# Short or open?

1. Jul 17, 2013

### peripatein

Hi,
Would someone care to please explain why no current passes through R1 (see attachment) in case Vs is replaced with a short and the inductor is considered a short too (in the steady state, for DC)? Why would the R1 branch be open and not short? Is it not in parallel with short under the above circumstances?

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2. Jul 17, 2013

### skeptic2

If Vs is replaced with a short, what is the voltage across R1?

3. Jul 17, 2013

### szynkasz

Yes, it is parallel with short so both termination of the resistor have the same potential. Therefore there is no current.

4. Jul 17, 2013

### peripatein

Why open? Doesn't zero potential mean short, rather than open? You can measure a potential between two terminations of an open branch, and the answer would not necessarily be zero. Whereas open means infinite resistance, hence current is zero.

5. Jul 17, 2013

### szynkasz

It isn't open but bypassed by short.

6. Jul 17, 2013

### peripatein

Alright, so let me ask this then: when may a component in parallel with short be rendered a short?

7. Jul 17, 2013

### szynkasz

This is not short nor open. Just: no potential difference = no current.

8. Jul 17, 2013

### peripatein

I got that, I am now asking a more general question, namely: when does a short turn a component in parallel to it into a short? In other words, if I take a component and connect it in parallel to a short - under what circumstances will that component, for calculation purposes, may be considered a short?

9. Jul 17, 2013

### szynkasz

Under any circumstances.

10. Jul 17, 2013

### peripatein

Well, why isn't that then what we have in the circuit in the attachment? Isn't R1 in parallel to a short? Following that reasoning why won't it then be considered a short itself, for calculation purposes?

11. Jul 17, 2013

### Floid

I think there is a slight misconception thinking of the situation as either a short or an open. An open can have a potential difference but no current flows, a short has no potential difference but has current flow.

In the R1 branch you have both no potential difference (because the two ends are connected by a short) and no current flow (because all the current flow will take the short due to it having less resistance).

12. Jul 17, 2013

### szynkasz

For calculation purposes you can replace it with a short. This branch doesn't affect rest of the circuit.

13. Jul 17, 2013

### peripatein

But if the resistor is considered a short (or may be considered a short), then both the branch with the inductor is a short and the branch with R1. Therefore, why would the current "prefer" flowing through the branch of the inductor? I mean, are they not both merely wires (for calculation purposes)?

14. Jul 17, 2013

### szynkasz

No, you treat whole part of circuit with inductor and resistor as a short. Current flows through inductor - because it is a short for DC - and doesn't flow through resistor.

15. Jul 17, 2013

### peripatein

Please don't get annoyed by all my questions (in case you are); am simply trying to understand this properly.
Didn't you just write that the resistor may be considered a short as well? Why won't the current consider both the resistor and the inductor to be a short? If it does, why then would the inductor (now a short, but so is the resistor) get current and not the resistor (if they are both a short then both of them have zero resistance)?

16. Jul 17, 2013

### szynkasz

I'm not annoyed Inductor has no DC resistance so it is a real short. This causes no voltage on resistor and thus no current. So you can get rid of resistor and nothing changes in the rest of the circuit. This is what I mean.

17. Jul 17, 2013

### peripatein

I think I truly got it now. Thank you very much! :-)

18. Jul 17, 2013

### psparky

When using superposition in this case, the AC voltage source is shorted. The inductor in that branch also turns into a short in DC steady state.

So when current travels from the DC source, it first sees the cap which is an "open" in steady state DC. So it ignores it and travels to VA. At VA it has a choice between a non resistive wire and a resistor. It surely takes the path of zero resistance and has no interest in the resistor. So the short we just created has zero resistance, so there is no voltage there. Since it's parrallel to that resistor, there is zero voltage acrross resistor, therefore zero current thru resistor R4 according to V=IR.

Lets do current division as well. We'll call your inductor and VS sources (now shorted for superpostion) R0.

So we have R4=10 ohms and R0=0 ohms.

Let's just say that the current heading into node VA from DC source is 5 amps (just random guess for this exercise)

So current division thru R4 would be:
5*(0/(10+0)= 0 current thru R4.

So that's the DC case. The VAC source is going to put current thru each and every branch!

Last edited: Jul 17, 2013
19. Jul 17, 2013

### peripatein

That was a very clear explanation. Thank you, psparky!

20. Jul 17, 2013

### psparky

I should have figured the exact amps coming from DC source....is very simple in the case.

It's just gonna be 10 volts/50 ohms which equals .2 amps leaving DC source......and .2 amps going back to source. (KCL)

50 ohms is the only resistance it encounters in this loop.

21. Jul 17, 2013