Simple Diode Problem: Solving for Unknown Voltages using KCL and KVL

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The discussion revolves around solving a diode circuit problem using Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL). The user identifies that diode 2 is not conducting due to its anode voltage being zero, while they seek to determine the voltage at the node where the 12V and 16V lines intersect to assess diode 1's state. They encounter challenges with KVL due to a lack of resistance in the path to the 12V source, leading to division by zero, and find KCL unhelpful due to the absence of closed loops. Suggestions include assuming diode 1 is conducting to establish voltage across a resistor, allowing for current calculations. The discussion highlights the complexities of analyzing circuits with ideal diodes and the limitations of KCL and KVL in certain scenarios.
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Homework Statement


[PLAIN]http://img27.imageshack.us/img27/4475/q12r.png

Homework Equations



KCL/KVL

Mesh current, node voltage

ohm's law

The Attempt at a Solution



for b)
I understand how diodes work, and i'v deduced that diode 2 is definitely not conducting as the voltage on it's anode is 0, and the voltage on it's cathode MUST be higher than 0.

To find weather diode 1 is off or not, i need to figure out the voltage at the node where the 12v and 16v lines meet. I tried using KVL, however the path to the 12 volt source has no resistance, and therefore results in me dividing by 0.

KCL can't be used as there's no useful closed loops in this.

Could anyone give me an idea on how to determine the voltage at that node?

I have similar problems with part a), I need to find the voltage of the node on the left, but one of it's arms have no resistance so i can't use node voltage or KCL. there is really not loops to use KVL.
 
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I would begin by assuming that D1 is conducting. If D1 is conducting then you immediately know the voltage across R2 since the voltage across an ideal diode is constant. From there you can calculate the current in each node. If D1 is not conducting, then the resulting currents and voltage drops will not obey KCL/KVL and you'll know something is wrong.
 

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