Need Help in equivalent resistance for a circut

AI Thread Summary
The discussion focuses on finding the equivalent resistance of a circuit using Kirchhoff's rules, with specific resistor values provided. The user seeks clarification on how to apply mesh and nodal analysis to derive the equivalent resistance, which is calculated to be approximately 1.89 ohms. Participants suggest introducing a voltage or current source to analyze the circuit and recommend using delta-star transformations for simplification. The importance of labeling nodes and applying Kirchhoff's Current Law (KCL) to set up equations is emphasized. Ultimately, the user expresses gratitude for the assistance received in understanding the solution process.
Aksand
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Hi,

I got a question in my exam ask for the equivalent resistance,

The Question is:

Find the equivalent resistance for this circuit using Kirchoff Rules

I build the Image by circuit maker.

http://lookpic.com/i/965/ud67CMW6.jpeg R1=2 Ohms
R2=2 Ohms
R3=2 Ohms
R4=4 Ohms
R5=1 Ohms

i know that the answer is 36/19 Approx 1.89

but i want to know how to get it,and i will not mind if their is another ways to solve it, i like to know more

and i wish to have the answer soon as possible..
 
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Basically introduce a voltage source V (1 Volt to make it easier) at the input, and figure out the current I that would flow into it.

The resistance is then R = V/I
 
i know that before , but how to get the last current
 
Aksand said:
i know that before , but how to get the last current

By mesh analysis. You introduce the source across the terminals where you want to find the resistance and analyse it as (in this case) a three mesh circuit (three circuating mesh currents and three KVL equations).

The other method (which I usually prefer) is to introduce a 1A current source at the terminal and then use nodal analysis to find the voltages. In this case the resistance is numerically equal to the solved voltage.

Lastly there is one other alternative for this particular type of circuit (called a H bridge). You do a delta-star transformation on one half of the circuit after which the circuit is amenable to a simple ladder network type collapse to a single resistor. In this case it's easiest to delta-star the lefthand half of the ciruit (because they're all 2 ohms which make the transform particularly simple).

After delta-star transform the resistance is, by inspection : 2/3 + 5/3 || 14/3 = 36/19
 
you can use wye-delta transformation which will result in parallel and series connections.
 
Sorry , i didn't get the answer, i am not brilliant

when i introduce 1A, i don't know the up circuit(R1,R4,R3) take what and the down what...

about wye or delta , what is the rules that i have to follow??
 
Aksand said:
Sorry , i didn't get the answer, i am not brilliant

when i introduce 1A, i don't know the up circuit(R1,R4,R3) take what and the down what...

about wye or delta , what is the rules that i have to follow??

Say what course you're studing and what level you are at. Have you done either "nodal" or "mesh" analysis yet? Do you know both KVL and KCL, and how to apply them to solve for unknown currents/voltages?
 
uart said:
Say what course you're studing and what level you are at. Have you done either "nodal" or "mesh" analysis yet? Do you know both KVL and KCL, and how to apply them to solve for unknown currents/voltages?



I am in the last year in the High school,i have taken the Kirchoff rules and,but in question , i don't know in the R3 if there a current and how to found it, it is always get false

i think that Delta or wye will solve my problem , but i want to know how to make my circuit as delta or wye?
 
  • #10
Since the question said "find ... using kirchhoff rules" you would be best to do it as follows.

1. Label the three nodes and nodal voltages (relative to the one marked as ground). For example, label the node at the junction of R4, R5 as V1, label the junction of R1, R3, R4 as V2 and label the junction of R2, R3, R5 as V3. These node voltages, V1, V2 and V3 are the three unknowns which you will now attempt to solve for.

2. Draw a 1 Amp current source directed into node 1 (and the other end of current source connected to ground).

3. Write KCL in terms of the three unknown voltages. This will give you three linear equations in three unknowns, making it fairly straight forward to solve. For example in the above circuit (with nodes as defined in step one) the set of equations is :

1/R4 (V1 – V2) + 1/R5 (V1 – V3) = 1
1/R1 (V2 – 0) + 1/R3 (V2 – V3) + 1/R4 (V2 – V1) = 0
1/R2 (V3 – 0) + 1/R3 (V3 – V2) + 1/R5 (V3 – V1) = 0

Just expand and solve as you would any simultaneous equations. When you solve for V1 you have found the equivalent resistance. That's it.
 
  • #11
Thanks a lot , it is solved
 
  • #12
in equation Nu.1 and Nu.2
Why the V=0 i mean "V2-0" and "V3-0"?,what the zero refer to??
 
  • #13
Aksand said:
in equation Nu.1 and Nu.2
Why the V=0 i mean "V2-0" and "V3-0"?,what the zero refer to??

Each of the bracketed expressions represents a voltage difference between adjacent nodes. Since one of the nodes in the circuit (the reference node or "ground") is taken as zero voltage then you get a zero when one of the adjacent nodes is the "ground". Obviously I could have left the zero out of those expressions without effecting the results, I just included them to help you follow the technique of applying KCL at each node.
 
  • #14
Aha,, Thanks a lot , you helped me a lot , and thanks to everyone who tried to help me and put a post
 

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