In the given figure,the voltage of point A is | vedclass

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(A) The diode is connected such that its $p$-side is at $-3 \ V$ and its $n$-side is connected to point $A$,which is connected to ground through a $5

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(A) The diode is connected such that its $p$-side is at $-3 \ V$ and its $n$-side is connected to point $A$,which is connected to ground through a $5 \ \Omega$ resistor.Since the potential at the $p$-side $(-3 \ V)$ is lower than the potential at the $n$-side (which is $0 \ V$ due to the ground connection),the diode is reverse-biased.In a reverse-biased condition,an ideal diode acts as an open circuit,meaning no current flows through the circuit.Since no current flows through the $5 \ \Omega$ resistor,there is no voltage drop across it $(V = IR = 0 	imes 5 = 0 \ V)$.Therefore,the potential at point $A$ remains equal to the ground potential,which is $0 \ V$.

In the given figure,the voltage of point $A$ is ............ $V$.

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