# Current w 0 R is non-0?

Discussion in 'Homework Help' started by DYLH, Sep 18, 2013.

1. ### DYLH Thread Starter New Member

Aug 13, 2013
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Io has to be 0 right? book answer is 800mA?!

I did get 32 for Vo

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Why does Io have to be zero?

3. ### DYLH Thread Starter New Member

Aug 13, 2013
28
1
From ohm's law... i = v/r
There is no resistence between the two 'ends' of the arrow... and the voltage on both ends is the same

The 'line' the arrow is asking about can be collapsed to a point/node... nothing to this point in the book as discussed anything about voltages/currents within a node.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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So in my attachment I have two unequal sources with different series resistances connected by a wire. The wire has the same voltage at both ends and has no resistance. Is there any current in the wire?

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5. ### WBahn Moderator

Mar 31, 2012
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By your reasoning, the Io in the following problem would have to be zero. Right?

If not, then what is different?

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6. ### WBahn Moderator

Mar 31, 2012
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Sure it has. The whole concept of KCL is based on the assertion that, absent net charge accumulation, current that enters a region has to equal the current that leaves a region. That region can be an entire node or, say, the left side of a node that connects two resistors with a horizontal piece of wire.

7. ### BrianH Active Member

Mar 21, 2007
43
0
The easiest way to solve this problem is as follows.

If 32V is across V0, then 32V is also across the 20R resistor. This means that 1.6A must be flowing through the 20 ohm resistor.

Next, if 32V is across the 20R then it follows that the remaining 168V must be being dropped across the 70R. With 168V dropped across the 70R 2.4A will flow through it.

If 2.4A flows through the 70R, but only 1.6A flows through the 20R, then the difference must flow through the path of I0. The difference is 800mA.

This, of course, is an approximation. In practice the piece of wire you mention will not be equal to zero ohms. There will be a small resistance, and this will create a slight potential difference between the two ends of the wire. This will allow the (approximately) 800mA to flow.

Last edited: Sep 18, 2013
8. ### anhnha Active Member

Apr 19, 2012
776
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But 0/0 is indeterminate.

9. ### WBahn Moderator

Mar 31, 2012
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Solving the problem for the OP and dishing it up on a silver platter may well be the easiest way for them to get the solution, but here in Homework Help we try to help the OP learn by guiding them toward a solution while having them do the work.

Consider that, most of the time, they have seen these concepts in class as well as in their textbook and have very likely seen examples worked in class and in the textbook. There is no reason to think that seeing yet one more example worked by someone else is suddenly going to make everything crystal clear. It might, but it usually won't. Instead, the person has some kind of a mental block and the best way for them to get past it is to run straight into it and have to deal with it. We can usually best help them by steering them along a path that will force them to run straight into it in such a way that it is obvious to them what it is that they have to contend with.

10. ### WBahn Moderator

Mar 31, 2012
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Yeah. So?

That just means that you have to look elsewhere to see if a solution can be found.

Again, consider my modified problem Post #5. The exact same reasoning leads to the exact same quandry in that you have 0V/0Ω. Does that mean either that there is no current flowing in that wire or that we can't figure out what it is?

11. ### anhnha Active Member

Apr 19, 2012
776
48
No, it is clearly that there is a current flowing through the wire.
$I = \frac{V}{ R_{tot} }$
In your modified picture: V = 200V and $R_{tot} = (70 \Omega || 30 \Omega ) + (20 \Omega || 5 \Omega )$
I just can't figure out how to calculate how to apply Ohm's law to calculate current for an ideal wire.
With Ohm's law: V = I.R
As R = 0Ω => V = 0 and the current I is determined by other components in the circuit.
Maybe, there is another form of Ohm's law be used to determine I but I don't know how to find out E here: $j = \sigma .E$

12. ### WBahn Moderator

Mar 31, 2012
18,087
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The OP's problem was summarized as follows:

"There is no resistence between the two 'ends' of the arrow... and the voltage on both ends is the same"

Well, how is that ANY different than the modified problem? Is there any resistance between the 'ends' of the arrow? Isn't the voltage on both ends the same?

So we have a 0Ω wire with 0V across across it, yet you managed to figure out how much current was flowing in that wire with no difficulty at all and without resorting to some other form of Ohm's law.

The same with the original problem.

How much current is flowing downward in the 70Ω resistor. How much current is flowing downward in the 20Ω resistor? If those two numbers are not the same, where does KCL insist that the rest of the current HAS to go?

13. ### anhnha Active Member

Apr 19, 2012
776
48
I have another question not sure.
Does the voltage across 70Ω resistor cause charges to move and a current formed flowing through it ?
I think this is not true, otherwise the same reasoning can be applied for Io and if so then Io would be zero.

14. ### #12 Expert

Nov 30, 2010
16,685
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Since you have the right answer at 32 volts, figure the current through all 4 resistors. When you see the answers, that should lead you to the next step.

15. ### WBahn Moderator

Mar 31, 2012
18,087
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The thing that you need to understand is that 0/0 is "indeterminate", which simply means that, based on that knowledge alone, we cannot determine what the answer is. May it is zero, maybe it is infinite, maybe it is 42, may it could take on any value in the range 50 to 2000 that is the ratio of two prime numbers. We don't know. We can't determine it based on just knowing that we have 0/0.

That does not mean, however, that we can't determine the answer at all. We can use other information and other constraints to determine the answer. We do this all the time. You did it when you gave an expression for finding the current using the voltage source's voltage and the equivalent resistance of the four resistors.

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16. ### DYLH Thread Starter New Member

Aug 13, 2013
28
1
Thanks, t_n_k, just got back to the forum after reading only your post earlier. After chewing on it a bit while working on other things, I figgered it out.

17. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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It is well worth cultivating the skill of redrawing a circuit to bring out the necessary relations.

I have done so in two ways in the attachment.

The important thing to realise is that the strapwire and all its connections are one single node for the purposes of circuit analysis. This is shown inside the dotted boundary on my first sketch.

So there is no question of 0/0 or infinity in the calculation.
Nor is there any voltage issue because the whole of the node is defined as being at the same potential. (yes 32 volts)

This is the first redrawing in my sketch. I have contracted everything inside the dotted boundary to a point node..

The second one shows the node expanded again but differently configured to bring out the two parallel paths conneced in series across the source.

This sort of question confuses many.

It may be worth looking at this thread where even the experienced have been bamboozled.

Edit I have noticed that I inadvertantly transposed some of the resistor values.

I have updated the sketches to correct this and to show the analysis method.

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Last edited: Sep 21, 2013
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18. ### anhnha Active Member

Apr 19, 2012
776
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Yes, I can do it but sometimes I think why we can't apply Ohm's law for an ideal wire (zero resistance). And that is confusing.

19. ### WBahn Moderator

Mar 31, 2012
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The reason you can't apply Ohm's law for an ideal wire is because Ohm's law only applies to ohmic materials -- namely materials in which the flow of current through them is proportional to the voltage across them. This isn't the case for lots of devices. It's not the case for a capacitor, it's not the case for an inductor, it's not the case for a diode, and it's not the case for an ideal wire. So stop trying to apply Ohm's law to devices for which it simply doesn't apply.

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20. ### BrianH Active Member

Mar 21, 2007
43
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My apologies; I felt that my post served an explanatory purpose. I didn't just give some mathematical answer, I explained the reasoning and logic behind this kind of problem. I felt that this was the best way to provide understanding. However, since you are 'moderating' this thread on the forum I will respect your opinion and simply avoid this thread from now on.

Thanks,

Brian.

Last edited: Sep 21, 2013