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Tricky circuit

by DaTario
Tags: circuit, tricky
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DaTario
#1
Mar4-14, 07:31 AM
P: 555
Hi All,
Concerning the first circuit in the attached figure:

By applying Kirchhoff laws we arrive at two different currents in the central resistor. But if we consider the second circuit with its auxiliar resistances we see no problem with the definition of the central current.

Is it due to the fact that batteries always have some resistance?
Infinite currents will appear in the side branches of this first circuit, but even so is there any way to get the appropriate answer to the question of what is the current in the central resistor?

I am not sure I have understood well the superposition principle which may well be the method to be used here (in the first circuit).

Thanks,

Best Wishes,

DaTario
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tricky electric circuit.jpg  
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DaleSpam
#2
Mar4-14, 07:38 AM
Mentor
P: 17,213
Quote Quote by DaTario View Post
Infinite currents will appear in the side branches of this first circuit, but even so is there any way to get the appropriate answer to the question of what is the current in the central resistor?
No. The first circuit has no solution. It is a self contradiction. You cannot have one node that is both 10 V and 20 V.

Superposition doesn't help. If you use the principle of superposition then you get two self contradictory circuits. The first where one node is both 10 V and 0 V and the second where one node is both 20 V and 0 V.
maajdl
#3
Mar4-14, 07:56 AM
PF Gold
P: 370
By applying the Kirchhoff law to the first circuit, you do not get two different currents.
Instead you come to an impossibility (which is 20V=10V).
Analyzing what happens makes sense only by comparing to a similar circuit that has been modified to avoid this.
The second circuit allows you to think so.

Applying the superposition principle to the first circuit leads twice to impossibilities (like 0V = 10V).

To use the superposition principle:

calculate the current in the central resistor when the left battery is replaced by a 0V battery,
calculate the current in the central resistor when the right battery is replaced by a 0V battery,
add the two and you will get the result with both batteries in the circuit

make sure you understand why this principle holds

DaTario
#4
Mar4-14, 08:35 AM
P: 555
Tricky circuit

Thanks Mr. DaleSpan, but what is the reason to this status of having no solution?




Quote Quote by DaleSpam View Post
No. The first circuit has no solution. It is a self contradiction. You cannot have one node that is both 10 V and 20 V.

Superposition doesn't help. If you use the principle of superposition then you get two self contradictory circuits. The first where one node is both 10 V and 0 V and the second where one node is both 20 V and 0 V.
DaleSpam
#5
Mar4-14, 08:42 AM
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Quote Quote by DaTario View Post
Thanks Mr. DaleSpan, but what is the reason to this status of having no solution?
Let v be the voltage at the top and let i be the current through the resistor. Then this circuit has the following equations:
v=20
v=2i
v=10

This is three linear equations in two unknowns. The system is over determined. Over determined systems of equations either have no solution or the equations are not linearly independent. In this case the first and third equation have no solution, so the system has no solution.
DaTario
#6
Mar4-14, 08:42 AM
P: 555
Hi maajdl,

Quote Quote by maajdl View Post
By applying the Kirchhoff law to the first circuit, you do not get two different currents.
Instead you come to an impossibility (which is 20V=10V).
I was considering the kind of separation used in applications of the superposition principle.



Quote Quote by maajdl View Post
Analyzing what happens makes sense only by comparing to a similar circuit that has been modified to avoid this.
The second circuit allows you to think so.
I thought we could legitimate the first circuit by accepting those infinite currents while focusing in the fact that the limiting case of the second circuit when the auxiliary resistances go to zero yields a well defined value for the central current (current in the central resistor).

Best Wishes,

DaTario
DaTario
#7
Mar4-14, 08:52 AM
P: 555
Quote Quote by DaleSpam View Post
Let v be the voltage at the top and let i be the current through the resistor. Then this circuit has the following equations:
v=20
v=2i
v=10

This is three linear equations in two unknowns. The system is over determined. Over determined systems of equations either have no solution or the equations are not linearly independent. In this case the first and third equation have no solution, so the system has no solution.
Now it seems prety well supported. Thank you.
DaTario
#8
Mar4-14, 09:01 AM
P: 555
Consider the attached graphic. It is current versus the value of the auxiliary resistance (both r1 and r2 were taken as equal). The red curve is the central current, the blue and the green are the currents in the side branches. The blue one corresponds to the branch with the smaller battery (10 V). Observe:

1) the convergence to 7.5 A when we take the limit of r -> 0,

and

2) ther fact that the current in the left side branch inverts its direction at r = 2.

Best wishes,

DaTario
Attached Thumbnails
tricky circuit 2.jpg  
dauto
#9
Mar4-14, 09:14 AM
Thanks
P: 1,948
The first circuit is an example of a short circuit. That's why it is potentially dangerous to connect different batteries in parallel with each other.
sophiecentaur
#10
Mar4-14, 11:19 AM
Sci Advisor
Thanks
PF Gold
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P: 12,132
Quote Quote by DaTario View Post
Hi maajdl,



I was considering the kind of separation used in applications of the superposition principle.





I thought we could legitimate the first circuit by accepting those infinite currents while focusing in the fact that the limiting case of the second circuit when the auxiliary resistances go to zero yields a well defined value for the central current (current in the central resistor).

Best Wishes,

DaTario
This is an example of a situation where the mathematics produces a result that does not fit the real world. There are many of these, where you divide by zero, add +∞ and -∞ or some other invalid step. You need not be surprised that the answer you get doesn't make Physical sense.
DaleSpam
#11
Mar4-14, 12:41 PM
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P: 17,213
Quote Quote by DaTario View Post
Consider the attached graphic. It is current versus the value of the auxiliary resistance (both r1 and r2 were taken as equal). The red curve is the central current, the blue and the green are the currents in the side branches. The blue one corresponds to the branch with the smaller battery (10 V). Observe:

1) the convergence to 7.5 A when we take the limit of r -> 0,

and

2) ther fact that the current in the left side branch inverts its direction at r = 2.
Yes, the second circuit has a solution, no question about that. And you might be tempted to take that limit as r→0 as the solution to the first circuit, but that would be a mistake.

That limit depends strongly on the assumption that r1 and r2 are equal. If you violate that assumption then you do not get the same limit. In fact, if r1 is any amount larger than r2 then you get i = 5 A, and if r2 is any amount larger than r1 you get i = 10 A, both as the smaller resistance approaches 0.

This kind of instability in the solutions of limiting cases is typical of circuits that do not have a solution.
CWatters
#12
Mar4-14, 04:56 PM
P: 3,135
Aside: You shouldn't connect two ideal current sources in series either :-)
sophiecentaur
#13
Mar4-14, 05:09 PM
Sci Advisor
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PF Gold
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P: 12,132
Quote Quote by CWatters View Post
Aside: You shouldn't connect two ideal current sources in series either :-)
And never leave the output terminals of a current transformer on a HV distribution system un-terminated. I really didn't get it when my Dad used to tell me that. It was only years afterwards that it struck me why.
DaTario
#14
Mar4-14, 08:05 PM
P: 555
Thank you all, I guess a sufficiently good explanation was given here.

Best Regards,

DaTario


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