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A $5.0 \mathrm{~V}$ stabilized power supply is required to be designed using a $12 \mathrm{~V} \mathrm{DC}$ power supply as input source. The maximum power rating of zener diode is $2.0 \mathrm{~W}$. The minimum value of resistance $\mathrm{R}_{\mathrm{s}}$ in $\Omega$ connected in series with zener diode will be
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The correct answer is:
$17.5$
Using the series resistance formula for a zener diode
$\begin{aligned}
\mathrm{R}_{\mathrm{S}} & =\frac{\left(\mathrm{V}_{\mathrm{S}}-\mathrm{V}_{\mathrm{Z}}\right)}{\mathrm{I}_{\mathrm{Z}_{\max }}} \\
\mathrm{I}_{\mathrm{Z} \max } & =\frac{\mathrm{P}_{\mathrm{Z}}}{\mathrm{V}_{\mathrm{Z}}}=\frac{2}{5}=400 \mathrm{~mA} \\
\therefore \quad \mathrm{R}_{\mathrm{S}} & =\frac{(12-5)}{400 \times 10^{-3}}=\frac{7}{400} \times 10^3 \\
& =17.5 \Omega
\end{aligned}$
$\begin{aligned}
\mathrm{R}_{\mathrm{S}} & =\frac{\left(\mathrm{V}_{\mathrm{S}}-\mathrm{V}_{\mathrm{Z}}\right)}{\mathrm{I}_{\mathrm{Z}_{\max }}} \\
\mathrm{I}_{\mathrm{Z} \max } & =\frac{\mathrm{P}_{\mathrm{Z}}}{\mathrm{V}_{\mathrm{Z}}}=\frac{2}{5}=400 \mathrm{~mA} \\
\therefore \quad \mathrm{R}_{\mathrm{S}} & =\frac{(12-5)}{400 \times 10^{-3}}=\frac{7}{400} \times 10^3 \\
& =17.5 \Omega
\end{aligned}$
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