The current through an ideal PN-junction diod | vedclass

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(A) In the given circuit,the $P$-terminal of the diode is connected to a potential of $1 \ V$ and the $N$-terminal is connected to a potential of $2 \

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$0$

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(A) In the given circuit,the $P$-terminal of the diode is connected to a potential of $1 \ V$ and the $N$-terminal is connected to a potential of $2 \ V$ through a $100 \ \Omega$ resistor.For a diode to be forward biased,the potential at the $P$-terminal must be higher than the potential at the $N$-terminal.Here,the potential at the $P$-terminal $(V_P = 1 \ V)$ is less than the potential at the $N$-terminal $(V_N = 2 \ V)$.Therefore,the diode is in reverse biasing.An ideal diode in reverse biasing acts as an open circuit,meaning it offers infinite resistance.Consequently,no current flows through the circuit.Thus,the current through the diode is $0 \ mA$.

The current through an ideal $PN$-junction diode shown in the following circuit diagram will be.....$mA$.

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