From ebcc273914c213612f41afa2fd9a035e9849b5c2 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Fri, 13 May 2011 12:47:04 -0400 Subject: [PATCH] Fix spelling Kirchoff -> Kirchhoff. --- 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 | 6 +++--- latex/problems/Giancoli_6/problem19.58.tex | 2 +- latex/problems/Giancoli_6/question19.07.tex | 2 +- latex/problems/Giancoli_6/question19.13.tex | 2 +- latex/problems/Young_and_Freedman_12/problem25.36.tex | 4 ++-- latex/problems/Young_and_Freedman_12/problem26.21.tex | 2 +- latex/problems/Young_and_Freedman_12/problem26.22.tex | 2 +- 10 files changed, 13 insertions(+), 13 deletions(-) diff --git a/latex/problems/Giancoli_6/problem19.07.tex b/latex/problems/Giancoli_6/problem19.07.tex index 189a1f2..ed609d6 100644 --- a/latex/problems/Giancoli_6/problem19.07.tex +++ b/latex/problems/Giancoli_6/problem19.07.tex @@ -7,7 +7,7 @@ $2200\U{\Ohm0}$ resistor? \begin{solution} First we find the total current in the circuit. The two resistances, $R_1 = 650\U{\Ohm}$ and $R_2 = 2200\U{\Ohm}$, in series provide an -effective resistance of $R_e = R_1 + R_2$. By Kirchoff's loop rule +effective resistance of $R_e = R_1 + R_2$. By Kirchhoff's loop rule \begin{align*} V - I R_e &= 0 \\ I &= \frac{V}{R_e} = \frac{V}{R_1 + R_2} diff --git a/latex/problems/Giancoli_6/problem19.15.tex b/latex/problems/Giancoli_6/problem19.15.tex index c636cbe..be50da1 100644 --- a/latex/problems/Giancoli_6/problem19.15.tex +++ b/latex/problems/Giancoli_6/problem19.15.tex @@ -7,7 +7,7 @@ bulb. \begin{solution} Let $V = 110\U{V}$ be the source voltage, $P_1 = 7.0\U{W}$ be the power of one bulb, and $R_1$ be the resistance of one bulb. By -Kirchoff's loop rule +Kirchhoff's loop rule \begin{align*} V - 8IR_1 &= 0 \\ I &= \frac{V}{8R_1} diff --git a/latex/problems/Giancoli_6/problem19.24.tex b/latex/problems/Giancoli_6/problem19.24.tex index 27083ee..15bc710 100644 --- a/latex/problems/Giancoli_6/problem19.24.tex +++ b/latex/problems/Giancoli_6/problem19.24.tex @@ -17,7 +17,7 @@ Determine the terminal voltage of each battery in Fig.~19-44. \end{problem*} \begin{solution} -From Kirchoff's loop rule +From Kirchhoff's loop rule \begin{align*} \mathcal{E}_1 - IR - \mathcal{E}_2 - Ir_2 - Ir_1 &= 0 \\ I(R+r_1+R_2) &= \mathcal{E}_1-\mathcal{E}_2 \\ diff --git a/latex/problems/Giancoli_6/problem19.31.tex b/latex/problems/Giancoli_6/problem19.31.tex index ff58077..28a539f 100644 --- a/latex/problems/Giancoli_6/problem19.31.tex +++ b/latex/problems/Giancoli_6/problem19.31.tex @@ -53,15 +53,15 @@ Label the resistors from left to right: $R_1 = 12\U{\Ohm}$, $R_2 = Label the batteries from left to right: $V_1 = 6.0\U{V}$ and $V_2 = 3.0\U{V}$. -Applying Kirchoff's junction rule to junction $a$ we have +Applying Kirchhoff's junction rule to junction $a$ we have $$I_1 + I_2 - I_3 = 0$$ -Applying Kirchoff's loop rule to the left-hand loop we have +Applying Kirchhoff's loop rule to the left-hand loop we have $$V_1 - I_1 (R_1 + R_2) + R_3 I_2 = 0$$ where we \emph{add} the voltage change over $R_3$ because we cross it \emph{against} the direction of the current $I_2$. -Applying Kirchoff's loop rule to the right-hand loop we have +Applying Kirchhoff's loop rule to the right-hand loop we have $$V_2 - R_4 I_3 - R_3 I_2 - R_5 I_5 = V_2 - I_3 (R_4 + R_5) - R_3 I_2 = 0$$ We now have three equations for three unknowns (the $I_i$). diff --git a/latex/problems/Giancoli_6/problem19.58.tex b/latex/problems/Giancoli_6/problem19.58.tex index a3d53c5..437ca5c 100644 --- a/latex/problems/Giancoli_6/problem19.58.tex +++ b/latex/problems/Giancoli_6/problem19.58.tex @@ -25,7 +25,7 @@ wire(b, B.beg, nsq); wire(a, B.end, nsq); \end{asy} \end{center} -Using Kirchoff's loop rule +Using Kirchhoff's loop rule \begin{align*} V - IR_1 - IR_2 &= 0 \\ I &= \frac{V}{R_1+R_2} diff --git a/latex/problems/Giancoli_6/question19.07.tex b/latex/problems/Giancoli_6/question19.07.tex index 2e679b5..b413cb5 100644 --- a/latex/problems/Giancoli_6/question19.07.tex +++ b/latex/problems/Giancoli_6/question19.07.tex @@ -9,7 +9,7 @@ We know that the power provided by the battery is given by $$P = IV$$ so the power supplied increases if the current $I$ increases (because $V$ remains constant for batteries). -From Kirchoff's loop rule, we know the voltage drop across the +From Kirchhoff's loop rule, we know the voltage drop across the resistors is the same as the voltage gain across the battery. $$V_b = V_R$$ We also know that the voltage across the resistors relates to the current via Ohm's law diff --git a/latex/problems/Giancoli_6/question19.13.tex b/latex/problems/Giancoli_6/question19.13.tex index 76338d3..feae864 100644 --- a/latex/problems/Giancoli_6/question19.13.tex +++ b/latex/problems/Giancoli_6/question19.13.tex @@ -20,7 +20,7 @@ battery. Make a circuit using a known resistance $R$ to connect the two terminals of the battery, and measure the current $I$. -From Kirchoff's loop rule +From Kirchhoff's loop rule \begin{align*} V - Ir - IR &= 0 \\ Ir &= V - IR \\ diff --git a/latex/problems/Young_and_Freedman_12/problem25.36.tex b/latex/problems/Young_and_Freedman_12/problem25.36.tex index ba63780..0a65e73 100644 --- a/latex/problems/Young_and_Freedman_12/problem25.36.tex +++ b/latex/problems/Young_and_Freedman_12/problem25.36.tex @@ -23,7 +23,7 @@ Resistances (\verb+ZZ+ and \verb+/\/\/+) are in \Ohm. Battery \begin{solution} \Part{a} -We're going to use Kirchoff's loop rule: the sum of voltage changes +We're going to use Kirchhoff's loop rule: the sum of voltage changes around a loop is $0$. \begin{equation} \sum_\text{loop} \Delta V = 0 @@ -36,7 +36,7 @@ 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 +Kirchhoff'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 diff --git a/latex/problems/Young_and_Freedman_12/problem26.21.tex b/latex/problems/Young_and_Freedman_12/problem26.21.tex index 29013b2..741586d 100644 --- a/latex/problems/Young_and_Freedman_12/problem26.21.tex +++ b/latex/problems/Young_and_Freedman_12/problem26.21.tex @@ -30,7 +30,7 @@ current in resistor $R$? \end{verbatim} \end{center} \Part{a} -Labeling the currents as above, we use Kirchoff's junction rule +Labeling the currents as above, we use Kirchhoff's junction rule summing the currents entering node $a$. \begin{align} 0 &= -I_1 - I_2 + I_3 & diff --git a/latex/problems/Young_and_Freedman_12/problem26.22.tex b/latex/problems/Young_and_Freedman_12/problem26.22.tex index c2ec360..c882d79 100644 --- a/latex/problems/Young_and_Freedman_12/problem26.22.tex +++ b/latex/problems/Young_and_Freedman_12/problem26.22.tex @@ -40,7 +40,7 @@ looping clockwise from $a$ through 3 and 2. \EMF_2 &= -(V_{ba}+I_2R_2) = \ans{7.00\U{V}} \end{align} -To find $\EMF_1$ we'll need $I_1$. Applying Kirchoff's junction rule +To find $\EMF_1$ we'll need $I_1$. Applying Kirchhoff's junction rule to $a$ \begin{align} 0 &= I_3 + I_1 - I_2 \\ -- 2.26.2