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enter image description here

If we considered the Diodes states to be:
D1 : OFF
D2: ON
D3:OFF
Now if we apply KVL on the closed loop, will it look like this?

-Vout +Vout -4 v+ Vin =0
Knowing that the volatge across R3 is +Vout

My confusion is that I2 has to pass through the positive terminal of R2 (since a resistor absorbs energy and if current passes through negative terminal of an element this means the element is producing energy which is not possible for the resistor to do since it is a passive component and not an active one) Thus, if I2 is passing in the direction indicated in the image, it has passed through the negative terminal of the V of R3 which is Vout which cancels out Vout from the KVL. Is there anything wrong with my thinking/approach?

Thank you

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    \$\begingroup\$ Assuming \$D_1\$ and \$D_3\$ are off and \$D_2\$ is on, then the most negative point in the loop can only be the place between \$V_B\$ and \$D_2\$. It is probably convenient to start at ground (because that's the 0 reference) and either go clockwise or else counter-clockwise around the loop. Clockwise for KVL I get: \$0\:\text{V}+v_{in}-V_B+V_{DIODE}+I_2 R_2 + I_2 R_3=0\:\text{V}\$. \$\endgroup\$ – jonk Jan 22 '18 at 1:52
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Equivalent ckt looks like -

enter image description here

Going around the closed loop in clock wise direction gives -

$$V_{in} - V_B + V_{D2} + I_2R_2 - V_{out} = 0 $$ where $$V_{out} = -I_2R_3$$

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