update ex 1.20

frequency is 60Hz and times 2 for full-wave

Chapter1.tex

@@ -253,7 +253,7 @@
     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
-    \[C \ge \frac{10\m\A}{2\times 120\Hz \times 0.1\V} = \mans{417 \u\F}\]
+    \[C \ge \frac{10\m\A}{2\times 60\Hz \times 0.1\V} = \mans{833 \u\F}\]
     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
     \[V_{\in, \pp} = 10\V + 2(0.6\V) = \mans{11.2\V}\]
 
@@ -263,4 +263,4 @@
     The magnitude of this expression can be found by multiplying by the complex conjugate and taking the square root.
     \[sqrt{V_\out V_\out^*} = \frac{1}{\sqrt{1 + \omega^2R^2C^2}}V_\in\]
 
-\end{document}
+\end{document}