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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?enter image description here

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  • \$\begingroup\$ Your source voltage is a clue. \$\endgroup\$ – KalleMP Apr 30 '17 at 23:38
  • \$\begingroup\$ @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. \$\endgroup\$ – TonyM Apr 30 '17 at 23:47
  • \$\begingroup\$ 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. \$\endgroup\$ – Sunnyskyguy EE75 May 1 '17 at 5:12
  • \$\begingroup\$ Why would you build this circuit in "the real world"? \$\endgroup\$ – Finbarr May 1 '17 at 7:28
  • \$\begingroup\$ @Finbarr, it's an academic exercise; 'real world' implies non-ideal diodes. \$\endgroup\$ – Chu May 1 '17 at 8:21
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The total voltage drop across R1 and the diode is equal to V2 = 10.6V The forward biased diode accounts for a 0.6V drop, so the remaining 10V must be dropped in R1.

The same goes for the 2nd branch. In this case, the 10V are dropped across two equal resistors in series, R2 and R3, so 5V at each one.

The third branch is different, though. The diode is reverse biased, no current flows, and thus no voltage is dropped in the resistor.

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  1. for R1, the diode is forward biased so current flows through R1. The voltage drop over R1 is 10.6v - 0.6v = ...;

  2. for R2/R3, the diode is also forward biased so the voltage drop over R2 + R3 is 10.6v - 0.6v, and R2/R3 drops 5v each.

  3. for R4, the diode is reverse biased so zero current flows through R4 and its voltage drop is 0v.

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