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(D) The circuit consists of a series resistor ${R_1}$ and a parallel resistor ${R_2}$ with a diode in series with ${R_1}$.For the positive half-cycle

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(D) The circuit consists of a series resistor ${R_1}$ and a parallel resistor ${R_2}$ with a diode in series with ${R_1}$.For the positive half-cycle of the input voltage $(+5V)$, the diode is reverse-biased and acts as an open circuit. Therefore, no current flows through ${R_2}$, and the output voltage ${V_0} = 0V$.For the negative half-cycle of the input voltage $(-5V)$, the diode is forward-biased and acts as a short circuit. The circuit becomes a voltage divider with ${R_1}$ and ${R_2}$ in series across the $-5V$ source.Since ${R_1} = {R_2}$, the voltage across ${R_2}$ is given by the voltage divider rule: ${V_0} = V_{in} 	imes \frac{R_2}{R_1 + R_2} = -5V 	imes \frac{R_2}{R_2 + R_2} = -5V 	imes \frac{1}{2} = -2.5V$.Thus, the output waveform will show $0V$ during the positive cycle and $-2.5V$ during the negative cycle.

$A$ waveform is applied to the following circuit. Assuming an ideal diode and ${R_1} = {R_2}$, which of the following output waveforms will be produced?

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