A junction diode has a resistance of 25 Omega | vedclass

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(B) In the given circuit,the $p$-side of the diode is connected to a higher potential $(5 	ext{ V})$ and the $n$-side is connected to a lower potenti

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$\frac{1}{7} 	ext{ A}$

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(B) In the given circuit,the $p$-side of the diode is connected to a higher potential $(5 	ext{ V})$ and the $n$-side is connected to a lower potential $(0 	ext{ V})$.Therefore,the diode is forward biased.The forward bias resistance of the diode is $R_d = 25 \Omega$.The external resistor is $R = 10 \Omega$.The total resistance of the circuit is $R_{eq} = R_d + R = 25 \Omega + 10 \Omega = 35 \Omega$.The potential difference across the circuit is $V = 5 	ext{ V} - 0 	ext{ V} = 5 	ext{ V}$.Using Ohm's law,the current $I$ in the circuit is given by $I = \frac{V}{R_{eq}} = \frac{5 	ext{ V}}{35 \Omega} = \frac{1}{7} 	ext{ A}$.

$A$ 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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