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(D) The diode is connected such that its anode is towards $R_1$ and its cathode is towards $R_2$. For the diode to be forward-biased, the potential at

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(D) The diode is connected such that its anode is towards $R_1$ and its cathode is towards $R_2$. For the diode to be forward-biased, the potential at the anode must be higher than the potential at the cathode. Given the input waveform, when the input is $-5\,V$, the anode is at $-5\,V$. Assuming the output $V_0$ is initially $0\,V$, the diode becomes forward-biased. When the diode conducts, it acts as a short circuit (ideal diode). The circuit becomes a voltage divider with $R_1$ and $R_2$ in series across the $-5\,V$ source. Since $R_1 = R_2$, the output voltage $V_0$ across $R_2$ is given by the voltage divider rule: $V_0 = V_{in} 	imes \frac{R_2}{R_1 + R_2} = -5\,V 	imes \frac{R_2}{R_2 + R_2} = -5\,V 	imes \frac{1}{2} = -2.5\,V$. When the input is $+5\,V$, the diode is reverse-biased and acts as an open circuit, so no current flows through $R_2$, and $V_0 = 0\,V$. Thus, the output waveform will show a negative pulse of $-2.5\,V$ during the interval where the input is $-5\,V$.

$A$ waveform 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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