--- /dev/null
+\begin{problem*}{24.53}
+Electronic flash units for cameras contain a capacitor for storing the
+energy used to produce the flash. In one such unit, the flash last
+for $\frac{1}{675}\U{s}$ with an average light power output of
+$2.70\E{5}\U{W}$. \Part{a} If the conversion of electrical energy to
+light is 95\% efficient (the rest of the energy goes to thermal
+energy), how much energy must be stored in the capacitor for one
+flash? \Part{b} The capacitor has a potential difference between its
+plates of $125\U{V}$ when the stored energy equals the value
+calculated in \Part{a}. What is the capacitance?
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+The total light energy released is
+\begin{equation}
+ U_\text{light} = P_\text{light}\cdot\Delta t \;.
+\end{equation}
+That is 95\% of the total energy released from the capacitor, so
+\begin{align}
+ 0.95 U_C &= U_\text{light} \\
+ U_C = \frac{U_\text{light}}{0.95}
+ = \frac{P_\text{light}\cdot\Delta t}{0.95}
+ = \frac{2.70\E{5}\cdot}{0.95\cdot675}\U{J}
+ = \ans{421\U{J}} \;.
+\end{align}
+
+\Part{b}
+The capacitance is given by
+\begin{align}
+ U_C &= \frac{1}{2} C V^2 \\
+ C &= \frac{2U_C}{V^2} = \frac{2\cdot421\U{J}}{(125\U{V})^2}
+ = \ans{53.9\U{mF}} \;. % typo in book solution, reported
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem*}{24.56}
+In Fig.~24.9a, let $C_1=9.0\U{$\mu$F}$, $C_2=4.0\U{$\mu$F}$, and
+$V_{ab}=28\U{V}$. Suppose the charged capacitors are disconnected
+from the source and from each other, and then reconnected to each
+other with plates of \emph{opposite} sign together. By how much does
+the energy of the system decrease?
+\begin{center}
+\begin{verbatim}
+ a-----+---------+
+ +|+ +|+
+Vab C1 === Q1 C2 === Q2
+ -|- -|-
+ b-----+---------+
+\end{verbatim}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+The initial energy is
+\begin{equation}
+ U_i = \frac{1}{2}(C_1+C_2)V_{ab}^2 = 5.10\U{mJ} \;.
+\end{equation}
+
+To find the final energy, we need to find the final voltage or charge.
+The initial charges are
+\begin{align}
+ Q_1 &= C_1V_{ab} = 252\U{$\mu$C} \\
+ Q_2 &= C_2V_{ab} = 112\U{$\mu$C} \;.
+\end{align}
+Flipping $C_2$ around gives a total charge in the top wire of
+\begin{equation}
+ Q_f = Q_1 - Q_2 = 140\U{$\mu$C} \;.
+\end{equation}
+The charge in the bottom wire is $-Q_1 + Q_2 = -Q_f$, as you'd expect
+since capacitors should have equal and opposite charges on either
+plate.
+
+This yields a final energy of
+\begin{equation}
+ U_f = \frac{Q_f^2}{2(C_1+C_2)}
+ = \frac{(140\U{$\mu$C})^2}{2\cdot14.0\U{$\mu$F}}
+ = 700\U{$\mu$J} \;,
+\end{equation}
+for a net decrease of
+\begin{equation}
+ -\Delta U = U_i - U_f = \ans{4.40\U{mJ}} \;.
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{24.57}
+For the capacitor network show in Fig.~24.31, the potential difference
+across $ab$ is $12.0\U{V}$. Find \Part{a} the total energy stored in
+this network and \Part{b} the energy stored in the $4.80\U{$\mu$F}$
+capacitor.
+\begin{center}
+\begin{verbatim}
+ 6.20 11.8
+ +--||--||--+
+ 8.60 4.80 | |
+a--||--||--+ +--b
+ | |
+ +----||----+
+ 3.50
+\end{verbatim}
+(all capacitances are given in $\mu$F)
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+Bundling down to a single equivalent capacitance, we see that the
+$8.60$ and $4.80\U{$\mu$F}$ capacitors are in series, as are the
+$6.20$ and $11.8\U{$\mu$F}$ capacitory. The circuit is thus
+equivalent to
+\begin{center}
+\begin{verbatim}
+ 4.06
+ +--||--+
+ 3.08 | |
+a--||--+ +--b
+ | |
+ +--||--+
+ 3.50
+\end{verbatim}
+\end{center}
+where
+\begin{align}
+ C_L &= \p({\frac{1}{8.60}+\frac{1}{4.80}})^{-1}\U{$\mu$F} = 3.08\U{$\mu$F} \\
+ C_T &= \p({\frac{1}{6.20}+\frac{1}{11.8}})^{-1}\U{$\mu$F} = 4.06\U{$\mu$F}\;.
+\end{align}
+
+Then we see that the $4.06$ and $3.50$ capacitors are in parallel, so
+the circuit is equivalent to
+\begin{center}
+\begin{verbatim}
+ 3.08 7.56
+a--||----||--b
+\end{verbatim}
+\end{center}
+where
+\begin{equation}
+ C_R = (4.06+3.50)\U{$\mu$F} = 7.56\U{$\mu$F} \;.
+\end{equation}
+
+These last two capacitors are themselves in series, so we can move
+down to a single equivalent capacitance
+\begin{equation}
+ C = \p({\frac{1}{3.08}+\frac{1}{7.56}})^{-1}\U{$\mu$F} = 2.19\U{$\mu$F} \;.
+\end{equation}
+
+We use the capacitor energy equation on this effective capacitance to
+find the total energy
+\begin{equation}
+ U = \frac{1}{2}CV_{ab}^2
+ = \frac{1}{2}\cdot 10.6\U{$\mu$F} \cdot (12.0\U{V})^2
+ = \ans{158\U{$\mu$J}} \;.
+\end{equation}
+
+\Part{b}
+Stepping back out into our more detailed picture, we remember that
+capacitors in series share the same charge (because the wire between
+them remains neutral as a whole), so both the $C_L=3.08$ and
+$C_R=7.65\U{$\mu$F}$ capacitors have the same charge $Q$ as the
+single equivalent capacitance $C_T$. Similarly, the two series
+capacitors that make up $C_L$ \emph{also} have the same charge $Q$ as
+$C_L$ (which is the same $Q$ as $C_T$). Therefore, the charge on the
+$4.80\U{$\mu$F}$ capacitor is \emph{the same} as the charge on $C_T$.
+
+Because the charge is the same on both $C_T$ and $C_{4.8}$, we'd like
+to find a formula for energy that involves charge. Plugging the
+capacitor formula $Q=CV$ into our usual energy formula yields
+\begin{equation}
+ U = \frac{1}{2}CV^2 = \frac{1}{2}C\p({\frac{Q}{C}})^2 = \frac{Q^2}{2C} \;.
+\end{equation}
+Isolating the constant $Q$ and applying the formula to both $C_T$ and
+$C_{4.8}$ yields
+\begin{align}
+ Q^2 &= 2C_TU_T = 2C_{4.8}U_{4.8} \\
+ U_{4.8} &= \frac{C_T}{C_{4.8}}U_T
+ = \frac{2.19\U{$\mu$F}}{4.80\U{$\mu$F}} 158\U{$\mu$J}
+ = \frac{2.19}{4.80} 158\U{$\mu$J}
+ = \ans{71.9\U{$\mu$J}}
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem*}{25.17}
+In household wiring, copper wire $2.05\U{mm}$ in diameter is often
+used. Find the resistance of a $24.0\U{m}$ length of this wire.
+\end{problem*}
+
+\begin{solution}
+We find the resistivity for copper $\rho=1.72\E{-8}\U{\Ohm$\cdot$m}$
+in Table 25.1 (page 851). The resistance is then given by
+Equation~25.10 (on page 853):
+\begin{equation}
+ R = \frac{\rho L}{A} = \frac{\rho L}{\pi r^2}
+ = \frac{(1.72\E{-8}\U{\Ohm$\cdot$m})\cdot(24.0\U{m})}{\pi(2.05\U{mm}/2)^2}
+ = \ans{125\U{m\Ohm}} \;.
+\end{equation}
+\end{solution}
--- /dev/null
+\begin{problem*}{25.24}
+You apply a potential difference of $4.50\U{V}$ between the ends of a
+wire that is $2.50\U{m}$ in length and $0.654\U{mm}$ in radius. The
+resulting current through the wire is $17.6\U{A}$. What is the
+resistivity of the wire?
+\end{problem*}
+
+\begin{solution}
+Juggling the resistivity/resistance relationship (Equation~25.10)
+\begin{equation}
+ R = \frac{\rho L}{A}
+\end{equation}
+and Ohm's law (Equation~25.11)
+\begin{equation}
+ V = IR \;,
+\end{equation}
+we get
+\begin{equation}
+ \rho = \frac{AR}{L} = \frac{AV}{LI} = \frac{\pi r^2 V}{LI}
+ = \frac{\pi(0.654\U{mm})^2\cdot(4.50\U{V})}{(2.50\U{m})\cdot(17.6\U{A})}
+ = \ans{1.37\E{-7}\U{\Ohm$\cdot$m}} \;.
+\end{equation}
+This puts the resistivity between that of tungsten and steel in Table~25.1.
+\end{solution}
--- /dev/null
+\begin{problem*}{25.36}
+The circuit shown in Fig.~25.37 contains two batteries, each with an
+emf and an internal resistance, and two resistors. Find \Part{a} the
+current in the circuit (magnitude \emph{and} direction); \Part{b} the
+terminal voltage $V_{ab}$ of the $16.0\U{V}$ battery.; \Part{c} the
+potential difference $V_{ac}$ of point $a$ with respect to point
+$c$. \Part{d} Using Fig.~25.21 as a model, graph the potential rises
+and drops in this circuit.
+\begin{center}
+\begin{verbatim}
+ 1.6 16.0
+ +---a---/\/\/---|i---b---+
+ | |
+ 5.0 Z Z 9.0
+ Z Z
+ | 1.4 8.0 |
+ +-------/\/\/---|i---c---+
+\end{verbatim}
+Resistances (\verb+ZZ+ and \verb+/\/\/+) are in \Ohm. Battery
+(\verb+|i+) voltages are in volts.
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+We're going to use Kirchoff's loop rule: the sum of voltage changes
+around a loop is $0$.
+\begin{equation}
+ \sum_\text{loop} \Delta V = 0
+\end{equation}
+The book doesn't name this rule until Chapter~26 (see Equation~26.6 on
+page~887), but we've been using since the capacitor chapter (see
+page~821 ``Capacitors in Parallel''), and this resistance chapter (see
+page~862, ``Potential changes around a circuit''). I think it's best
+to just get the name out in the open :p.
+
+Anyhow lets travel the loop along the path $a\rightarrow b\rightarrow
+c\rightarrow d\rightarrow a$, adding up voltage changes as we go. By
+Kirchoff's loop rule, the total voltage change must be $0$. We'll
+need to pick a direction for the current $I$ to be flowing as well, so
+we know whether the potential increases or decreases going across
+resistors. Let's pick the same direction (clockwise) as the path
+we're taking. The total voltage change is then
+\begin{equation}
+ 0 = -1.6\U{\Ohm}I-16.0\U{V}-9.0\U{\Ohm}I+8.0\U{V}-1.4\U{\Ohm}I-5.0\U{\Ohm}I
+\end{equation}
+where we used Ohm's law $V=IR$ to find the voltage change across each
+resistor, and gave it a negative sign (drop) because we were traveling
+\emph{in the same direction} as $I$. Now it's just algebra:
+\begin{align}
+ 0 &= -(1.6+9.0+1.4+5.0)\U{\Ohm}I + (-16.0+8.0)\U{V} \\
+ 17.0\U{\Ohm}I &= -8.0\U{V} \\
+ I &= \ans{-470\U{mA}} \;.
+\end{align}
+The $-$ sign shows us that the current actually flows in the
+\emph{opposite} direction to the one we picked at the beginning, so
+there is $\ans{470\U{mA}}$ of current flowing \ans{counterclockwise}.
+
+\Part{b}
+This just an excerpt from \Part{a}. Starting from $b$ and moving
+counterclockwise,
+\begin{equation}
+ V_{ab} = 16.0\U{V} - 1.6\U{\Ohm}\cdot470\U{mA} = \ans{15.2\U{V}}
+\end{equation}
+
+\Part{c}
+Another excerpt from \Part{a}. Starting from $c$ and moving
+clockwise,
+\begin{equation}
+ V_{ac} = 8.0\U{V} + 1.4\U{\Ohm}\cdot470\U{mA} + 5.0\U{\Ohm}\cdot470\U{mA}
+ = 11\U{V}
+\end{equation}
+
+\Part{d}
+Setting point $V_b=0\U{V}$, we have
+\begin{align}
+ V_b &= 0\U{V} \\
+ V_{ab/2} &= 16\U{V} \\
+ V_{a} &= 15.2\U{V} \\
+ V_{ac/3} &= 12.9\U{V} \\
+ V_{2ac/3} &= 12.2\U{V} \\
+ V_{c} &= 4.2\U{V} \\
+\end{align}
+\end{solution}
--- /dev/null
+\begin{problem*}{25.53}
+In the circuit in Fig.~25.39, find \Part{a} the rate of conversion of
+internal (chemical) energy to electrical energy within the
+battery; \Part{b} the rate of dissipation of electrical energy in the
+battery; \Part{c} the rate of dissipation of the electrical enegy in
+the external resistor.
+\begin{center}
+\begin{verbatim}
+ 1.0 12.0
+ +---a---/\/\/---|i---d---+
+ | |
+ | |
+ +---b-------/\/\/----c---+
+ 5.0
+\end{verbatim}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+First we need to find the current. Going counterclockwise around the
+loop from $d$ and using only SI units
+\begin{align}
+ 0 &= 12.0 - 1.0I - 5.0I \\
+ I &= \frac{12.0}{6.0} = 2.0\U{A}
+\end{align}
+
+Power through any element is given by $P=IV$ (Equation~25.18,
+page~864), and all the electrical energy from the battery is being
+created by chemical energy creating the internal $12\U{V}$ emf, so
+\begin{equation}
+ P_\text{chem} = 2.0\U{A} \cdot 12.0 \U{V} = \ans{24.0\U{W}} \;.
+\end{equation}
+
+\Part{b}
+The battery dissipates some of this energy into heat through it's
+internal resisitance
+\begin{equation}
+ P_\text{i} = IV = I(IR) = I^2R = (2.0\U{A})^2\cdot1.0\U{\Ohm}
+ = \ans{4.0\U{W}} \;.
+\end{equation}
+
+\Part{c}
+Similarly to \Part{b}, the external resistor dissipates
+\begin{equation}
+ P_\text{e} = I^2R = (2.0\U{A})^2\cdot5.0\U{\Ohm}
+ = \ans{20.0\U{W}} \;.
+\end{equation}
+
+Note that all $24.0\U{W}$ of electrical power from the reaction in the
+battery is being dissipated into heat by the two resistors
+($4+20=24\U{W}$).
+\end{solution}
projects / course.git / commitdiff