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\begin{document}
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\heading{Flash Current}
\auname{I.D. 036553790}
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{\bf The problem:}
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In a circuit of flash lamp current flows through the lamp only if the
voltage on it reaches the outbreak voltage V$_{L}$. When this happens the capacitor discharges through the lamp quickly (which mean, the
time of discharge negligible), to the voltage V$_{G}$ (not to zero). The lamp is required to give three flashes in one second. What should be the resistance $R$?
\includegraphics[width=0.45\hsize]{e_42_4_086_p0.jpg}
$V_{L}=80V,V_{G}=30V,\varepsilon=100V,C=0.2\mu F$
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\Dn
{\bf The solution:}
\Dn
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1.
\begin{eqnarray}
\varepsilon-IR-q/c=0
\end{eqnarray}
Substitute $I = dq/dt$:
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\begin{eqnarray}
\frac{dq}{dt}=\frac{\varepsilon}{R}-\frac{q}{RC}%
\end{eqnarray}
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Integrating this expression, using the fact that $q = 0$ at $t = 0$, we obtain:%
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\begin{eqnarray}
q &=& C\varepsilon(1-e^{-t/RC})+q_{0}e^{^{-t/RC}}\\
V &=& \frac{q}{C}=\varepsilon(1-e^{-t/RC})+V_{0}e^{-t/RC}
\end{eqnarray}
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$V_{0}=V_{G}$ because time of discharge negligible so:%
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\begin{eqnarray}
V_{L}&=&\varepsilon(1-e^{-t/RC})+V_{G} \\
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R &=& -\frac{t}{C \ln(1-\frac{V_{L}-V_{G}}{\varepsilon})}
\end{eqnarray}
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Substituting $t=\frac{1}{3}\sec$ and the numeric values of all the given
parameters we obtain:
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\begin{eqnarray}
R= 2.4\times 10^6\ \Omega
\end{eqnarray}
\bigskip
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\end{document}