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(A) The $p-n$ junction diode is forward biased because the $p$-terminal is at a higher potential $(+4\;V)$ than the $n$-terminal $(-6\;V)$.An ideal di

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$10^{-2}\;A$

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(A) The $p-n$ junction diode is forward biased because the $p$-terminal is at a higher potential $(+4\;V)$ than the $n$-terminal $(-6\;V)$.An ideal diode in forward bias acts as a short circuit (zero resistance).Therefore,the current $I_{AB}$ flowing through the circuit is given by Ohm's law:$I_{AB} = \frac{V_A - V_B}{R} = \frac{4\;V - (-6\;V)}{1\;k\Omega} = \frac{10\;V}{1000\;\Omega} = 10^{-2}\;A$.

Consider the junction diode as ideal. The value of current flowing through $AB$ is

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