# equivalent resistance

#### qrious

Joined Sep 16, 2009
22
Please tell me the value of the equivalent resistance between the points A & B for the diagram drawn above in the attachment..all the resistances are equal in value=R ohms .
Also please tell me if i can replace the diagram above with the one drawn below?..are both of these ckts equivalent? if not, then why?..

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#### ELECTRONERD

Joined May 26, 2009
1,146
This might help you, notice that there are resistors in a triangular configuration.

I wasn't able to place the resistors in a triangular format, but you can draw it on paper.

Ra = (R1xR2) / (R1+R2+R3)

Rb = (R1xR2) / (R1+R2+R3)

Rc = (R2xR3) / (R1+R2+R3)

Example, in the attached schematic, we want to have the resistance from point Ra to the other point (A & B). Therefore, we solve using the Ra formula. If all the resistors are 1Ω, we can equate the following:

Ra= (1x1) / (1+1+1) = 0.33Ω

Now, to solve for R4 and R5, we can replace our triangular resistive circuit with a simple "Ra." Notice that R4 and R5 are parallel. So 1 / (1/1 + 1/1) = 0.5Ω Thus, the total resistance is 0.33+0.5 = Approximately 0.83Ω.

Hope this helps!

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#### KL7AJ

Joined Nov 4, 2008
2,208
This might help you, notice that there are resistors in a triangular configuration.

I wasn't able to place the resistors in a triangular format, but you can draw it on paper.

Ra = (R1xR2) / (R1+R2+R3)

Rb = (R1xR2) / (R1+R2+R3)

Rc = (R2xR3) / (R1+R2+R3)

Example, in the attached schematic, we want to have the resistance from point Ra to the other point (A & B). Therefore, we solve using the Ra formula. If all the resistors are 1Ω, we can equate the following:

Ra= (1x1) / (1+1+1) = 0.33Ω

Now, to solve for R4 and R5, we can replace our triangular resistive circuit with a simple "Ra." Notice that R4 and R5 are parallel. So 1 / (1/1 + 1/1) = 0.5Ω Thus, the total resistance is 0.33+0.5 = Approximately 0.83Ω.

Hope this helps!

Good show! This is also a good application for the principle of "virtual shorts" which works ONLY if all values of resistance are equal. In this case R2 has NO effect on the equivalent resistance. All bridge circuits work this way. It's a little harder to prove than it is to demonstrate...but as a thought experiment, replace R2 with both an open circuit and with a short circuit....you come out with the same answer.

Eric

#### The Electrician

Joined Oct 9, 2007
2,724
Please tell me the value of the equivalent resistance between the points A & B for the diagram drawn above in the attachment..all the resistances are equal in value=R ohms .
Also please tell me if i can replace the diagram above with the one drawn below?..are both of these ckts equivalent? if not, then why?..
Due to the extreme symmetry of the network, you can do the steps shown in the attachment.

Fold the top half of the circuit down onto the bottom half, and split one of the resistors in two. You should be able to figure it out then.

I get 4R/5 ohms.

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#### KL7AJ

Joined Nov 4, 2008
2,208
Due to the extreme symmetry of the network, you can do the steps shown in the attachment.

Fold the top half of the circuit down onto the bottom half, and split one of the resistors in two. You should be able to figure it out then.

I get 4R/5 ohms.

Yet another great approach. I'm not sure if this works with unequal resistances.....something to mull over this weekend~
Eric

#### ELECTRONERD

Joined May 26, 2009
1,146
Good show! This is also a good application for the principle of "virtual shorts" which works ONLY if all values of resistance are equal. In this case R2 has NO effect on the equivalent resistance. All bridge circuits work this way. It's a little harder to prove than it is to demonstrate...but as a thought experiment, replace R2 with both an open circuit and with a short circuit....you come out with the same answer.

Eric
While it is a good application for "virtual shorts," which works ONLY when resistance values are equal, it is also a good application for different values as well. Got the method out of this old electronic book I found at a hamfest.

#### qrious

Joined Sep 16, 2009
22
realy sorry..but i should have mentioned that i do not want a star-delta transform...i want a solution using only kcl/kvl ..4R/5 is indeed the answer..thanks a lot for solving the question..but i stil dont get electrician's solution..can't proceed from where you left off!! sorry!!!....and please tell me you guys whats wrong with that 2nd diagram i made ????

#### qrious

Joined Sep 16, 2009
22
how do u understand when to fold a circuit? and like..what does it mean??.and stuff like "the resistance of the network looking through a pair of terminals" n stuff...what does it mean? this looking through??..i dont understand such terms..please can someone help!

#### qrious

Joined Sep 16, 2009
22
can anyone please solve my 2nd diagram and tell me what the resstance is between a & b...im all confused..help please

#### ELECTRONERD

Joined May 26, 2009
1,146
qrious,

We won't solve any more answers for you, you have to do the work yourself.

Did you pay attention in class?

#### qrious

Joined Sep 16, 2009
22
ohh come on plzzz..its just the last one!.. i did the classes but didn't help..im all confused! help me man~!!

#### ELECTRONERD

Joined May 26, 2009
1,146
It is just a simple matter of parallel and series resistive adding. Starting from left to right, both the 2R would become (1)R and in the center R + R/2 would be (1)R 1/2. You should be able to go from there. Do they have an answer that they give?

#### The Electrician

Joined Oct 9, 2007
2,724
realy sorry..but i should have mentioned that i do not want a star-delta transform...i want a solution using only kcl/kvl .
If you want a solution using only kcl/kvl then you will need to write 6 mesh equations or 6 node equations.

Do you know how to do that?

#### qrious

Joined Sep 16, 2009
22
no i dont...and i dont even wana do it with kcl/kvl anymore..i think your method is really cool...if only i could grasp it....please just solve that sum using symmetry..when can i use symmetry? and what is the usefulness of a circuit being symmetric...and when is a circuit considered symmentric????..im a true nit wit..please help me out..how do you get 4r/5???

#### The Electrician

Joined Oct 9, 2007
2,724
no i dont...and i dont even wana do it with kcl/kvl anymore..i think your method is really cool...if only i could grasp it....please just solve that sum using symmetry..when can i use symmetry? and what is the usefulness of a circuit being symmetric...and when is a circuit considered symmentric????..im a true nit wit..please help me out..how do you get 4r/5???
Look at top circuit in the attachment. The network is symmetrical because the part of the circuit in red is identical to the part in green (as long as the resistors are all the same value).

I've added a battery at the left side and grounded the right. You could connect a voltmeter between ground and various nodes of the circuit and measure the voltages that appear at those nodes.

Consider the nodes labeled X and X'; the voltages measured there would be identical because of the symmetry. The same for nodes Y and Y'.

When you have two nodes in a resistor network that are at the same voltage, you can connect them together without changing the operation of the circuit. In this case, that includes the resistance between the left and right ends.

Connecting X to X', and Y to Y' is the same as folding the circuit as shown in the second circuit. Folding the circuit puts identical resistors on top of each other; they then become connected in parallel. This is the same as allowing the red and green circuits to be replaced with one circuit, but with resistor values 1/2 of what they were originally, as shown in the third circuit.

Then make the change shown in the fourth circuit, replacing the R/2 resistor with two R/4 resistors in series. Again, because of the symmetry, the voltage at the junction (this is a node of the circuit now) of the two R/4 resistors is the same as the voltage at the top middle node of the circuit. That means that those nodes can be connected together with no change in the resistance of the network.

Now you have R/2 in parallel with R/4. Look at the subcircuit at the very lower right. That is half of the total circuit, which has half the total resistance.

You should be able to figure out the resistance of that half; it's 2R/5, so the total resistance is 4R/5.

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#### qrious

Joined Sep 16, 2009
22
thank you good sir for taking such a lot of trouble for me..im really grateful..i understnd bettr nw....bt i was reading up on symmetry and i found stuff like rotational symmetry and revolutional symmetry...which are used in maths to describe congruency..will these work in ckt theory???...i mean if a ckt looks the same..if i revolve it about an axis or rotate it etc.. through various angles of the form 360/n where n is discrete..then can i say the ckt is symmetric? and the sub ckts are congruent??..if the sub ckts are congruent then will the corresponding node voltages and branch currents always be the same?

#### The Electrician

Joined Oct 9, 2007
2,724
thank you good sir for taking such a lot of trouble for me..im really grateful..i understnd bettr nw....bt i was reading up on symmetry and i found stuff like rotational symmetry and revolutional symmetry...which are used in maths to describe congruency..will these work in ckt theory???...i mean if a ckt looks the same..if i revolve it about an axis or rotate it etc.. through various angles of the form 360/n where n is discrete..then can i say the ckt is symmetric? and the sub ckts are congruent??..if the sub ckts are congruent then will the corresponding node voltages and branch currents always be the same?
Whether various kinds of symmetry allow for simplified analysis of a network will depend on whether the excitation (application of a voltage or current) is such as to excite the network symmetrically. You would have to consider this on a case by case basis, I think.

For example, in the top circuit shown in the attachment to post #15, if you wanted to know the resistance measured from the node I labeled X to the right end of the circuit, symmetry won't help you, because the circuit is then not excited symmetrically.

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#### qrious

Joined Sep 16, 2009
22
ok consider the ckt in the attachment...if i label the junctions of the ckt starting from the top left corner as 1,2,3,4 going clockwise...then in this case is the excitation symmetrical?..this ckt surely cannot be folded..but if i rotate the square by 180 degrees then the junctions become 1->3, 2->4, 3->1 & 4->2....under this transformation the ckt remains similar to the original.then can i say this ckt is symmetrical ? is the current distribution shown correct?..if the ckt is symmetrical then current thru the 10 ohm resistors shud be same and current thru 5 ohm resistors shud be same. Is my analysis correct? plz help

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#### The Electrician

Joined Oct 9, 2007
2,724
This is a classic unbalanced wheatstone bridge.

The current in the top left 10Ω resistor is not the same as the current in the bottom left 5Ω resistor. The excitation is not sufficiently symmetric.

If you were to exchange the 10 and 5 ohm resistors on the top edge of the circuit, then there would be sufficient symmetry; the bridge would be balanced.

It's true that the current in the 10Ω resistors is the same, as is the current in the 5Ω resistors, but that's not enough.

Remember, I described your circuit in the first post of this thread as possessed of "extreme symmetry".