DC circuit analysis using Reciprocity Theorem

[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.

 

[Hesch]

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.

 

[gruba]

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?

 

[Hesch]

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) ?

 

[NascentOxygen]

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.

 

[SammyS]

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

 

[Hesch]

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

? ?
E1 = 30V in the second case.

 

[SammyS]

? ?
E1 = 30V in the second case.

I was able to solve for i₁ without using the fact that E₁ = 30V .

 

[Hesch]

I was able to solve for i1 without using the fact that E1 = 30V .

Show me.

 

[SammyS]

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.)

 

[SammyS]

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 .

 

[Hesch]

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 ).

 

[gruba]

@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.

 

[Hesch]

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?

 

[gruba]

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?

 

[SammyS]

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.

 

[SammyS]

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]

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 ).

 

[Hesch]

Correction:

Example, L1:

-( 300Ω +R1 )L1 - 300Ω*L4 - V1 = 30V

Should have been:

- 300Ω * L1 - 300Ω * L4 - V1 = 30V

 

[NascentOxygen]

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?

 

[gruba]

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.

Sorry for late reply.

 

[NascentOxygen]

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.