C$.

\noindent
The current is
$$I(t) = \frac{dQ}{dt} = -\frac{\ems}{R}e^{-t/RC}$$
We see from the figure that we chose the "wrong" direction for the current $I$:
Positive charge will flow from the positively charged plate, and therefore give
a positive current counterclockwise. However, this is taken into account since
the calculated current came out with a negative sign.

\vspace{0.2cm}
\noindent
{\em Note} that is we had chosen the opposite direction for $I$ in the figure,
we could no longer write $I = dQ/dt$, but rather $I = -dQ/dt$, since a positive
current in that case would correspond to a {\em reduction} in the charge on
the capacitor. In other words: $dQ/dt$ will then be negative for positive $I$,
and we must write $I = -dQ/dt$ to have the same sign on both sides of the equation.

\vspace{0.2cm}
\noindent
I recommend to choose the direction of $I$ into the plate with charge $Q$, as done above.
Then we may stick to the relation $I = dQ/dt$, i.e., positive $I$ corresponds
to a positive change in the charge $Q$. The initial condition for each particular
problem will make sure that the sign of $I$ is correct in the end!

\vspace{2.0cm}
\noindent
Next week:

\vspace{0.2cm}
\noindent
Magnetic interactions! We start (in the "ekstratimen" on Tuesday) by showing, or
at least indicating, that the magnetic field and magnetic forces are a direct
consequence of electrostatics (i.e., that charges at rest influence each other
with Coulomb forces) {\em and} Einstein's special theory of relativity. In other
words, we may conclude that {\em magnetism is a relativistic effect}.

\vspace{0.3cm}
\noindent
Next, we will consider the movement of a charged particle in a magnetic field,
and we will also introduce the {\em Biot--Savart law}, which gives us the recipe
for calculating the magnetic field $\bold{B}$ when we know the electric current(s)
in our system. The Biot--Savart law is to magnetostatics what Coulomb's law is to
electrostatics, how to calculate the electric field $\bold{E}$ when we know the
electric charges in our system. And, if we know the fields $\bold{E}$ and $\bold{B}$,
we may proceed to evalute the force on a given charge $q$ with velocity $\bold{v}$:
$$\bold{F} = q \bold{E} + q \bold{v} \times \bold{B}$$
which is the famous {\em Lorentz force}.

\end{document}
