In the circuit shown here, assuming the thres | vedclass

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(D) The circuit consists of an $AC$ source $V = V_{0} \sin \omega t$ in series with a diode and a resistor $R$. The output voltage $V_{A B}$ is measur

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(D) The circuit consists of an $AC$ source $V = V_{0} \sin \omega t$ in series with a diode and a resistor $R$. The output voltage $V_{A B}$ is measured across the resistor $R$.During the positive half-cycle of the input $AC$, the diode is reverse-biased ($R.B.$). Since the diode acts as an open circuit in reverse bias, no current flows through the resistor $R$, and thus $V_{A B} = 0$.During the negative half-cycle of the input $AC$, the diode is forward-biased ($F.B.$). Since the threshold voltage is negligible, the diode acts as a short circuit. The entire input voltage appears across the resistor $R$. However, because the input voltage is negative during this half-cycle, $V_{A B}$ will also be negative.Therefore, the output voltage $V_{A B}$ will show negative half-cycles corresponding to the negative half-cycles of the input, and zero during the positive half-cycles. This matches the graph shown in option $D$.

In the circuit shown here, assuming the threshold voltage of the diode is negligibly small, the voltage $V_{A B}$ is correctly represented by which of the following graphs?

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