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Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003 Elektrisitet og magnetisme

\noindent
V{\aa}r 2005

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Solution to {\o}ving 11
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\noindent
Guidance March 31 and April 1


\vspace{0.5cm}
\noindent
{\em Exercise 1}

\vspace{0.2cm}
\noindent
$a)$ At first, we should try to realize that what we have here is the following
circuit: [a parallel connection of
$R_1$, $R_2$ and $R_3$] coupled in series with [a parallel connection of $R_4$ and $R_0 = 0$]
in series with [$R_5$]. In other words, the resistance $R_4$ is "cut short", so that no
current passes through $R_4$.
(Alternatively: We have the same value for the potential on each side of $R_4$. Then, no
current can pass through it.)
Thus, the total resistance is
$$R = \left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right)^{-1} + R_5$$

\vspace{0.2cm}
\noindent
$b)$ It should be clear that the total current $I$ in the circuit must be the same
as the current $I_5$ passing through $R_5$. Further, it should also be clear that
$I$ must distribute itself on the three currents passing through
$R_1$, $R_2$ and $R_3$: $I = I_1 + I_2 + I_3$. In $a)$ above, we have already
concluded that no current passes through $R_4$: $I_4 = 0$.

\noindent
The voltage drop across the three upper resistances is the same:
$$V' = R_1 I_1 = R_2 I_2 = R_3 I_3$$
The voltage drop across $R_5$ is
$$V'' = R_5 I_5 = R_5 I = R_5 \frac{\cal E}{R}$$
These two together must equal the value of the voltage source:
$${\cal E} = V' + V''$$
Thus
$$V' = {\cal E} - V'' = {\cal E} - R_5 \frac{\cal E}{R} = {\cal E} \left(1 - \frac{R_5}{R} \right)$$
And finally,
$$I_1 = \frac{V'}{R_1}$$
$$I_2 = \frac{V'}{R_2}$$
$$I_3 = \frac{V'}{R_3}$$

\vspace{0.2cm}
\noindent
$c)$ With the given numerical values, we have
$$R = \left(1 + \frac{1}{2} + \frac{1}{3} \right)^{-1} + 5 = \frac{61}{11} \; \Omega$$
Thus,
$$V' = 9 \cdot \left(1 - \frac{5}{61/11} \right) = \frac{54}{61} \; {\rm V}$$
and
$$V'' = 9 - \frac{54}{61} = \frac{495}{61} \; {\rm V}$$
The various currents are
$$I_1 = \frac{54}{61\cdot1} = \frac{54}{61} \; {\rm A} \simeq 0.885 \; {\rm A}$$
$$I_2 = \frac{54}{61\cdot2} = \frac{27}{61} \; {\rm A} \simeq 0.443 \; {\rm A}$$
$$I_3 = \frac{54}{61\cdot3} = \frac{18}{61} \; {\rm A} \simeq 0.295 \; {\rm A}$$
$$I_5 = I = \frac{9\cdot11}{61} = \frac{99}{61} \; {\rm A} \simeq 1.623 \; {\rm A}$$


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\noindent
{\em Exercise 2}

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\noindent
$a)$ Bulb 1 will be brightest in circuit B and weakest in circuit A. In circuit A, the voltage drop
across bulb 1 is only 1/3 of the emf of the applied voltage source. In B, the voltage drop across
all the bulbs is equal to the emf of the applied voltage source. In C, the parallel connection
of 2 and 3 constitutes a resistance $(1/R + 1/R)^{-1} = R/2$, if the resistance in one bulb is $R$.
Then, the voltage drop over bulb 1 becomes 2/3 of the emf of the applied voltage source, and the
light intensity somewhere between that in A and B.

\vspace{0.2cm}
\noindent
$b)$ In A, we end up with an open ("broken") circuit, and therefore zero current, i.e., bulbs 1
(and 2) go out. In B, bulb 3 does not at all influence the voltage across bulb 1, so the light
intensity is unchanged. In C, we end up with two resistances $R$ coupled in series, so the
voltage drop across bulb 1 must be half of the emf of the applied voltage source. I.e., smaller
than what it was when bulb 3 was in place. Thus, the light intensity becomes smaller in circuit C.

\vspace{0.5cm}
\noindent
{\em Exercise 3}

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\noindent
Let us put time $t=0$ when the student cuts the circuit in A. Before this, we have
a stationary situation, with voltage drop
$$V_0 = V_R = V_C$$
over both the resistance $R$ and the capacitance $C$. At $t=0$, the circuit is broken
("opened") in A, and we are left with the loop with $R$ and $C$ (and no voltage source
anymore). As derived in the lectures, we now get a discharge of the capacitor, with
a time dependence
$$Q(t) = Q_0 e^{-t/RC}$$
where
$$Q_0 = Q(0) = C V_0$$
After a certain time $t_1$, the circuit is broken at B, which means that the discharge stops.
Then we have a charge
$$Q(t_1) = Q_0 e^{-t_1/RC}$$
on the capacitor. This corresponds to a voltage drop
$$V(t_1) = \frac{Q(t_1)}{C} = \frac{Q_0}{C} e^{-t_1/RC}$$
across the capacitor. During this time, the student has moved a distance $d$, so
the velocity must be
$$v = d/t_1$$
We solve for $t_1$ in the expression for $V(t_1)$:
$$t_1 = RC \ln \frac{V_0}{V(t_1)}$$
where we have used $Q_0 = V_0 C$ and $\ln x = -\ln(1/x)$. With numerical values inserted:
$$t_1 = 150 \cdot 10^{-3} \cdot \ln \frac{9.00}{3.58} = 0.138 \; {\rm s}$$
The velocity becomes
$$v = d/t_1 = 1.00/0.138 = 7.23 \; {\rm m/s} \, \simeq 26 \, {\rm km/h}$$



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