Need Help in equivalent resistance for a circut

In summary: V1, the bracketed expression V1-0 represents the voltage difference between V1 and V2, or in other words the voltage at V1 minus the voltage at V2.
  • #1
Aksand
14
0
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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  • #2
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
 
  • #3
i know that before , but how to get the last current
 
  • #4
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
 
  • #5
you can use wye-delta transformation which will result in parallel and series connections.
 
  • #6
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??
 
  • #7
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?
 
  • #9
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
 

Related to Need Help in equivalent resistance for a circut

What is equivalent resistance?

Equivalent resistance is the total resistance of a circuit when all resistors are combined into one. It represents the overall resistance that an external power source experiences in a circuit.

Why is finding equivalent resistance important?

Finding equivalent resistance is important because it allows us to simplify complex circuits and make calculations easier. It also helps us understand the overall behavior of a circuit.

How do you calculate equivalent resistance?

To calculate equivalent resistance, you can use the formula R = R1 + R2 + R3 + ... + Rn. This means that you add up the resistance values of all the resistors in the circuit to get the total equivalent resistance.

What happens to equivalent resistance in a series circuit?

In a series circuit, the equivalent resistance increases as more resistors are added. This is because the current has to flow through each resistor, resulting in a higher overall resistance.

What happens to equivalent resistance in a parallel circuit?

In a parallel circuit, the equivalent resistance decreases as more resistors are added. This is because the current has multiple paths to flow through, resulting in a lower overall resistance.

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