# Help understanding circuit with diodes in it

I'm having trouble figuring out how this circuit works in the real world. I know theoretically the diodes should have a resistance of 0, 0 and infinite respectively, which makes the question quite easy to solve.

However I'm more confused when you factor in a voltage drop of say 0.6V across the first two diodes. How does that change the voltage drop across the resistors and how do I factor it into ohm's law? • Your source voltage is a clue. – KalleMP Apr 30 '17 at 23:38
• @KalleMP, how is the source voltage a clue into how to factor diode drops into Ohm's Law? I know the answer to the OP question and I can't see that as a clue. – TonyM Apr 30 '17 at 23:47
• Diode switch a logical part with 0.6V drop in the 1mA range. If reversed the logic is open circuit. If forward biased it will be treated as a low resistance junction << 100 ohms @1mA thus the solution here is trivial. (10.6-0.6V)/Rload if diode is forward biased for each string. – Tony Stewart Sunnyskyguy EE75 May 1 '17 at 5:12
• Why would you build this circuit in "the real world"? – Finbarr May 1 '17 at 7:28
• @Finbarr, it's an academic exercise; 'real world' implies non-ideal diodes. – Chu May 1 '17 at 8:21