Applying Kirchoff's law for a circuit with 3 DC supplies

[EddieH]

Homework Statement

The question is from DC circuits:
Calculate the current through the 11V battery, and the dissipated power in R1, R2, and R4.

I have made an attempt at the question and would appreciate feedback before I submit please.

Relevant Equations

Kirchoff's laws.
P=I^2R

See my attached working out.

Thanks in advance

[Images of attached PDF pasted into post by the Mentors]

 

[berkeman]

I didn't go through the detailed calculations, but it looks like your approach is okay. Can you back-annotate the voltages in the circuit and check yourself that way? Or another good trick is to re-solve the problem using KCL node equations to see if you get the same answers.

 

[DeBangis21]

OP, your traverse isn't consistent...

 

[scottdave]

One thing you can do to check your currents: start at the bottom node (ground) and work through each path to the top and check that the voltage at the node at top middle is the same regardless of the path taken.

 

[Gavran]

Homework Statement: The question is from DC circuits:
Calculate the current through the 11V battery, and the dissipated power in R1, R2, and R4.

I have made an attempt at the question and would appreciate feedback before I submit please.
Relevant Equations: Kirchoff's laws.
P=I^2R

See my attached working out.

Thanks in advance

View attachment 346685
View attachment 346686
View attachment 346687
[Images of attached PDF pasted into post by the Mentors]

The calculations and the method, both of them are correct.

 

[DeBangis21]

OP, your traverse isn't consistent...

Seeing other comments, this means I did not get the basics, or the circuit above is a different case, for what I learnt was 'when traversing around loops, the direction should be same.'

Pls enlight me more.

 

[DaveE]

Seeing other comments, this means I did not get the basics, or the circuit above is a different case, for what I learnt was 'when traversing around loops, the direction should be same.'

Pls enlight me more.

These polarity questions don't really have a correct answer. You can assign polarities (current directions, etc.) at random if you like. What you then have to do, every time, is make sure that the equations you derive are consistent with your definitions, as well as how the real world works.

 

[EddieH]

Hi all,

Thanks for your input, it cheers me up knowing that not only am I using the correct interpretation of the laws, I somehow managed to get the correct answers.

And I was flapping to think I was bashing my face against the keyboard, course notes and google for nothing lol




Ed.

 

[DeBangis21]

These polarity questions don't really have a correct answer. You can assign polarities (current directions, etc.) at random if you like. What you then have to do, every time, is make sure that the equations you derive are consistent with your definitions, as well as how the real world works.

Thank you very much.

 

[Babadag]

If we take in consideration only I1 and I2 we get these 2 relations:
I1*(R1+R4)+I2*R4=V1-V3
I2*(R2+R4+R3)+I1*R4=V2-V3
Then we make a general determinant:
R1+R4 R4
R4 R2+R3+R4
Dg=26
Then we change column 1 with V1-V3
V2-V3
and we get D1=16 I1=16/26=0.6153846
Then we change column 2 with V1-V3
V2-V3
and we get D2=2 I2=2/26=0.076923

 

[LvW]

Seeing other comments, this means I did not get the basics, or the circuit above is a different case, for what I learnt was 'when traversing around loops, the direction should be same.'

Pls enlight me more.

As you probably know, there is not only one single approach (method) to anylyze such a circuit.
For my opinion, the best and quickest and also most logical (what is a mesh current?) method is superposition.
We have a linear network with resistors and three voltage sources - and each voltage source contributes to the currents within the network.
Why not find separately the contribution of each source?
This is the principle of superposition:
* Replace two sources with a short (that means: Set it to zero) and find the wanted current (resp. produced output voltage) as it is caused by one source only
* Apply this method for each of the three voltage sources.
* Finally, add all three portions (with correct sign) to find the final solution.

 

[DeBangis21]

@LvW , that is very intuitive. We treated the theorem in our PHY202 last semester.
Thanks for your input.

 

[The Electrician]

If we take in consideration only I1 and I2 we get these 2 relations:
I1*(R1+R4)+I2*R4=V1-V3
I2*(R2+R4+R3)+I1*R4=V2-V3
Then we make a general determinant:
R1+R4 R4
R4 R2+R3+R4
Dg=26
Then we change column 1 with V1-V3
V2-V3
and we get D1=16 I1=16/26=0.6153846
Then we change column 2 with V1-V3
V2-V3
and we get D2=2 I2=2/26=0.076923

Babadag is showing how to setup the loop equations in matrix form using a text description.
Using standard linear algebra math symbols and using matrix inverse rather than Cramer's rule:

 

[DaveE]

Babadag is showing how to setup the loop equations in matrix form using a text description.
Using standard linear algebra math symbols and using matrix inverse rather than Cramer's rule:

View attachment 347371

Yes, the right way for a computer to solve this.
OTOH, analog EEs will reduce this sort of problem with superposition and/or source transformations. Very few equations required, just some back of the envelope sketches.

 

[bob012345]

As you probably know, there is not only one single approach (method) to anylyze such a circuit.
For my opinion, the best and quickest and also most logical (what is a mesh current?) method is superposition.
We have a linear network with resistors and three voltage sources - and each voltage source contributes to the currents within the network.
Why not find separately the contribution of each source?
This is the principle of superposition:
* Replace two sources with a short (that means: Set it to zero) and find the wanted current (resp. produced output voltage) as it is caused by one source only
* Apply this method for each of the three voltage sources.
* Finally, add all three portions (with correct sign) to find the final solution.

I tried this several times. It’s three times the work and I never got the right answers! I think it’s better to include the sources in the original loop equations.

 

[DaveE]

I tried this several times. It’s three times the work and I never got the right answers! I think it’s better to include the sources in the original loop equations.

There are several different methods to solve circuits like this. So EEs use the ones they are most comfortable with. You may need to practice other methods to become proficient. The "best" method depends on the problem, the methods you know, and your preferences.

The superposition method is shown below. It looks a bit laborious because I drew out every step. In practice we would do most of this in our head since each case is simply a voltage divider. Especially in this case since we can combine R2 and R3 and move the voltage sources all to the bottom which results in symbolically identical branches. You really only need to solve 1 case and then interchange the component names for the others (always look for and exploit symmetry if you find it).

I usually prefer source transformations, depending on the network, of course. That method looks like this. I've drawn this one the way I actually would in practice, skipping steps.

I usually prefer these more graphical transformations to purely algebraic solutions because shorter equations are less error prone, common terms often appear that make calculation easier, and (most importantly) you get a little insight into how the network works, not just what the answers are. This latter point is important in the real world of design where the process works in reverse; you know what you need and have to choose the network/values to achieve that.

For example, this network is part of the most common summing amplifier circuits and these methods yield the easy approach to understanding and designing them. So how hard would it be if I asked you to add a 6th input to the summer below, or increase the gain in channel 4 by 20% without changing the other channels? Would you have to start over, or just adjust a few things?

 

[The Electrician]

Your source transformation method didn't find I1, I2 and I3.

 

[DaveE]

Your source transformation method didn't find I1, I2 and I3.

Really???
It's trivial once you know the voltages. I'd say it's left as an exercise for the reader, except that it's exactly the same as what I did in the previous example. I also didn't tell you that 179/13 = 13.78. I kind of lost interest at that point.

It's not a contest. You can do it the way you like. I'm OK with that.

PS: This does illuminate an important point in circuit solutions. Often you can (should) look at a network and try to identify the most important missing data, often the hardest to solve for. So, if you can say "if I knew that voltage (current, whatever), the rest of the network would be easy to solve". V_A, in this example. Then it's often best to use the simplest method to just solve for those key parameters. Then you can go back to the original network and switch methods, or just do simple solutions for the others. This is especially important with source transformations because you often lose or obscure the other parameters when you simplify between steps. The idea is to first reduce the complexity of the problem, perhaps by splitting it into separate smaller problems, before you solve for everything. For big networks the rote KCL, KVL equations can get really big and unwieldly.

 

[Wana Banda]

Seeing other comments, this means I did not get the basics, or the circuit above is a different case, for what I learnt was 'when traversing around loops, the direction should be same.'

Pls enlight me more.

No it's actually alright
It's up to you to maintain the same direction or not