# DC circuit analysis using Reciprocity Theorem

• Engineering
• gruba
In summary: I don't think so. If the value of E1 becomes too low, the value of R1 becomes negative. I don't know if that makes...sense?
gruba

## Homework Statement

In the following circuit I need to find current I1 if the given parameters are E1=0V,E2=15V,E3=60V
In the second case, given parameters are I2=100mA,I3=40mA,E1=30V,E2=E3=0V
In both cases, R2=R3=R4=R5=300 ohm

## Homework Equations

Using reciprocity theorem for DC circuits.

## The Attempt at a Solution

I used reciprocity theorem on the second case, moving the source E1 between the nodes where I3 flows. From there I1=-I3=-40mA. Is this correct? This is the second case. Could someone help how to find current I1 in the first case?

Thanks for replies.

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gruba said:
Could someone help how to find current I1 in the first case?
Draw 4 loop currents in the diagram: L1 . . L4.
Make 4 equations, using Kirchhoffs voltage law ( KVL ).
Solve the 4 equation, and you have found I1.

Hesch said:
Draw 4 loop currents in the diagram: L1 . . L4.
Make 4 equations, using Kirchhoffs voltage law ( KVL ).
Solve the 4 equation, and you have found I1.
Resistors R1 and R6 are not given. Is it possible to find them, and then the current?

gruba said:
Resistors R1 and R6 are not given. Is it possible to find them, and then the current?
Something is wrong here.
In the first case you know: E1=0V,E2=15V,E3=60V and R2=R3=R4=R5=300 ohm. That's all.

You could find a possible solution by:

Assume R1 = infinit → R1 and E1 can be removed. It doesn't change anything.
Assume R6 = infinit → R6 can be removed.
Now R3 ends blind and can be removed.

Find I2 and I3. I1 = 0.

That's a possible ( and easy ) solution.

Or do you have to find I1 as a function of R1 and R6? : I1(R1,R6) ?

Last edited:
In case (1) you are given the values of three voltage sources and asked to find the current through a resistor of unknown value? With the information provided, I'm sure that can't be done.

In case (2) where there is only one voltage source, then the current from that source cannot be less than 100mA. You don't need any special theorem to show this; if you have a resistance network and a single voltage source, you can't have greater current through any resistor than is being supplied by the source.

It would have been far better had you presented just a single problem in this thread; mixing two like you have is thoroughly confusing.

I'm pretty sure that the second case can be solved, the value of E1 not being needed.

SammyS said:
I'm pretty sure that the second case can be solved, the value of E1 not being needed.
? ?
E1 = 30V in the second case.

Hesch said:
? ?
E1 = 30V in the second case.
I was able to solve for i1 without using the fact that E1 = 30V .

SammyS said:
I was able to solve for i1 without using the fact that E1 = 30V .
Show me.

gruba said:

## Homework Statement

In the second case, given parameters are I2=100mA,I3=40mA,E1=30V,E2=E3=0V
In both cases, R2=R3=R4=R5=300 ohm
2. Homework Equations

Using reciprocity theorem for DC circuits.

## The Attempt at a Solution

I used reciprocity theorem on the second case, moving the source E1 between the nodes where I3 flows. From there I1=-I3=-40mA. Is this correct? This is the second case.

Thanks for replies.
@gruba

I didn't use reciprocity. I got a different answer for the second problem.

(I should have mentioned this earlier.)

Hesch said:
Show me.
@Hesch

Since OP has not solved this, I won't give the full solution.

The current though R2 is ##\ i_1+i_2\ ## downward.

The current though R4 is ##\ i_1+i_2+i_3\ ## upward.

Furthermore, E2 = 0 so R2 and R4 are in parallel .

SammyS said:
The current though R2 is i1+i2 \ i_1+i_2\ downward.
The current though R4 is i1+i2+i3 \ i_1+i_2+i_3\ upward.
Ok, thank you.
I'll have a look at it ( maybe using my 4 equations in #2 ).

SammyS said:
@Hesch

Since OP has not solved this, I won't give the full solution.

The current though R2 is ##\ i_1+i_2\ ## downward.

The current though R4 is ##\ i_1+i_2+i_3\ ## upward.

Furthermore, E2 = 0 so R2 and R4 are in parallel .

How do you know that current through R3 flows upward?
If you use Kirchhoff's law for nodes, current through R4 (I3 - current though R2), (again how to determine direction).
How did you get I1+I2+I3
These currents are not in relation by any node.
I think R2 and R4 not in parallel.

Could you show your solution, I can't figure it out.

SammyS said:
Since OP has not solved this, I won't give the full solution.
I just wondered if some currents would be altered due to altering E1, but they won't. Only the value of R1 will be changed, everything else is constant.

And well, I had to use 5 equations to solve the problem, ( are you reading this gruba? 5 equations! ). But then I also know the values of R1 and R6 .
Hint: R6 = 2 * R1 at E1 = 30V.

If the value of E1 becomes too low, the value of R1 becomes negative. I don't know if that makes sense?

Hesch said:
I just wondered if some currents would be altered due to altering E1, but they won't. Only the value of R1 will be changed, everything else is constant.

And well, I had to use 5 equations to solve the problem, ( are you reading this gruba? 5 equations! ). But then I also know the values of R1 and R6 .
Hint: R6 = 2 * R1 at E1 = 30V.

Can you show how you have done it?

gruba said:
How do you know that current through R3 flows upward?
If you use Kirchhoff's law for nodes, current through R4 (I3 - current though R2), (again how to determine direction).
How did you get I1+I2+I3
These currents are not in relation by any node.
I think R2 and R4 not in parallel.

Could you show your solution, I can't figure it out.

I stated that current through R4 is upward. (That's using the direction of the currents as given in the figure.)

This is for the second problem so that I2=100mA, I3=40mA, E1=30V, E2=E3=0V .

E2 = 0 , therefore, R2 and R4 have the same voltage drop across them and they are in parallel.

Furthermore, E3 = 0 which leads to R4 and R5 also being in parallel. That's not needed for finding ##\ i_1\ ##.

The junction law tells you that the current through R4 is upward.

Let me state again, that these directions are for the assumed current directions given in the figure. It may well be that the value found for ##\ i_1\ ## is a negative value of sufficient size so that the resulting current through R4 is downward.

As a further note: I am not using reciprocity for this solution, so I have not switch location of sources as you did.

gruba said:
Can you show how you have done it?
That's not how we operate here at PF .

We try to help you solve problems, but it's you that solves.

Hesch said:
Draw 4 loop currents in the diagram: L1 . . L4.
Make 4 equations, using Kirchhoffs voltage law ( KVL ).
Solve the 4 equation, and you have found I1.
I've attached the diagram with 4 loops drawn. From the 4 loops you can setup 4 equations, but the loop-current L3 is known ( 0.04A ), so you need only equations as for L1, L2, L4. Then you need two equations:

L3 = 0.04A
L2 - L1 = 0.1A

Now, the voltage drop across R6 = L4 * R6 and across R1 = L1 * R1. Just call these voltage drops V6 and V1. Then you have 5 equations with 5 unknown:
L1, L2, L4, V1, V6.

Example, L1:

-( 300Ω +R1 )L1 - 300Ω*L4 - V1 = 30V

Setup the last 4 equations and solve them.

To find the current through R3, you just sketch an arrow for example upward. Now the current will be L4+L1. Maybe the result is negative, but it's because the current actually goes downward ( positive ).

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gruba
Correction:
Hesch said:
Example, L1:

-( 300Ω +R1 )L1 - 300Ω*L4 - V1 = 30V

Should have been:

- 300Ω * L1 - 300Ω * L4 - V1 = 30V

It seems that no one has been able to see how to analyze by applying the Reciprocity Theorem, specifically, to this problem.

gruba, has your course provided you with the Reciprocity Theorem solution?

NascentOxygen said:
It seems that no one has been able to see how to analyze by applying the Reciprocity Theorem, specifically, to this problem.

gruba, has your course provided you with the Reciprocity Theorem solution?

I don't know how to to solve this problem using reciprocity theorem. I tried using it, but didn't work.

I may have it figured out.

Analyse the circuit below, determining the reading on each ammeter in terms of V and the resistances. Using this result, calculate the reading on A1 with V equal to 15V, and then swap around that source and ammeter A1. The Theorem of Reciprocity says reading on A1 remains unchanged, so it now tells you the contribution of current in wire a-b due to the 15V source when that source is repositioned to where A1 originally was.

https://www.physicsforums.com/attachments/img_20150822_155400-1-jpg.87664

Next, go back to where you analyzed the basic circuit and this time set V to be 60V and calculate reading A2, then swap around this 60V source with ammeter A2. This ammeter reading is unchanged by the swap, and it now tells you the contribution to current in wire a-b due to that 60V source relocated to where A2 was initially.

In the final arrangement with those two sources, by superposition the total current in path a-b will be the sum of those two contributions.

## 1. What is the Reciprocity Theorem in DC circuit analysis?

The Reciprocity Theorem is a fundamental concept in DC circuit analysis that states the ratio of voltage to current at any point in a circuit is the same as the ratio of current to voltage at the same point, regardless of the source locations. In other words, the effect of a source on a circuit can be determined by considering the position of the source and the circuit element it is connected to, regardless of the other sources in the circuit.

## 2. How is the Reciprocity Theorem applied in DC circuit analysis?

The Reciprocity Theorem is applied by swapping the positions of the source and the circuit element it is connected to. This means that the voltage and current values are also swapped. By doing this, the effect of the source on the circuit can be determined. This method is particularly useful when analyzing complex circuits with multiple sources.

## 3. What are the benefits of using the Reciprocity Theorem in DC circuit analysis?

One of the main benefits of using the Reciprocity Theorem is that it simplifies the analysis of complex circuits. By swapping the positions of the source and the circuit element, the analysis can be reduced to simpler circuits, making it easier to determine the effect of the source. Additionally, the theorem is applicable to both linear and nonlinear circuits, making it a versatile tool for circuit analysis.

## 4. Are there any limitations to using the Reciprocity Theorem in DC circuit analysis?

While the Reciprocity Theorem is a powerful tool, it does have its limitations. It can only be used for circuits with linear elements, and it assumes that the circuit is in a steady-state condition. Additionally, it may not provide accurate results for circuits with sources that are not in close proximity to the circuit elements.

## 5. How is the Reciprocity Theorem different from other circuit analysis methods?

The Reciprocity Theorem is different from other circuit analysis methods, such as Kirchhoff's laws or Thevenin's theorem, because it focuses on the effects of individual sources in a circuit, rather than the overall behavior of the circuit. It also allows for the analysis of circuits with multiple sources without having to consider them all at once, making it a more efficient method for complex circuits.

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