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(D) The $P-N$ junction diode will conduct only when it is forward-biased. Looking at the circuit, the diode is oriented such that it conducts when the

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(D) The $P-N$ junction diode will conduct only when it is forward-biased. Looking at the circuit, the diode is oriented such that it conducts when the input voltage $V_{in}$ is negative (specifically $-5\,V$).When $V_{in} = -5\,V$, the diode acts as a short circuit (ideal).The circuit then becomes a voltage divider with two equal resistors $R_1$ and $R_2$ in series across the source voltage of $5\,V$ (magnitude).The output voltage $V_0$ across $R_2$ is given by the voltage divider rule:$V_0 = V_{in} 	imes \frac{R_2}{R_1 + R_2}$Since $R_1 = R_2$, we have:$V_0 = -5\,V 	imes \frac{R_2}{R_2 + R_2} = -5\,V 	imes \frac{R_2}{2R_2} = -5\,V 	imes 0.5 = -2.5\,V$.Therefore, the output waveform will show a negative pulse of $-2.5\,V$ during the interval when the input is $-5\,V$.

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

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