P4 Jun2008 IACR
posted by Anonymous
Apartado 3:

\mbox{Divisor de tensión:}\\

V_{R2} = \frac{V_{g}R_{2}}{R_{1}+R_{2}} = 9.75 V\\

W_{e}^{md}(C_{2}) = \frac{C_{2}|V|^{2}}{4} = 2.37\  10^{-5} J \ \ \  \mbox{Demasié?}
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