# Calculate Equivalent Impedance

#### dileepchacko

Joined May 13, 2008
102
Hello

How to calculate the equivalent resistance of the network from the V1 voltage source. Can I reduce R1, R2, R3 and R4 as a equivalent circuit? I have attached the circuit in pdf format for your reference.

Thanks and Regards
Dileep Chacko

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#### steveb

Joined Jul 3, 2008
2,436
Hello

How to calculate the equivalent resistance of the network from the V1 voltage source. Can I reduce R1, R2, R3 and R4 as a equivalent circuit? I have attached the circuit in pdf format for your reference.

Thanks and Regards
Dileep Chacko
For me, the fastest way would be to simply calculate the current supplied by the source. Then V1/I is the equivalent resistance. But, I'm fast at doing that calculation, so others may not like that approach as much as I do.

I think you can also use a type of delta-to-wye equivalent transform and maybe even wye-to-delta later in the process. This should allow you to transform the circuit in steps and combine resistances. With this topology it may not be proper to use the terms delta and wye, because it's not a 3-phase setup, but the same basic idea can be used. Maybe, pi to tee transform is a better name.

http://www.paulcarlson.info/electronics/electric-15.html

Obviously, a first step is to combine those parallel resistances (R1//R2 and R6//R7.

#### dileepchacko

Joined May 13, 2008
102

#### t_n_k

Joined Mar 6, 2009
5,455
Since all resistor values are equal, another approach is to use symmetry to split the circuit into two equivalent halves. The line of symmetry is equivalent to the common ground rail potential.

Along the line of symmetry the source V is split into two sources of V/2 volts.

Above the line of symmetry one would have
R_equiv=(R1||R2)||[R3+(R4||{R5/2})].

The "trick" is splitting R5 into two equal parts around the line (potential) of symmetry. The process is pretty simple from then on because R1=R2=R3=R4=R5=R

This value R_equiv would equal half the total resistance seen by the source.

#### steveb

Joined Jul 3, 2008
2,436
Since all resistor values are equal, another approach is to use symmetry to split the circuit into two equivalent halves. The line of symmetry is equivalent to the common ground rail potential.

Along the line of symmetry the source V is split into two sources of V/2 volts.

Above the line of symmetry one would have
R_equiv=(R1||R2)||[R3+(R4||{R5/2})].

The "trick" is splitting R5 into two equal parts around the line (potential) of symmetry. The process is pretty simple from then on because R1=R2=R3=R4=R5=R

This value R_equiv would equal half the total resistance seen by the source.
That's an excellent observation and suggestion!

But, why did you do it out for him? He asked "How to calculate ..." and "Can I reduce ..." not "please calculate or reduce for me". Your idea makes it too easy for him already.

Anyway, nice work! This highlights the importance to always take some time and think before diving in and solving a problem by brute force.

#### t_n_k

Joined Mar 6, 2009
5,455
But, why did you do it out for him? He asked "How to calculate ..." and "Can I reduce ..." not "please calculate or reduce for me". Your idea makes it too easy for him already.
You're right Steve - "my bad", as they say. I'll try to be more creative in getting the OP to do some useful work in the future. Thanks for the suggestions.

#### steveb

Joined Jul 3, 2008
2,436
You're right Steve - "my bad", as they say. I'll try to be more creative in getting the OP to do some useful work in the future. Thanks for the suggestions.
Ah, no harm in this case. I'm sure he could do it once the idea of symmetry was suggested.

The best thing the OP can do in this case is solve the problem several ways and prove all ways give the same answer. When I was young, I did this to learn effectively. Now I do it because my older mind tends to make mistakes too easily.