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I attended a workshop on troubleshooting and we were asked to find the flaw in the given circuit. A guy pointed out to the organizer that R3 was short ckt and R2 was open ckt.

However I was pretty sure that R2 was short ckt and R3 was open ckt because same voltage is appearing, but didnt point out ( for which I lost a 1 minute clapping).

So who was correct (or no one ) ?

Thank you.

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    \$\begingroup\$ Actually you only need R3 to be an open circuit. \$\endgroup\$ – Samuel Jan 28 '15 at 23:24
  • \$\begingroup\$ Samuel is probably right - either all of R1, R2, and R4 are shorted, or R3 is open. \$\endgroup\$ – Greg d'Eon Jan 28 '15 at 23:26
  • \$\begingroup\$ So is there a way I can narrow down the faulty part w/o actually testing ? \$\endgroup\$ – Plutonium smuggler Jan 28 '15 at 23:28
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    \$\begingroup\$ R3 must be very high resistance (open circuit), no testing required (aside from the measurements to get the node voltages), you would need more testing only to determine if the other resistors are shorts or not. \$\endgroup\$ – Samuel Jan 28 '15 at 23:35
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    \$\begingroup\$ @Plutoniumsmuggler - That's not what I meant. A "short" can be 100 ohms, if the adjacent "open" is 100megohms. An "open" can be 100 ohms if the adjacent "short" is 0.001 ohm. \$\endgroup\$ – Hot Licks Jan 29 '15 at 12:51
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Actually you only need R3 to be an open circuit. Then no current is flowing in the other three resistors, hence the given readings. You could say that the others are shorts, but it's not required (or knowable given R3 as open).

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There are two choices which would produce this result:

    1. R3 is open circuit.
      All points above the "break" are pulled to 4V.
      All points below the break are pulled to ground.
    1. R1, R2, R4 are all short circuit.
      R1 + R2 short apply 4V directly to the top of R3.
      R4 short applies ground to the bottom of R3.

Occam's Razor suggests the preferable conclusion is 1. - R3 is open circuit.
For 2. to be true 3 components would need to have failed - unless they were 0 Ohm links to start with. Unless there are good reasons to think that conditions have occurred that would cause 3 series resistors to go short circuit this seems unlikely. An overload in a resistor will usually cause and O/C rather than a S/C failure. It is possible that some forms of resistor may "melt" in some way and allow end to end continuity, but this is not usual.


Note that Occam's Razor does NOT say that the simplest solution is most likely to be correct - it rather says that when you have several possible solutions and none are obviously more or less correct than any oter that it makes sense to start with the simplest one.

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{(R3 IS OPEN) AND (ALL THE OTHER RESISTOR ARE NOT OPEN)} has to be true to get your output. And, NOT OPEN means finite resistance value (that could also be ZERO).

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There's no voltage difference across R1, so there's no current flowing through it, hence the ladder is broken somewhere.
Since you measure a voltage on both ends of R1 its connections are fine.

You can apply exactly the same reasoning for R2, so that's fine as well.

Since both ends of R4 are also at the same voltage (0V) its connections are fine too. (see note below)

Only R3 remaining, so the open circuit has to be there. Either R3 is broken (open-circuit) or it isn't soldered well.

note:
the reading at the R3/R4 node is 0V, which could mean one of two things:
- you're actually measuring the ground voltage. In that case R4 is fine, as mentioned above
- but there's also the possibility that you're not measuring anything, because R4 is broken as well. In that case the node R3/R4 is floating and will read )v too.

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I see the point between R2 and R3 as 0.4 volts, not 4.0, so R1 and R4 are shorts, and R2/R3 form a simple voltage divider. R2 and R3's values would be in a 9:1 ratio.

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  • \$\begingroup\$ Oh no...thats a simple line like this -> ..... dont misinterpret it as dot. Its 4 V only. \$\endgroup\$ – Plutonium smuggler Jan 29 '15 at 3:46

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