# Open v. Short Circuit understanding

#### 10th&Bennett

Joined Apr 15, 2018
27
Greeting AAC Family,

The next topic I am delving into is understanding open v. short circuits.

I'll get to the attached problem in a second, but the way I understand it is this. An open circuit is one where no current will be able to flow, so the resistance is infinite. A short is essentially a wire with no resistance. If there is an alternate way to understand this fundamentally - I am all ears!

So, with the attached, I have to find R_t when the a and b terminals are (a) open-circuited, and (b) short-circuited.

Game plan for (a):
Akin to the "walking from a to b" guidance that @WBahn provided, for an open circuit I realize that no current will flow over a and b, so we can treat the 3 Ohm resistor in parallel with the 4 Ohm resistor, and the 5 Ohm in parallel with the 8 Ohm. That result (~4.79 Ohms) would then be in parallel with the 10 Ohm resistor, and then that result (~3.23 Ohms) would be in series with the 7 Ohm and 8 Ohm resistor, for a grand total of 18.2 Ohms. That answer checks.

Game plan for (b):
For a short circuit I realize that current will flow over a and b and there won't be much resistance, so we can treat the 8 Ohm resistor in parallel with the 4 Ohm resistor, and the 3 Ohm in parallel with the 5 Ohm. That result (~4.535 Ohms) would then be in parallel with the 10 Ohm resistor, and then that result (~3.12 Ohms) would be in series with the 7 Ohm and 8 Ohm resistor, for a grand total of 18.1 Ohms. That answer checks.

What I am not seeing:
Ok, I see part (a) with 3 Ohm || 4 Ohm and 5 Ohm || 8 Ohm when no current flows from a to b since it's not connected. No question there. But for part (b), I am having a hard time visualizing how 8 Ohm || 4 Ohm and 3 Ohm || 5 Ohm given that the terminals would now be connected. At first glance, to be honest I thought they would all be in series if the middle wire were connected. Admittedly, my approach for part (b) was "brute force" to match the answer, but I want to understand "why."

Thank you again for your help. Oh, and yes, this is not homework, just to get back into it (self-study).

10th

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#### MrChips

Joined Oct 2, 2009
22,099
Sorry to burst your bubble but you have it all wrong.

(a) Open circuit. Two resistors are in series. Another two are in series. Then the two pairs are in parallel.

(b) Short circuit. Two resistors are in parallel. Another two are in parallel. Then the two pairs are in series.

#### WBahn

Joined Mar 31, 2012
26,301
Greeting AAC Family,

The next topic I am delving into is understanding open v. short circuits.

I'll get to the attached problem in a second, but the way I understand it is this. An open circuit is one where no current will be able to flow, so the resistance is infinite. A short is essentially a wire with no resistance. If there is an alternate way to understand this fundamentally - I am all ears!
That's basically it. From a fundamental aspect that emphasizes the symmetry between the two cases, an open circuit is when no restriction is placed on the voltage that appears across the terminals but the current that can flow between them is forced to be zero. A closed circuit, on the other hand, is when no restriction is placed on the current that can flow between the terminals, but the voltage that appears across them is forced to be zero.

So, with the attached, I have to find R_t when the a and b terminals are (a) open-circuited, and (b) short-circuited.

Game plan for (a):
Akin to the "walking from a to b" guidance that @WBahn provided, for an open circuit I realize that no current will flow over a and b, so we can treat the 3 Ohm resistor in parallel with the 4 Ohm resistor, and the 5 Ohm in parallel with the 8 Ohm. That result (~4.79 Ohms) would then be in parallel with the 10 Ohm resistor, and then that result (~3.23 Ohms) would be in series with the 7 Ohm and 8 Ohm resistor, for a grand total of 18.2 Ohms. That answer checks.
Remember that for two resistors to be in parallel, it must be true that whatever voltage appears across one MUST be the same voltage that appears across the other. The easiest way to see this is to color code all of the nodes. If two resistors are connected to the same two colors, then they are in parallel. If they aren't, then they aren't.

Game plan for (b):
For a short circuit I realize that current will flow over a and b and there won't be much resistance, so we can treat the 8 Ohm resistor in parallel with the 4 Ohm resistor, and the 3 Ohm in parallel with the 5 Ohm. That result (~4.535 Ohms) would then be in parallel with the 10 Ohm resistor, and then that result (~3.12 Ohms) would be in series with the 7 Ohm and 8 Ohm resistor, for a grand total of 18.1 Ohms. That answer checks.
You are getting numbers that a close enough by sheer coincidence. I could change the values of the resistors such that you would get numbers that are way off.

What I am not seeing:
Ok, I see part (a) with 3 Ohm || 4 Ohm and 5 Ohm || 8 Ohm when no current flows from a to b since it's not connected. No question there. But for part (b), I am having a hard time visualizing how 8 Ohm || 4 Ohm and 3 Ohm || 5 Ohm given that the terminals would now be connected. At first glance, to be honest I thought they would all be in series if the middle wire were connected. Admittedly, my approach for part (b) was "brute force" to match the answer, but I want to understand "why."

Thank you again for your help. Oh, and yes, this is not homework, just to get back into it (self-study).

10th

#### 10th&Bennett

Joined Apr 15, 2018
27
Ah! Thanks @WBahn and @MrChips! I see now where I went wrong. Thanks!

Next topic - Mesh and nodal analysis!

#### WBahn

Joined Mar 31, 2012
26,301
One thing to keep in mind when you are studying them is that mesh analysis is merely a systematic way of applying KVL such that the correct number of independent equations are produced and that KCL is automatically enforced, while nodal analysis is merely a systematic way of applying KCL such that the correct number of independent equations are produced and that KVL is automatically enforced.