From 862315a1430a21238f3f3015519ba77f1394dfc6 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Tue, 25 Aug 2009 02:37:28 -0400 Subject: [PATCH] Added topic tags to some Giancoli-6 problems + minor typos --- latex/problems/Giancoli_6/problem17.40.tex | 4 +- latex/problems/Giancoli_6/problem19.02.tex | 13 ++-- latex/problems/Giancoli_6/problem19.07.tex | 2 +- latex/problems/Giancoli_6/problem19.15.tex | 2 +- latex/problems/Giancoli_6/problem19.24.tex | 2 +- latex/problems/Giancoli_6/problem19.31.tex | 82 +++++++++++---------- latex/problems/Giancoli_6/problem19.58.tex | 70 +++++++++--------- latex/problems/Giancoli_6/question19.04.tex | 2 +- latex/problems/Giancoli_6/question19.07.tex | 2 +- latex/problems/Giancoli_6/question19.13.tex | 2 +- 10 files changed, 94 insertions(+), 87 deletions(-) diff --git a/latex/problems/Giancoli_6/problem17.40.tex b/latex/problems/Giancoli_6/problem17.40.tex index e18f780..b5f14eb 100644 --- a/latex/problems/Giancoli_6/problem17.40.tex +++ b/latex/problems/Giancoli_6/problem17.40.tex @@ -1,10 +1,10 @@ -\begin{problem*}{40} +\begin{problem*}{17.40} % capacitor circuits A $C_1 = 7.7\U{$\mu$F}$ capacitor is charged by a $V = 125\U{V}$ battery (Fig. 17-29a) and then is disconnected from the battery. When this capacitor ($C_1$) is then connected (Fig. 17-29b) to a second (initially uncharged) capacitor, $C_2$, the final voltage on each capacitor is $V_2 = 15\U{V}$. What is the value of $C_2$? -[\emph{Hint}: charge is conserved.] +[\emph{Hint:} charge is conserved.] \end{problem*} \empaddtoprelude{ diff --git a/latex/problems/Giancoli_6/problem19.02.tex b/latex/problems/Giancoli_6/problem19.02.tex index 05344df..a11b21a 100644 --- a/latex/problems/Giancoli_6/problem19.02.tex +++ b/latex/problems/Giancoli_6/problem19.02.tex @@ -1,4 +1,4 @@ -\begin{problem*}{2} +\begin{problem*}{19.2} % resistor networks Four $1.5\U{V}$ cells are connected in series to a $12\U{\Ohm}$ lightbulb. If the resulting current is $0.45\U{A}$, what is the internal resistance of each cell, assuming they are identical and @@ -6,11 +6,12 @@ neglecting the wires. \end{problem*} \begin{solution} -This is simply an application of the procedure outlined in Question 13. -The external resistance is the lightbulb $R_{ext}=12\U{\Ohm}$. -The total internal resistance is the sum of all the individual cell resistances $r_{int}=6r$. -The total voltage is the sum of all the individual cell voltages $V = 6\cdot 1.5\U{V} = 9\U{V}$. -Putting these together we have +This is simply an application of the procedure outlined in +Question~13. The external resistance is the lightbulb +$R_{ext}=12\U{\Ohm}$. The total internal resistance is the sum of all +the individual cell resistances $r_{int}=6r$. The total voltage is +the sum of all the individual cell voltages $V = 6\cdot 1.5\U{V} = +9\U{V}$. Putting these together we have \begin{align*} r_{int} = 6r &= \frac{V}{I} - R_{ext} = \frac{9\U{V}}{0.45\U{A}} - 12\U{\Ohm} \\ r &= \ans{1.3\U{\Ohm}} diff --git a/latex/problems/Giancoli_6/problem19.07.tex b/latex/problems/Giancoli_6/problem19.07.tex index f511132..189a1f2 100644 --- a/latex/problems/Giancoli_6/problem19.07.tex +++ b/latex/problems/Giancoli_6/problem19.07.tex @@ -1,4 +1,4 @@ -\begin{problem*}{7} +\begin{problem*}{19.7} % resistor networks A $650\U{\Ohm}$ and a $2200\U{\Ohm}$ resistor are connected in series with a $12\U{V}$ battery. What is the voltage across the $2200\U{\Ohm0}$ resistor? diff --git a/latex/problems/Giancoli_6/problem19.15.tex b/latex/problems/Giancoli_6/problem19.15.tex index 6dd4418..c636cbe 100644 --- a/latex/problems/Giancoli_6/problem19.15.tex +++ b/latex/problems/Giancoli_6/problem19.15.tex @@ -1,4 +1,4 @@ -\begin{problem*}{15} +\begin{problem*}{19.15} % resistor networks Eight $7.0\U{W}$ Christmas tree lights are connected in series to each other and to a $110\U{V}$ source. What is the resistance of each bulb. diff --git a/latex/problems/Giancoli_6/problem19.24.tex b/latex/problems/Giancoli_6/problem19.24.tex index 38fdf69..27083ee 100644 --- a/latex/problems/Giancoli_6/problem19.24.tex +++ b/latex/problems/Giancoli_6/problem19.24.tex @@ -1,4 +1,4 @@ -\begin{problem*}{24} +\begin{problem*}{19.24} % internal resistance Determine the terminal voltage of each battery in Fig.~19-44. \begin{center} \begin{asy} diff --git a/latex/problems/Giancoli_6/problem19.31.tex b/latex/problems/Giancoli_6/problem19.31.tex index 771032c..ff58077 100644 --- a/latex/problems/Giancoli_6/problem19.31.tex +++ b/latex/problems/Giancoli_6/problem19.31.tex @@ -1,24 +1,24 @@ -\begin{problem*}{31} - Calculate the currents in each resistor of Fig.~19-49. +\begin{problem*}{19.31} % resistor networks +Calculate the currents in each resistor of Fig.~19-49. \end{problem*} \begin{nosolution} \begin{center} \begin{asy} - import Circ; - real u = 3cm; - TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$"); - TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", ""); - TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$"); - pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y); - TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", ""); - TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$"); - TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", ""); - TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", ""); - wire(Ba.end, Raa.end, rlsq); - wire(Rab.beg, Jbot, nsq); - wire(Jbot, Rb.end, nsq); - wire(Jbot, Rcb.end, rlsq); +import Circ; +real u = 3cm; +TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$"); +TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", ""); +TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$"); +pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y); +TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", ""); +TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$"); +TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", ""); +TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", ""); +wire(Ba.end, Raa.end, rlsq); +wire(Rab.beg, Jbot, nsq); +wire(Jbot, Rb.end, nsq); +wire(Jbot, Rcb.end, rlsq); \end{asy} \end{center} \end{nosolution} @@ -26,24 +26,24 @@ \begin{solution} \begin{center} \begin{asy} - import Circ; - TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$"); - TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", ""); - TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$"); - pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y); - TwoTerminal Ic = current((Jbot+Rcb.end)/2, 0, "", "$I_3$"); - TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", ""); - TwoTerminal Ib = current(Rb.end, -90, "", "$I_2$"); - TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$"); - TwoTerminal Ia = current(Ba.end, 180, "$I_1$", ""); - TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", ""); - TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", ""); - wire(Ia.end, Raa.end, rlsq); - wire(Jbot, Ib.end, nsq); - wire(Jbot, Ic.beg, nsq); - wire(Ib.end, Rb.end, nsq); - wire(Ic.end, Rcb.end, rlsq); - dot("a", Jbot, S); +import Circ; +TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$"); +TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", ""); +TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$"); +pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y); +TwoTerminal Ic = current((Jbot+Rcb.end)/2, 0, "", "$I_3$"); +TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", ""); +TwoTerminal Ib = current(Rb.end, -90, "", "$I_2$"); +TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$"); +TwoTerminal Ia = current(Ba.end, 180, "$I_1$", ""); +TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", ""); +TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", ""); +wire(Ia.end, Raa.end, rlsq); +wire(Jbot, Ib.end, nsq); +wire(Jbot, Ic.beg, nsq); +wire(Ib.end, Rb.end, nsq); +wire(Ic.end, Rcb.end, rlsq); +dot("a", Jbot, S); \end{asy} \end{center} Label the resistors from left to right: $R_1 = 12\U{\Ohm}$, $R_2 = @@ -75,9 +75,12 @@ R_2$ and $R_{45} \equiv R_4 + R_5$ to save writing later. We can then plug those currents into the junction rule and solve for $I_2$ \begin{align*} \frac{V_1 + R_3 I_2}{R_{12}} + I_2 - \frac{V_2 - R_3 I_2}{R_{45}} &= 0 \\ - \frac{V_1}{R_{12}} + \frac{R_3}{R_{12}} I_2 + I_2 - \frac{V_2}{R_{45}} + \frac{R_3}{R_{45}}I_2 &= 0 \\ - \p({\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}})\cdot I_2 &= \frac{V_2}{R_{45}} - \frac{V_1}{R_{12}} \\ - I_2 &= \frac{\frac{V_2}{R_{45}} - \frac{V_1}{R_{12}}}{\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}} \\ + \frac{V_1}{R_{12}} + \frac{R_3}{R_{12}} I_2 + I_2 + - \frac{V_2}{R_{45}} + \frac{R_3}{R_{45}}I_2 &= 0 \\ + \p({\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}})\cdot I_2 + &= \frac{V_2}{R_{45}} - \frac{V_1}{R_{12}} \\ + I_2 &= \frac{\frac{V_2}{R_{45}} - \frac{V_1}{R_{12}}}{\frac{R_3}{R_{12}} + + 1 + \frac{R_3}{R_{45}}} \\ I_2 &= \ans{-28\U{mA}} \end{align*} Where the $-$ sign means the true current is in the opposite direction @@ -88,5 +91,8 @@ figure). We can now plug this current in to find $I_1$ and $I_3$. I_3 &= \frac{V_2 - R_3 I_2}{R_{45}} = \ans{264\U{mA}} \end{align*} -Double-checking our algebra, we see $I_1 + I_2 - I_3 = 292 - 27 - 264 = -1\U{mA} \approx 0$ where difference of $1\U{mA}$ is due to rounding errors from forcing our answers to milli-Volt precision. +Double-checking our algebra, we see + $I_1 + I_2 - I_3 = 292 - 27 - 264 = -1\U{mA} \approx 0$ +where difference of $1\U{mA}$ is due to rounding errors from forcing +our answers to milli-Volt precision. \end{solution} diff --git a/latex/problems/Giancoli_6/problem19.58.tex b/latex/problems/Giancoli_6/problem19.58.tex index be0fa0f..a3d53c5 100644 --- a/latex/problems/Giancoli_6/problem19.58.tex +++ b/latex/problems/Giancoli_6/problem19.58.tex @@ -1,9 +1,9 @@ -\begin{problem*}{58} - A $45\U{V}$ battery of negligable internal resistance is connected - to a $38\U{k\Ohm}$ and a $27\U{k\Ohm}$ resistor in series. What - reading will a voltmeter, of internal resistance $95\U{k\Ohm}$, - give when used to measure the voltage across each resistor? What is - the percent inaccuracy due to meter resistance for each case? +\begin{problem*}{19.58} % internal resistance +A $45\U{V}$ battery of negligable internal resistance is connected to +a $38\U{k\Ohm}$ and a $27\U{k\Ohm}$ resistor in series. What reading +will a voltmeter, of internal resistance $95\U{k\Ohm}$, give when used +to measure the voltage across each resistor? What is the percent +inaccuracy due to meter resistance for each case? \end{problem*} \begin{solution} @@ -11,18 +11,18 @@ Case 1: \\ The original situation looks like \begin{center} \begin{asy} - import Circ; - real u = 0.5cm; - TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$"); - pair a = B.end+(0,u); - pair b = B.beg-(0,u); - TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$"); - TwoTerminal Rb = resistor(Ra.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$"); - TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I$"); - wire(Rb.end, I.beg, nsq); - wire(I.end, b, udsq); - wire(b, B.beg, nsq); - wire(a, B.end, nsq); +import Circ; +real u = 0.5cm; +TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$"); +pair a = B.end+(0,u); +pair b = B.beg-(0,u); +TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$"); +TwoTerminal Rb = resistor(Ra.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$"); +TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I$"); +wire(Rb.end, I.beg, nsq); +wire(I.end, b, udsq); +wire(b, B.beg, nsq); +wire(a, B.end, nsq); \end{asy} \end{center} Using Kirchoff's loop rule @@ -40,23 +40,23 @@ Case 2: \\ With the voltmeter across $R_1$ we have \begin{center} \begin{asy} - import Circ; - real u = 0.5cm; - TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$"); - pair a = B.end+(0,u); - pair b = B.beg-(0,u); - TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$"); - TwoTerminal Ia = current(Ra.end, 0, "", "$I_1$"); - TwoTerminal Rv = resistor(a+(0,4u), normal, 0, "$95\U{k\Ohm}$", "$R_v$"); - TwoTerminal Iv = current(Rv.end, 0, "", "$I_v$"); - TwoTerminal Rb = resistor(Ia.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$"); - TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I_T$"); - wire(Rb.end, I.beg, nsq); - wire(I.end, b, udsq); - wire(b, B.beg, nsq); - wire(a, B.end, nsq); - wire(a, Rv.beg, nsq); - wire(Iv.end, Ia.end, rlsq); +import Circ; +real u = 0.5cm; +TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$"); +pair a = B.end+(0,u); +pair b = B.beg-(0,u); +TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$"); +TwoTerminal Ia = current(Ra.end, 0, "", "$I_1$"); +TwoTerminal Rv = resistor(a+(0,4u), normal, 0, "$95\U{k\Ohm}$", "$R_v$"); +TwoTerminal Iv = current(Rv.end, 0, "", "$I_v$"); +TwoTerminal Rb = resistor(Ia.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$"); +TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I_T$"); +wire(Rb.end, I.beg, nsq); +wire(I.end, b, udsq); +wire(b, B.beg, nsq); +wire(a, B.end, nsq); +wire(a, Rv.beg, nsq); +wire(Iv.end, Ia.end, rlsq); \end{asy} \end{center} Using our formula for resistors in parallel, we can bundle $R_v$ and $R_1$ into a single resistor $R_1'$, where diff --git a/latex/problems/Giancoli_6/question19.04.tex b/latex/problems/Giancoli_6/question19.04.tex index 2d29fcf..492ae00 100644 --- a/latex/problems/Giancoli_6/question19.04.tex +++ b/latex/problems/Giancoli_6/question19.04.tex @@ -1,4 +1,4 @@ -\begin{problem*}{Q4} +\begin{problem*}{Q19.4} % resistor networks Two lightbulbs of resistance $R_1$ and $R_2$ ($R_2 > R_1$) are connected in series. Which is brighter? What if they are connected in parallel? Explain. diff --git a/latex/problems/Giancoli_6/question19.07.tex b/latex/problems/Giancoli_6/question19.07.tex index e395d38..2e679b5 100644 --- a/latex/problems/Giancoli_6/question19.07.tex +++ b/latex/problems/Giancoli_6/question19.07.tex @@ -1,4 +1,4 @@ -\begin{problem*}{Q7} +\begin{problem*}{Q19.7} % resistor networks If two identical resistors are connected in series to a battery, does the battery have to supply more power or less power than when only one of the resistors is connected? Explain. diff --git a/latex/problems/Giancoli_6/question19.13.tex b/latex/problems/Giancoli_6/question19.13.tex index 0143b6f..76338d3 100644 --- a/latex/problems/Giancoli_6/question19.13.tex +++ b/latex/problems/Giancoli_6/question19.13.tex @@ -1,4 +1,4 @@ -\begin{problem*}{Q13} +\begin{problem*}{Q19.13} % internal resistance Explain in detail how you could measure the internal resistance of a battery. \end{problem*} -- 2.26.2