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I am having a bit of difficulty figuring out when is a resistor actually short circuited. Take the following 2 examples:

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In here, current flows through both 8k and 3k ohm resistors. Why is the current not bypassing them completely?

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In this example, the 8k, 3k and 1k ohm resistors are completely ignored and all the current goes through the 4k and 6k ohm resistors. I understand why the current is bypassing the 2k ohm resistor (short parallel with resistor), but how to tell exactly when a resistor is shorted?

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  • \$\begingroup\$ Anytime there is a wire in parallel with a resistor (or series resistors) it's shorted. In your second example, there is a wire that goes between the same points as the 8k, 3k and 1k ohm resistors. In your first example there is no wire that bypasses any resistors. \$\endgroup\$ – Samuel Feb 8 '18 at 20:07
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    \$\begingroup\$ NOTE: Your question would be much more understandable if you used the word 'bypassed' instead of shorted. By itself it implies a defective resistor, not a good resistor that has been bypassed. \$\endgroup\$ – user105652 Feb 8 '18 at 22:36
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If its two terminals are connected to the same node, the resistor is short-circuited.

In practical circuits, we might also say a resistor is short-circuited if a much lower value resistor is connected in parallel with it. In this case, the same potential will be across the two resistors, but the lower-value one will carry much more current than the higher value one.

In [the first example], current flows through both 8k and 3k ohm resistors. Why is the current not bypassing them completely?

Because of Ohm's law. If there's a potential difference across the resistor, a current will flow through it.

In this case, the two terminals of these resistors are connected to different nodes, so there is (very likely) a potential difference across them, and thus a current will flow through them.

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  • \$\begingroup\$ I hope I understood you right. Is this what you're saying? That because there is a potential difference between the 2 highlighted nodes means the resistor is not short circuited? \$\endgroup\$ – user2342352 Feb 8 '18 at 19:58
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    \$\begingroup\$ Yes. Just to add confusion, technically a resistor can still be "not shorted" even if the voltage at each node is the same. \$\endgroup\$ – BeB00 Feb 8 '18 at 20:07
  • \$\begingroup\$ Technically also even a short circuit has "some" finite resistance. But if you replace a wire a 0R0 resistor for your understanding, that's ok too. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 8 '18 at 20:10
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Consider a simplified case:

schematic

simulate this circuit – Schematic created using CircuitLab

Current flows in both R1 and R2. Why does the current not bypass R2 completely?

This is just an illustration to highlight and reinforce what other answers already say. I1 is a current source, so by definition, there must be 10mA of current flowing through it. In order for that current to flow, it must flow through either one or both of the resistors because there is no other path.

By Ohm's law, if there is current flowing in a resistor, then there must be a non-zero voltage drop across it.

But, the resistors are parallel. They must both see the same non-zero voltage.

By Ohm's law again, if there is a non-zero voltage across a resistor, then current must flow in the resistor.

In the example above, I made both resistors the same value to highlight the flaw in the question. But everything I said above would still be true even if R2 was a thousand Ohms, or even a million Ohms.

schematic

simulate this circuit

In the second case, almost all but not quite all of the current will flow in R1. The voltage is going to be pretty close to 1V. But if you put 1V across R2, some current must flow (about one microAmpere).

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    \$\begingroup\$ You should add some math for a more complete answer. For instance 10 mA/100 ohm = 1 volt. 1 volt across 1 meg ohm = 1 uA of current flow, so R1 current is 10 mA - 1 uA. Getting picky I know, but the math helps an OP understand better. Many of these questions are homework assignments. \$\endgroup\$ – user105652 Feb 8 '18 at 22:43
  • \$\begingroup\$ Actually, R1 is not getting 10 mA - 1 uA. V = 10 mA * (100//1e6) = 0.99990000999 V thus I_R1 = 9.9990000999 mA != 10 mA - 1 uA. But yeah, your numbers look good too :P \$\endgroup\$ – Vladimir Cravero Feb 8 '18 at 23:02

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