Added 1.20

Chapter1.tex

@@ -248,5 +248,13 @@
     \ex{1.19}
     The magnetic flux produced within the coil is proportional to the number of turns. Now, because the inductance of the inductor is proportional to the amount of magnetic flux that passes through all the coils, it is proportional to the product of the magnetic flux and the number of coils. Thus the inductance is propotional to the square of the number of turns.
     \todo{Check/clarify this answer}
+
+    \ex{1.20}
+    We can use the formula for the full-wave rectifier ripple voltage to find the capacitance.
+    \[\frac{I_\load}{2fC} = \Delta V \le 0.1 \V_\pp\]
+    The maximum load current is 10mA and assuming a standard wall outlet frequency of 60 Hz, we have
+    \[C \ge \frac{10\m\A}{2\times 120\Hz \times 0.1\V} = \mans{417 \u\F}\]
+    Now we need to find the AC input voltage. The peak voltage after rectification must be 10V (per the requirements). Since each phase of the AC signal must pass through 2 diode drops, we have to add this to find out what our AC peak-to-peak voltage must be. Thus we have
+    \[V_{\in, \pp} = 10\V + 2(0.6\V) = \mans{11.2\V}\]
 
 \end{document}

taoesolutions.sty

@@ -27,13 +27,16 @@
 \newcommand{\W}{\ensuremath{\text{W}}}
 \newcommand{\A}{\ensuremath{\text{A}}}
 \newcommand{\F}{\ensuremath{\text{F}}}
-\renewcommand{\H}{\ensuremath{|text{H}}}
+\renewcommand{\H}{\ensuremath{\text{H}}}
+\newcommand{\Hz}{\ensuremath{\text{Hz}}}
 
 \newcommand{\Th}{\ensuremath{\text{Th}}} % Thevenin
 \renewcommand{\in}{\ensuremath{\text{in}}} % In (for V_{In})
 \newcommand{\out}{\ensuremath{\text{out}}}
 \newcommand{\load}{\ensuremath{\text{load}}}
 \newcommand{\source}{\ensuremath{\text{source}}}
+\newcommand{\pp}{\ensuremath{\text{p-p}}}
+\newcommand{\rms}{\ensuremath{\text{rms}}}
 
 \newenvironment{circuit}[2]% param1=label, param2=caption
     {