Why does voltage stay the same regardless of resistor connect to the source
Probably the easiest way to see this is like:







simulate this circuit &ndash; Schematic created using CircuitLab

Now, just think about that sine wave. It's oscillating between \$V_\text{IN}=-12\:\text{V}\$ and \$V_\text{IN}=+12\:\text{V}\$. When  \$V_\text{IN}\ge +10\:\text{V}\$ (a short period) then the diode doesn't conduct because it is reverse-biased. So there is no current through it and therefore also no current in \$R_1\$, either. Since there is no current in \$R_1\$ there is no voltage drop across \$R_1\$. Therefore, \$V_\text{OUT}=+10\:\text{V}\$ for the period of time while \$V_\text{IN}\ge +10\:\text{V}\$.

When \$V_\text{IN}\lt +10\:\text{V}\$ (most of the period of time), then the diode is forward-biased. Now the best way to imagine is to realize that the diode's cathode is essentially nailed to \$V_\text{IN}\$, so its anode will be one diode drop above it. In short, the anode of \$D_1\$ will "follow" the sine wave source during this period, but will follow it biased upward by one diode drop away. The anode has to follow the cathode. It has no choice. So as a consequence, \$V_\text{OUT}\$ during this period will be a diode drop above \$V_\text{IN}\$.

That's it.