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.

We've seen this problem before:

http://www.physicsforums.com/showthread.php?t=513513

 

Thank you Electrician,

Did you also look at the thread I linked to in post#17 here?
This discusses the matter further.

Your link suggests much the same as I have done - that there is only one node in play.

However it is not necessary to invoke Kirchoff at all to solve this problem as they imply. A simple application of parallel and series combinations formula for resistors and Ohm's law correctly applied will do, or the potential divider formula will get you 32 volts. You can, of course use K if you wish and even set up simultaneous equations for the loops and solve them. You will still come to the same conclusion.

 

This question falls into the same trap as the one posted before, i.e. a short circuit across a power supply:

http://forum.allaboutcircuits.com/showthread.php?t=80545

except in this case the subtlety is better hidden.

You cannot have zero ohms across a voltage source and still maintain a non-zero voltage.

The answer is indeterminate. You cannot apply Ohm's Law or Kirchhoff's Laws to find the current through the short.

Add resistance to the short and you have a solution.

The book answer is wrong.

 

Mr Chips.

There is no short circuit in this question. It is legitimate.

I agree the other that you linked to represented an impossible or ill posed arrangement of circuit elements.

 

I understand your argument. You can use Ohm's Law to calculate the current through the series combination of 70R and 5R.

Then you can do the same to calculate the current through 30R and 20R.

Subtracting the two currents gives one answer.

But there are a million other solutions of which no one is the correct answer.

 

Mr C, would you like to review your thoughts here, you usually talk extreme sense?

Thank you though for making me check my sketch with the original question. I seem to have transposed a couple of resistors

OOPs.

The principle remains the same however.


Edit
I have edited the sketch so it now conforms to the original circuit.

 

Ok, I did review.

You can find a solution by inserting a resistance R between the two nodes.
Then work out the answer as R approaches the limit of zero ohms.

I have to run now so I can't do the calculation.

Someone please do this.

 

Nonsense.

The (correct) network reduces to 70 ohms in parallel with 30 ohms = 21 ohms in series with 20 ohms in parallel with 5 ohms = 4 ohms.

4 ohms in series with 21 ohms has 4/25 * 200 = 32volts across it.

32 volts across 4 ohms produces 8 amps.

go well

 

Ok, I did review.

You can find a solution by inserting a resistance R between the two nodes.
Then work out the answer as R approaches the limit of zero ohms.

I have to run now so I can't do the calculation.

Someone please do this.

This method will work but it requires a large amount of work. I tried to do this but finally I got a set of equations 3x3 with a parameter R. I gave up

 

Thank you Electrician,

Did you also look at the thread I linked to in post#17 here?
This discusses the matter further.

Yes.

Your link suggests much the same as I have done - that there is only one node in play.

Did you notice that you were a participant in that thread? As was I.

It's exactly the same situation as the current thread, except for the applied voltage.

However it is not necessary to invoke Kirchoff at all to solve this problem as they imply. A simple application of parallel and series combinations formula for resistors and Ohm's law correctly applied will do, or the potential divider formula will get you 32 volts. You can, of course use K if you wish and even set up simultaneous equations for the loops and solve them. You will still come to the same conclusion.

 

Ok, I did review.

You can find a solution by inserting a resistance R between the two nodes.
Then work out the answer as R approaches the limit of zero ohms.

I have to run now so I can't do the calculation.

Someone please do this.

This method will work but it requires a large amount of work. I tried to do this but finally I got a set of equations 3x3 with a parameter R. I gave up

I worked this out in post #50 of this thread:

http://www.physicsforums.com/showthread.php?t=513513

Just change the applied voltage from 50 volts to 200 volts.

 

Electrician,

You are right, I did post briefly in the middle of that long thread elsewhere.

 

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.

One might adopt an unsymmetrical point of view about Ohm's law and ideal wires.

As has been discussed in a couple of other threads, division by zero is undefined, but multiplication by zero is well defined.

Ohms law can be put in a couple of forms:

E=I*R, or I = E/R

For an ideal wire of zero ohms, if we know the current through the wire, we can calculate the voltage across the wire with Ohms law:

E = I*R, so if R=0 then E = I*0 = 0. That is, 0 = I*0; perfectly well defined.

The voltage across an ideal wire of zero ohms, carrying I amps is zero; the voltage across the ideal wire is zero for any finite current.

On the other hand, even though we know that the voltage across an ideal wire carrying a finite current is zero volts, we can't calculate the current using Ohm's law because:

I = E/R so if R=0, I = 0/0. The current could be any finite value.

If you know the finite current in an ideal wire, you can use Ohm's law to calculate the voltage across the wire, a trivial, determinate, result. But you can't use Ohm's law to calculate the current through the ideal wire by knowing the voltage (zero) across the ideal wire.

 

OK so if you must find the current in the wire then proceed as follows

There are 168 volts across the top two resistors and 32 volts across the bottom two.

So resistor currents are

168/70 = 2.4A
168/30 = 5.6A
32/20 = 1.6A
32/5 = 6.4A

I have shown these in the attached sketch, from where it is clear that 0.8A of the current passing through the 70Ω resistor is diverted through the 5Ω resistor.

You can also get this result by the Electrician's Thevenin method he posted in his link.

 

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.

I'm most certainly not 'moderating' anything. I'm not a moderator and don't have the ability to moderate anything even if I wanted to.

I've stating my opinion and belief, that is all. You are certainly free to agree, disagree, change, not change, or do whatever else you wish (short of doing something that gets an actualy moderator involved, of course).

 

This question falls into the same trap as the one posted before, i.e. a short circuit across a power supply:

http://forum.allaboutcircuits.com/showthread.php?t=80545

except in this case the subtlety is better hidden.

You cannot have zero ohms across a voltage source and still maintain a non-zero voltage.

The answer is indeterminate. You cannot apply Ohm's Law or Kirchhoff's Laws to find the current through the short.

Add resistance to the short and you have a solution.

The book answer is wrong.

Huh?

This is a perfectly valid question with a very clear answer (and, as a bonus, the book answer is even correct) that is easily arrived at through any of a number of methods, none of which involve indeterminate zero voltages across zero ohm resistors. Look at the modified question I posed and you will certainly see no problem with it. Yet it is identical to the question posed in every relevant way.

 

I understand your argument. You can use Ohm's Law to calculate the current through the series combination of 70R and 5R.

Then you can do the same to calculate the current through 30R and 20R.

Subtracting the two currents gives one answer.

But there are a million other solutions of which no one is the correct answer.

Please provide just one other example that satifies KVL, KCL, and Ohm's Law for this circuit that has anything other than 800mA flowing left to right in the indicated branch.

 

Ok, I did review.

You can find a solution by inserting a resistance R between the two nodes.
Then work out the answer as R approaches the limit of zero ohms.

I have to run now so I can't do the calculation.

Someone please do this.

Why?

Are you REALLY saying that if I want to know the current in a wire connecting a 12V battery to a 24Ω lamp filament that I have to insert into the wire that connects two together and take the limit as the resistance goes to zero. Because that is the inevitable conclusion that must be drawn if we have to insert a resistor and let it go to 0Ω in this case.

If nothing else, do straightfoward Mesh Analysis on the darn thing. Define the mesh currents as I1 in left loop, I2 in the top right loop, and I3 in the bottom right loop. All going counter clockwise, makeing the Io we are looking for equal to I2-I3, and the equations representing the sum of the voltage drops around the mesh in the direction of the mesh current.

\(\begin{array}{rcrcrcr}
90\Omega I_1 &-& 70\Omega I_2 &-& 20\Omega I_3 &=& -200V
-\ 70\Omega I_1 &+& 100\Omega I_2 && &=& 0
-\ 20\Omega I_1 && &+& 25\Omega I_3 &=& 0
\end{array}
\)

Which immediately reduce to

\(\begin{array}{rcrcrcr}
9 I_1 &-& 7 I_2 &-& 2 I_3 &=& -20A
-\ 7 I_1 &+& 10 I_2 && &=& 0
-\ 4 I_1 && &+& 5 I_3 &=& 0
\end{array}
\)

Solve these using any technique you like, and you will get

\(
I_1 = -8A
I_2 = -5.6A
I_3 = -6.4A
\
I_0=I_2-I_3=-5.6A\ -\ -6.4A\ =\ 800mA
\)

 

I am a bit confused about Ohm's law. Which form is Ohm's law:
1. \(I = \frac{V}{R} \)
2. \( V = I.R\)
3. \(R = \frac{V}{I} \)

Does Ohm's law only apply for ohmic materials (as WBahn said) not ideal wire?

As my understanding till now is that if we only apply Ohm's law for ohmic materials (R ≠ 0) then all three forms above are equivalent and they are all called Ohm's law.

...
For an ideal wire of zero ohms, if we know the current through the wire, we can calculate the voltage across the wire with Ohms law:

E = I*R, so if R=0 then E = I*0 = 0. That is, 0 = I*0; perfectly well defined.

The voltage across an ideal wire of zero ohms, carrying I amps is zero; the voltage across the ideal wire is zero for any finite current.

On the other hand, even though we know that the voltage across an ideal wire carrying a finite current is zero volts, we can't calculate the current using Ohm's law because:

I = E/R so if R=0, I = 0/0. The current could be any finite value.

If you know the finite current in an ideal wire, you can use Ohm's law to calculate the voltage across the wire, a trivial, determinate, result. But you can't use Ohm's law to calculate the current through the ideal wire by knowing the voltage (zero) across the ideal wire.

If I understand you correctly, you are saying that Ohm's law can be used for ideal wire but the form you are using is V=IR. I think that form is what you mean by Ohm's law.

I think if we consider ideal wires are not ohmic materials and don't apply Ohm's law for it then the problem will be easier.
But how do you know that the voltage drop across an ideal wire is zero?
1. By using Ohm's law: V = IR with R = 0 then V = 0
2. Maybe, based on experiment or something else.

 

Why?

Are you REALLY saying that if I want to know the current in a wire connecting a 12V battery to a 24Ω lamp filament that I have to insert into the wire that connects two together and take the limit as the resistance goes to zero. Because that is the inevitable conclusion that must be drawn if we have to insert a resistor and let it go to 0Ω in this case.

If nothing else, do straightfoward Mesh Analysis on the darn thing. Define the mesh currents as I1 in left loop, I2 in the top right loop, and I3 in the bottom right loop. All going counter clockwise, makeing the Io we are looking for equal to I2-I3, and the equations representing the sum of the voltage drops around the mesh in the direction of the mesh current.

\(\begin{array}{rcrcrcr}
90\Omega I_1 &-& 70\Omega I_2 &-& 20\Omega I_3 &=& -200V
-\ 70\Omega I_1 &+& 100\Omega I_2 && &=& 0
-\ 20\Omega I_1 && &+& 25\Omega I_3 &=& 0
\end{array}
\)

Which immediately reduce to

\(\begin{array}{rcrcrcr}
9 I_1 &-& 7 I_2 &-& 2 I_3 &=& -20A
-\ 7 I_1 &+& 10 I_2 && &=& 0
-\ 4 I_1 && &+& 5 I_3 &=& 0
\end{array}
\)

Solve these using any technique you like, and you will get

\(
I_1 = -8A
I_2 = -5.6A
I_3 = -6.4A
\
I_0=I_2-I_3=-5.6A\ -\ -6.4A\ =\ 800mA
\)

This is exactly what I did but I didn't replace R by 0 in set of equations before solving it.
The result in the case are same but I don't know if this method (replace parameter R = 0 before solving the set of equations) is valid in all cases.
I remember in an old thread The Electrician said that we have to solve set of equations with R then after get result, we find the limit R -> 0.

I tried to solve set of equations 3x3 with R but it took a lot of time and I gave up.