Parallel Diode with AC voltage and DC voltage
The diode only conducts when Anode is more positive then the Cathode. Since the Anode is fixed at 1.5V, the Cathode could not be lower than 0.8V during AC swing of the voltage source. Therefore Vx varies between ~0.8V and \$ \frac{R_B}{R_B+R} \$.4V =3.4V.

The voltage at the cathode actually follows the supply voltage change. For example, if the voltage source decreases, then does so the cathode's. The problem is that the cathode cannot be brought less than 0.8V due to the anode being held fixed at 1.5V. So during the negative cycle of the voltage source the voltage at the cathode, or across Rb, stays fixed at about 0.8V. In this case, the diode is forward biased.

But during the positive cycle of the voltage source the diode is reverse-biased because the cathode is being more positive than the anode. This shuts down the diode and no current would go through the middle branch. The circuit then reduces to a simple resistive divider. /end