Finding voltage and current in a somewhat unusual circuit

In summary: Kirchoffs current law.Yep.In summary, the back of the book agrees with you about the voltage, but not about the current. The answer given is 0.2 A. I think there's been a typo. They probably mean 2 A, because that's the total current I found; of course, the current can't be zero, because then the circuit would be open...
  • #1
chever
8
0
[Disclaimer: I'm pretty sure this doesn't belong in the HW questions forum. I'm a linguistics major not taking any EE courses and it's the summertime. I just happen to be using an undergrad EE text.]

For the life of me I cannot crack the following:

82qEk.png


Since I believe the 70 and 30 ohm resistors are in parallel, likewise the 20 and 5 ohm resistors, I tried finding the total current through the circuit by combining them and then combining the two equivalent in series but I'm not even sure I can do that and I really am just stuck at this point.

Your help would be greatly appreciated.
 
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  • #2
Yep, you can do that.
 
  • #3
Ok, I did. So I believe the total current is 2 amps. Is that right? Even if I've got that right, I can't seem to figure out how the current is divided up in the network. I tried to apply KCL with incorrect results. Stuck...
 
  • #4
Hmm... how to crack it?
Io is zero, because the line in which it "flows" is just connection between two nodes, You can simplify it by reducing it to one node.
How to get Vo ? (Vo is the voltage that appears at the same time at 20 ohm and 5 ohm resistors). Look at the circut, distribution of voltage is that: 50=I*(70||30+20||5)
The amount of voltage on resistors are proportional to the restistance so Vo=50*(R1/R1+R2) where R1=20||5=4ohm, R2=70||30=21ohm. Vo=8V
 
  • #5
gerbi said:
Hmm... how to crack it?
Io is zero, because the line in which it "flows" is just connection between two nodes, You can simplify it by reducing it to one node.
How to get Vo ? (Vo is the voltage that appears at the same time at 20 ohm and 5 ohm resistors). Look at the circut, distribution of voltage is that: 50=I*(70||30+20||5)
The amount of voltage on resistors are proportional to the restistance, so Vo=50*(R2/R1) where R1=70||30=21ohm and R2=20||5=4ohm. so Vo=8V

The back of the book agrees with you about the voltage, but not about the current. The answer given is 0.2 A. I think there's been a typo. They probably mean 2 A, because that's the total current I found; of course, the current can't be zero, because then the circuit would be open...

So this might be resolved?
 
  • #6
chever said:
The back of the book agrees with you about the voltage, but not about the current. The answer given is 0.2 A. I think there's been a typo. They probably mean 2 A, because that's the total current I found; of course, the current can't be zero, because then the circuit would be open...

So this might be resolved?

Are You sure about the current Io can't be zero ? The two nodes which are connected with this simply line (where Io "flows").. what is the resistance of this line ? (since ALL the resistances in the circut are represented by a resistor, no other resistances are allowed). Then what is the current Io? No resistance, no voltage drop. Infinity ? Zero ?
 
  • #7
In any case, something's not quite right with the answer given for this problem. Maybe I should tell the publisher.
 
  • #8
chever said:
In any case, something's not quite right with the answer given for this problem. Maybe I should tell the publisher.

Chever, the answer given in the book is not correct.. I find it not very unusual.

Anyway, this thread shouldn't be on this forum. Moderator will move it soon I guess
 
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  • #9
WAIT before you leave.
your book is wonderful. The answer is Indeed 0.2A.
 
  • #10
Here is How.
1. First 70 in parallel with 30 is reduced to 21 Ohm.
2. 20 in Parallel to 5 is reduced to 4 Ohm.
3. Now, We find total current is 50/(21+4) = 2A.
4. We Can now use current divisor formula to find current through all the resisters
Now, it happens that current through 70 Ohm is 0.6A and through 20 Ohm is 0.4A.
So, rest 0.2A flows through the Path marked Io.
Find the current through all resisters. It will be consistent with Kirchoffs current law.
 
  • #11
I agree. I solved circuit and got I0=0.2A which is correct answer-

If EE isn't your major, you probably can't solve circuit like this. We electrical engineer have our specific circuit analysis methods for circuits like that.
 
  • #12
@gerbi How did you come to believe that Io is zero simply because its a connection of two nodes?
And Its not unusual to find people blaming the book when they can't find the answer. Revise Yourself ten times before blaming on the author.
 
  • #13
I_am_learning said:
Here is How.
1. First 70 in parallel with 30 is reduced to 21 Ohm.
2. 20 in Parallel to 5 is reduced to 4 Ohm.
3. Now, We find total current is 50/(21+4) = 2A.
4. We Can now use current divisor formula to find current through all the resisters
Now, it happens that current through 70 Ohm is 0.6A and through 20 Ohm is 0.4A.
So, rest 0.2A flows through the Path marked Io.
Find the current through all resisters. It will be consistent with Kirchoffs current law.

Really strange way to solve problems. What is the voltage drop in the line where Io flows ? (since it is no resistance). You see, that's the diffrence.. When you will asume there is one node (not two, where Io "flows" between them), what You can surely do, there is no current.
 
  • #14
I_am_learning said:
@gerbi How did you come to believe that Io is zero simply because its a connection of two nodes?
And Its not unusual to find people blaming the book when they can't find the answer. Revise Yourself ten times before blaming on the author.

the reason is simple, all the resistances in this circut are represented by resistors. If branche have no resistance it can be simplified to one node. Would You agree ?

If You will look ath cicrut Your way, two nodes, which is incorect for me, then Your solution is good

Another edit of this post:I am electrical engineer. My answer might be strange for You guys because I'm from Europe.. and I suppose most of ppl here are US. Maybe you analyze circuts in some other way.. dunno
One more thing, I was little hasty saying the answer in the book is wrong. It's correct when You assume some.. really strange, and unacceptalbe (in my opinion), assumptions
 
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  • #15
I_am_learning said:
4. We Can now use current divisor formula to find current through all the resisters
Now, it happens that current through 70 Ohm is 0.6A and through 20 Ohm is 0.4A.
So, rest 0.2A flows through the Path marked Io.

I'm with you up to step 4

can you explain how current division was used here in more detail?

OkuuEE said:
If EE isn't your major, you probably can't solve circuit like this. We electrical engineer have our specific circuit analysis methods for circuits like that.

apparently Kirchoff's laws and Ohm's law are adequate for solving this circuit

and I've gotten everything else right up until now
 
  • #16
chever said:
I'm with you up to step 4

can you explain how current division was used here in more detail?



apparently Kirchoff's laws and Ohm's law are adequate for solving this circuit

and I've gotten everything else right up until now

1. If You have two parallel branches, like here 70 and 30 ohm and current 2A, the flowing current will divide into two branches like this I(70ohm)=2A*(30/70+30) and I(30ohm)=2A*(70/70+30) so it's 0,6 A and 1,4 A (2A in total).

2. Ohm's and two Kirhoff laws are enought for this.. but You still need to remember about some basic pronciples
 
  • #17
chever said:
apparently Kirchoff's laws and Ohm's law are adequate for solving this circuit

and I've gotten everything else right up until now
Ofcourse those methods come from those 3 basic laws.

Check http://en.wikipedia.org/wiki/Current_divider for current division. Total current of circuit is 2A so we can solve current flowing through 70 ohm resistor. Same can be done for 20 ohm resistor and then we just use KCL to solve I0.
 
  • #18
gerbi said:
1. If You have two parallel branches, like here 70 and 30 ohm and current 2A, the flowing current will divide into two branches like this I(70ohm)=2A*(30/70+30) and I(30ohm)=2A*(70/70+30) so it's 0,6 A and 1,4 A (2A in total).

I got that far. But I don't know how to proceed from there.

Remember, I just started learning circuits a week ago.
 
  • #19
chever said:
gerbi said:
1. If You have two parallel branches, like here 70 and 30 ohm and current 2A, the flowing current will divide into two branches like this I(70ohm)=2A*(30/70+30) and I(30ohm)=2A*(70/70+30) so it's 0,6 A and 1,4 A (2A in total).

I got that far. But I don't know how to proceed from there.

Remember, I just started learning circuits a week ago.

Then You have a long way to go ;)

What next ? If You want to go by way selected by I_am_learning ..
Current in the second parralel brach (20 and 5 ohm) is 2A in total (just like in the first one) You will need to find how the current flows, what is the distribution.
Like before I(5ohm)=2A*(20/20+5)=1,6A and I(20ohm)=2A*(5/20+5)=0,4A. Write it on the schematic of Your circrut. Now You see that some of the current in 70Ohm branch (it's 0,2 A) must flow to 5Ohm branch. So it must flow thru "Io line". Hope it's clear.

But like I said, for me it won't work. The reasons are like I stated before. For me Io=0A (How You would use Kirchoff law for this "Io nodes". It needs to be one node to use Kirchoff law!). But whatever.. the book says something else
 
  • #20
gerbi said:
the reason is simple, all the resistances in this circut are represented by resistors. If branche have no resistance it can be simplified to one node. Would You agree ?

If You will look ath cicrut Your way, two nodes, which is incorect for me, then Your solution is good

Another edit of this post:I am electrical engineer. My answer might be strange for You guys because I'm from Europe.. and I suppose most of ppl here are US. Maybe you analyze circuts in some other way.. dunno
One more thing, I was little hasty saying the answer in the book is wrong. It's correct when You assume some.. really strange, and unacceptalbe (in my opinion), assumptions

Yeah, I agree. There is only one node there. I am not assuming two nodes while solving the currents in the resisters. But in this case the node has been stretched out into to points. Still the sum of incoming current to this stretched node is equal to sum of out-going current from this stretched node. I am simply applying kirchhoffs current rule to the Point 1 (out of two points of stretched node) to find the current flowing from one point of the node to the other point of same node.
 
  • #21
I_am_learning said:
Yeah, I agree. There is only one node there. I am not assuming two nodes while solving the currents in the resisters. But in this case the node has been stretched out into to points. Still the sum of incoming current to this stretched node is equal to sum of out-going current from this stretched node. I am simply applying kirchhoffs current rule to the Point 1 (out of two points of stretched node) to find the current flowing from one point of the node to the other point of same node.

Ok. there is only one node.. then what is the value of Io current ? (since it's flowing between two nodes assumed as one). Ahh.. yes, first there are one node (to use Kirchoff law) and then there are two nodes ? You have to choose: one node Io=0A, use Kirchoffs.. or ? Or what ?
 
  • #22
gerbi said:
Ok. there is only one node.. then what is the value of Io current ? (since it's flowing between two nodes assumed as one). Ahh.. yes, first there are one node (to use Kirchoff law) and then there are two nodes ? You have to choose: one node Io=0A, use Kirchoffs.. or ? Or what ?
Did you not get the point of 'Stretched out' node?
Io is current inside the node. Not from one node to another node.
 
  • #23
I_am_learning said:
Did you not get the point of 'Stretched out' node?

I guess not, because there is no such a thing in Electrical Engineering. What is the scientific explanation behind that?

Whoah.. current in node ? NODE IS A POINT !
 
  • #24
Do you use simulation software? You may like to simulate it.
Also, see this.

On the original figure, mark a point on the top side of 70ohm res, name it A.
Also mark the top side of 30ohm res, name it B.
A and B points are the same node. Now are you telling me that no current flows in the wire segment AB?
 
  • #25
I_am_learning said:
Do you use simulation software? You may like to simulate it.
Also, see this.

On the original figure, mark a point on the top side of 70ohm res, name it A.
Also mark the top side of 30ohm res, name it B.
A and B points are the same node. Now are you telling me that no current flows in the wire segment AB?

Please, don't tell me about simulation. Computer can calculate anything if human is stupid enought to assume it's correct. You will need to remember some basic principles.

Segment A-B.. there is no A-B segment because A and B are the same. Node is a point where You can use Kirchoff law. Kirchoff law won't work for segment.

Dear Sir, please rethink all the things You are telling here. You seriously need to improve Your knowledge (don't take that as offence, I don't mean to).

I won't be answering any further (unless anyone other will post an answer).
 
  • #26
The mesh (loop) method can be used to solve this network.

If the fact that Io is carried in a wire and not a resistor seems confusing, try this:

Replace the wire with a resistor of value R and solve the network; obtain an expression for the current in the resistor R. I get 250/(63*R+1250). If the value of R is allowed to approach zero, then in the limit the current will be 250/1250 = 1/5 = .2
 
  • #27
gerbi said:
Segment A-B.. there is no A-B segment because A and B are the same. .
I was talking physically. If in the real world, I arranged circuit exactly like in the figure and attached paper labels A and B, then would you still deny the presence of wire segment AB.
Get practical.
Simulation was suggested to you to inform you that you are wrong from yet another source.
You don't need to reply if you don't want.
I have full rights to speak though.
 
  • #28
The Electrician said:
The mesh (loop) method can be used to solve this network.

If the fact that Io is carried in a wire and not a resistor seems confusing, try this:

Replace the wire with a resistor of value R and solve the network; obtain an expression for the current in the resistor R. I get 250/(63*R+1250). If the value of R is allowed to approach zero, then in the limit the current will be 250/1250 = 1/5 = .2

Guys.. we are talking here about circuts.. on paper. Kirhoff law is a nodal law. Nod means point. Basic principle of that kind of analysis is: line is only conecting, it means both ends of simply line is COMPLETLY the same.
 
  • #29
I_am_learning said:
I was talking physically. If in the real world, I arranged circuit exactly like in the figure and attached paper labels A and B, then would you still deny the presence of wire segment AB.
Get practical.
Simulation was suggested to you to inform you that you are wrong from yet another source.
You don't need to reply if you don't want.
I have full rights to speak though.

Physically.. Oh right. Physically both ends of A-B segment is completely the same (no resistance, no phase shift, no lenght, nothing.. segment it's just on the paper my friend to inform that there is connection).

Pracitical ? Know what ? there is no perfect conductor in practice. there is no node in practice. there are no lines in practice. How about that ? There is no real world here.
That kind of analysis is done by some assumptions, You violate one of them. Want to be real practical ? use some field soulution software. If You use circut methods respect assumptions.

Another source.. please.. software is set by You, it's still the same source.

You have right to speak, as we all.
 
  • #30
Goodness, folks.

Surely it is obvious this question was deceptively simple, though not to the level of a trick question.
That it was deliberately designed to highlight the issue now being debated. That blind use of kirchhoffs laws can get you into trouble.

You have two voltage dividers of unequal ratio in parallel across a common supply with a direct connection between their division points.

You can regard this as an unbalanced bridge or solve it as I am learning has done.
 
  • #31
gerbi said:
1. If You have two parallel branches, like here 70 and 30 ohm and current 2A, the flowing current will divide into two branches like this I(70ohm)=2A*(30/70+30) and I(30ohm)=2A*(70/70+30) so it's 0,6 A and 1,4 A (2A in total).

2. Ohm's and two Kirhoff laws are enought for this.. but You still need to remember about some basic pronciples

Im sorry sir but the current in that branch doesn't have to be 0. Especially when you have uneven distribution of resistors. If the circuit was symmetric maybe it could be zero.

I0=200 mA. I did it by applying those Kirchhoff's rules that you spoke of. V0 = 8V.

When you put those 2 points together you are essentially making them have same electric potential. So in order circuit to do that, "it sends some current through that blank wire to equalize those 2 points".

If there was no blank wire, those 2 points wouldn't be on the same electric potential and you could equivalent 70 and 20 in series, which you cannot do in this case.And just to be SURE of my results I simulated this circuit confirming my results.
 

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  • #32
Ok.. guys..

What means if there is a line between two points in schematic? Electrical connection, no resistance.. Points are meant to be the same. Correct ? It'a a node.

How currnet flows thru a node ? when there are no voltage drop. Please answer me.
 
  • #33
gerbi said:
Ok.. guys..

What means if there is a line between two points in schematic? Electrical connection, no resistance.. Points are meant to be the same. Correct ? It'a a node.

How currnet flows thru a node ? when there are no voltage drop. Please answer me.

Current has to go somewhere. And can't you see that problem has 2 nodes and not 1?

0,2 A will flow through that wire. And do you not see that you can't make 1 node out of 2? It changes the problem. That is not the same, problem ASKS to calculate the current through WIRE. Wire exists. If you remove the wire, of course current won't flow through something that doesn't exist. That is one node and its KR is:
1,4+0,6-0,4-1,6=0
 
  • #34
Bassalisk said:
Current has to go somewhere. And can't you see that problem has 2 nodes and not 1?

0,2 A will flow through that wire. And do you not see that you can't make 1 node out of 2? It changes the problem.

How I can't ? Is there any resistance ?

And again.. How current flows when there is no voltage drop ?

there is no answer like "current has to go somewhere"
 
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  • #35
If there is a wire with 1 micro-ohm of resistance, would you be able to solve the problem?

Suppose you then allow the resistance of the wire to become less and less until it becomes zero ohms as if it were a superconducting wire. Wouldn't there still be a current in that wire?

Go back and look at what I said in post #26,
 

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