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A junction diode has a resistance of $25 \Omega$ when forward biased and $2500 \Omega$ when reverse biased. The current in the diode, for the arrangement shown will be
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The correct answer is:
$\frac{1}{7} \mathrm{~A}$
$\mathrm{R}_{\mathrm{eq}}=25+10=35 \Omega$
Because diode is forward biased. So $I=\frac{V}{R_{e q}}=\frac{5}{35}=\frac{1}{7} \mathrm{~A}$
Because diode is forward biased. So $I=\frac{V}{R_{e q}}=\frac{5}{35}=\frac{1}{7} \mathrm{~A}$
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