Why Does the Bottom Diode Short First in a Multi-Diode Circuit?

[xcvxcvvc]

http://img38.imageshack.us/img38/6127/88280233.jpg
I circled the given answer. The problem asks for I and V.

I guess the bottom diode shorts first and then that makes the top two diodes be open circuited. Is there any good explanation on why the circuit behaves like that? I guess it comes from the fact that three different voltages relative to ground cannot be applied to the same node? Therefore, the diode with the lowest voltage on its anode "wins" and the other diodes shut off due to the negative $$V_D$$?

 

[vk6kro]

We had that problem a couple of weeks ago. Maybe that poster is in your class?

https://www.physicsforums.com/showthread.php?t=376290

See the second post for a working link.

 

[xcvxcvvc]

We had that problem a couple of weeks ago. Maybe that poster is in your class?

https://www.physicsforums.com/showthread.php?t=376290

See the second post for a working link.

So when I have multiple diodes tied to the same node that would all be in forward bias if they were alone, I say they're all open circuits. Next, I choose the diode that when short circuited makes the others reverse bias?

 

[vk6kro]

If that helps... not sure it is right though.


I picture the voltage at V rising from zero when the power is applied. When it gets to 1 volt, the bottom diode starts to conduct and stops the voltage V getting any higher.

If the voltage tried to rise higher, there would be a voltage across a perfect diode which has no resistance. So, there would have to be infinite current flowing.
But infinite current isn't available because of the 1 K resistor.
So the voltage can't rise.

The other diodes are reverse biased so they can't conduct, so the voltage at V stays at 1 volt.

This assumes that the 1 volt power source can conduct current in reverse, not just deliver it, and also that it has no internal resistance.