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(C) The reverse saturation current $I$ in a $p-n$ junction diode under reverse bias is given by Ohm's law:$I = \frac{V}{R}$Given:Voltage $V = 8 \, V$R

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$0.2$

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(C) The reverse saturation current $I$ in a $p-n$ junction diode under reverse bias is given by Ohm's law:$I = \frac{V}{R}$Given:Voltage $V = 8 \, V$Resistance $R = 4 	imes 10^7 \, \Omega$Substituting the values:$I = \frac{8}{4 	imes 10^7} \, A$$I = 2 	imes 10^{-7} \, A$To convert this into microamperes $(\mu A)$, we multiply by $10^6$:$I = 2 	imes 10^{-7} 	imes 10^6 \, \mu A$$I = 2 	imes 10^{-1} \, \mu A = 0.2 \, \mu A$Therefore, the correct option is $C$.

$A$ $p-n$ junction diode is reverse biased with a voltage of $8 \, V$. If the resistance of the diode is $4 \times 10^7 \, \Omega$, then the reverse saturation current is: (in $ \, \mu A$)

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