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How can the states of ideal diodes be determined in simple circuits with only DC sources and resistors without a trial and error approach?

I posted this question on Electronics SE and found out that there is none.

This is one of the situations which bugs me that something systematic is going on but I am not able to find it.For instance;

This in an incredibly simple circuit , the current has to circulate in certain directions and only one set of states for diodes satisfy the circulation of current. I frustrates me to solve this simple problem by guess and check.

Assumption 1

Assumption 2

Assumption 3

Assumption 4

Answer

It looks like solving $2x=5$ by trial and error!

So, what possible approaches can be used to analyze this type of circuits in a deterministic manner? Or at least how can I approach this particular problem in deterministic way?

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    $\begingroup$ Keywords: piece-wise linear (PWL) circuits, Katzenelson's algorithm, LCP $\endgroup$ – Fizz Sep 25 '15 at 4:47
  • $\begingroup$ I'm not sure I agree with the closure of this question. It is mostly a theoretical/algorithmic one although perhaps cs.stackexchange.com would have been the best place to ask. Hardly anyone in EE actually uses the algorithms I mentioned in practice... because they don't setup the problem like this but use a smooth function for the characteristic (Shokley's equation). $\endgroup$ – Fizz Oct 19 '15 at 18:08
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Let $D_1$ denote the left diode, $D_2$ the right diode, $R_1 = 4 \;k\Omega$, $R_2 = 6 \;k\Omega$, $U_1 = 10 \;V$, $U_2 = 3 \;V$.

Then we could write voltage drop on $R_2$, using voltage divisor formula: $$ U_{R_2} = {R_2 \cdot U_1 \over {R_1 + R_2}} = {6 \;k\Omega \cdot 10\;V \over {4 \;k\Omega + 6 \;k\Omega}} = {6\cdot 10^3 \cdot 10 \over {10 \cdot 10^3}}\;V = 6 \;V$$

This mean that right diode $D_2$ is blocked, because voltage drop on $R_2$:

$$U_{R_2} \gt U_2 \;, \text{where} U_{R_2} = 6\;V\ \text{and}\ U_2=3\;V$$

Now we could calculate current trough voltage divisor, that also is $D_1$ current: $$I_{R_{12}} = {U_1 \over {R_1 + R_2}} = {10\;V \over {4 \cdot 10^3\; \Omega + 6 \cdot 10^3\;\Omega}}= {10\;V \over {10 \cdot 10^3\;\Omega}} = {1\;mA}$$

Bibliography:

Voltage Divider Circuits : Divider Circuits And Kirchhoff's Laws, Electronics Textbook

Superposition Theorem : DC Network Analysis, Electronics Textbook

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    $\begingroup$ You assumed $D_1$ conducting, then it lead to the correct conclusion. Why did you assume it in the first place? $\endgroup$ – Hashir Omer Sep 23 '15 at 20:20
  • $\begingroup$ @HashirOmer Because $U_1 \gt U_2$, then I've checked if $U_{R_2} \gt U_2$ also. See this: Superposition Theorem : DC Network Analysis - Electronics Textbook tinyurl.com/p9vs543 $\endgroup$ – user103028 Sep 23 '15 at 20:33
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    $\begingroup$ @VitalieGhelbert Please slow down on the unnecessary and distracting formatting oddities. The less the better. $\endgroup$ – Did Sep 23 '15 at 21:57
  • $\begingroup$ @Did, I'm learning StackOverFlow. Reading this and testing on the fly. MathJax basic tutorial and quick reference - Mathematics Meta Stack Exchange $\endgroup$ – user103028 Sep 23 '15 at 22:07
  • $\begingroup$ I see you rolled back the revisions I made to your post. Did you at least understand why the version I arrived at was vastly superior? Answers on main are not made for "testing on the fly" whatever silly formatting idea you may have. meta.math.stackexchange.com/q/21504 meta.math.stackexchange.com/q/21501 $\endgroup$ – Did Sep 23 '15 at 22:56

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