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(B) The circuit shows that the $p-n$ junction diode is forward biased because the $p$-terminal is connected to a higher potential $(10 \, V)$ and the

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$\frac{1}{15} \, A$

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(B) The circuit shows that the $p-n$ junction diode is forward biased because the $p$-terminal is connected to a higher potential $(10 \, V)$ and the $n$-terminal is connected to a lower potential $(0 \, V)$.The forward resistance of the diode is $R_d = 50 \, \Omega$.The external resistor in the circuit is $R = 100 \, \Omega$.Since the diode and the resistor are in series,the total effective resistance of the circuit is:$R_{total} = R_d + R = 50 \, \Omega + 100 \, \Omega = 150 \, \Omega$The potential difference across the circuit is $V = 10 \, V - 0 \, V = 10 \, V$.Using Ohm's law,the current $I$ in the circuit is:$I = \frac{V}{R_{total}} = \frac{10 \, V}{150 \, \Omega} = \frac{1}{15} \, A$

$A$ junction diode has a resistance of $50 \, \Omega$ when forward biased and $5000 \, \Omega$ when reverse biased. The current in the arrangement shown in the figure will be

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