P4 Jun2008 IACR
posted by Anonymous
posted 1y 323d ago
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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