- #1
How to Transform a Circuit to One Resistor?
- Thread starter builder_user
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Homework Help Overview
The discussion revolves around transforming a circuit into an equivalent representation with a single resistor. The context involves circuit analysis techniques, specifically focusing on Delta-Y transformations and Thevenin equivalents.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants explore various methods for simplifying the circuit, including Delta-Y transformations and the application of voltage sources to analyze current flow. Questions arise regarding the implications of open circuit conditions and the relationships between different circuit elements.
Discussion Status
The discussion is active, with participants providing insights into different approaches and clarifying concepts. There is a focus on understanding the implications of circuit configurations and the behavior of current in open circuits, though no consensus has been reached on a single method for transformation.
Contextual Notes
Participants express uncertainty about the definitions and implications of certain terms, such as "imagining" a voltage source, and question the conditions under which current flows in the circuit. The conversation reflects a mix of theoretical exploration and practical application of circuit analysis principles.
- #2
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you haven't said anything about what you know or what you have tried. No one is going to just DO it for you, but I'm sure folks will be happy to help you learn how to do it if you say what you know and what you've tried. "Don't know squat and haven't tried anything" is not likely to get you any help.
- #3
- #4
gneill
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Yes, a Delta-Y transformation will get you there.
Another approach is to imagine a voltage source V across the terminals and solve the loop equations for the current in the voltage source loop. The ratio V/I is the equivalent resistance.
Another approach is to imagine a voltage source V across the terminals and solve the loop equations for the current in the voltage source loop. The ratio V/I is the equivalent resistance.
- #5
- #6
gneill
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Imagine: pretend; postulate; suppose
- #7
builder_user
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gneill said:
I mean in equatations.
Use U or some "image U(E)"?
- #8
gneill
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builder_user said:
I meant to suggest that you imagine that you've connected a voltage source V (or U, or E, or whatever variable name you wish) at the open terminals where you want to find the resistance. If you solve for the current that this voltage produces, then you can find the resistance as V/I.
- #9
- #10
gneill
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builder_user said:
Yes, exactly.
- #11
builder_user
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gneill said:
But how can it help?What loop includes E?
- #12
gneill
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builder_user said:
E --> R3 --> R4 --> back to E
- #13
builder_user
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A!I see now.gneill said:
What's with 5# scheme?
- #14
gneill
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builder_user said:
You could use a similar method, solving for the current in the final loop to find the voltage across R5. Once again it's just three loops to be concerned with.
As an alternative, I might be tempted to start from U1 and build Thevenin equivalents, absorbing more and more of the network as I progress across, until I'd be left with one Thevenin voltage and resistance.
- #15
gneill
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By the way, in your second circuit, does resistor R3 have a value?
- #16
- #17
gneill
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builder_user said:
What happened to diagonal resistor R4? It should make two loops out of the "square" R3-R2-R5-R6.
You don't want U2 to be in a loop: it's a 'measurement' that you want to make, not a source voltage. It happens to be the voltage across R5.
- #18
gneill
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- #19
builder_user
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gneill said:
Delta-y transformation
- #20
gneill
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builder_user said:
Ah. I see. So now you're down to a situation with just two loops to worry about.
- #21
builder_user
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When I find I1 && I2 then I can find I3.I3*R(30)=U2 right?
- #22
gneill
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builder_user said:
Since the terminals where U2 is measured are an open circuit, I3 must be zero.
The current through the 30 Ohm resistor (R5) will be determined by your I2.
- #23
- #24
gneill
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builder_user said:
I'm not sure that I follow what you are saying. Is U in this circuit a real voltage source? Can you draw the circuit you have in mind that does not have R1?
- #25
- #26
gneill
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builder_user said:
Well, if U is a voltage source, then the current i exists with or without R1. It will of course have different values for the two cases.
- #27
builder_user
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It is U like U2(or U1) in previous circuit
- #28
gneill
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builder_user said:
In the previous circuit you're looking for a relationship between U1 and U2. That means, if you put a real source (say V1) where U1 is, you want to find what the open-terminal voltage is at the terminals U2. In such a case the terminals at U2 are open -- not connected to anything (except perhaps a voltmeter!). So no current will flow through U2.
Alternatively, you could put a real voltage source (say V2) where U2 is, and measure the resulting voltage at the U1 terminals. Again, in such a case the terminals at U1 would be open -- not connected to anything, and no current would flow there.
- #29
builder_user
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gneill said:
А!There is no current if I have voltage source in the circuit?
- #30
gneill
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builder_user said:
All I am saying is, if the terminals at U2 are open (not connected to a source), then there is no loop and no current flows (through terminals U2).
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