# How to calculate the voltage across a resistor using assumed states method? [closed]

The diode in the next figure is considered ideal (i.e. works as a simple switch, being turned ON when $$U_D \geq 0$$ and OFF when $$U_D < 0$$). What is the voltage $$U_S$$ across $$R_S$$?

I've assumed that the diode is OFF and replaced it with an open circuit. The conditions are: $$U_D > 0, I_D = 0$$. I think I should be applying KVL now to find the voltage, but I'm not sure how. Also, I've also taken into consideration using Thevenin's Theorem, but I get stuck at a point where I need to find $$I_D$$ in order to calculate $$U_S$$. However, I think the assumed states method is better in this case. Could you give me an idea on how to continue this?

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First note that if you're using the German (or East European) convention for the voltage polarity, the diode is ON when $$U_D\ge 0$$ and OFF when $$U_D < 0$$.
If you assume that the diode is OFF, that is, you assume $$U_D < 0$$, then you can ignore the branch with the diode and the circuit is a simple voltage divider with $$U_S = E/2 = 6\,\mathrm{V}$$. But since $$U_D = U_S+E_0 = 9\,\mathrm{V} > 0$$, this contradicts the assumption, and the diode is instead ON.
• Thank you, could you please tell me how you got the 2 relations : $U_S = E/2$ and $U_D = U_S + E_0$? – Andrei0408 2 days ago
• @Andrei0408 The first one is a simple application of the voltage divider's formula, taking into account that the two resistances are equal, $R=R_S$; the second one can be obtained by writing the Kirchhoff's voltage law at the loop containing the diode, $E_0$ and $U_S$. – Massimo Ortolano 2 days ago