Solving Circuit Analysis Problems with Kirchhoff's Laws

[Elysium]

Hi, I'm wondering if I made a mistake in getting the Kirchhoff equations to find all the currents.

Here's what I got:

Loop 1:
10.0I_1 + 20.0I_2 + 30.0 I_4 = 15.0 V

Loop 2:
40.0I_3 + 30.0 I_4 = 15.0 V

Loop 3:
20.0I_2 + 30.0 I_4 = 10.0 V

Current Law:
I_2 + I_3 - I_4 = 0 A

Then I can use Cramer's Rule (heheh) to find each individual current right? I'm afraid I don't know how to make matrices in LaTeX yet.

 

[Cyrus]

Eek, why not just ref the matrix with your calculator??

\left(\begin{array}{ccccc}10&20&0&30&15\\0&0&40&30&15\\0&20&0&30&10\\0&1&1&-1&0\end{array}\right)

That column of mostly zeros looks interesting if one were to do it by hand...

 

[Elysium]

Eek, why not just ref the matrix with your calculator??

I don't really know how to use Maple either.

That column of zeros looks interesting...

It'll ceratinly make my life easier finding the determinant manually.

But is this system of equation compliant with the attached circuit drawing?

 

[Cyrus]

I can't see your drawing until it gets approved. But you seem to know what your doing.

Ref it on your Ti-83. Thats what I do.

 

[Elysium]

Here's a link to the image. Sorry it took so long.

[img=http://img234.imageshack.us/img234/2566/a5p4600010jx.th.jpg]

 

[Cyrus]

Did you use a supernode? Are you using mesh current analysis, or node voltage?

 

[Elysium]

Did you use a supernode? Are you using mesh current analysis, or node voltage?

Supernode? Sorry I'm not familiar with that term. :(

 

[Cyrus]

Your equations seem good to me, but your picture has your resistors labeled I2 and I4 with the +,- backwards. Based on how you have them drawn, they would be negative voltages.

Tell me what answers you get and I will compare it to what the computer simulation tells me. (im too lazy to do it by hand )

 

[Elysium]

Your equations seem good to me, but your picture has your resistors labeled I2 and I4 with the +,- backwards. Based on how you have them drawn, they would be negative voltages.

Whoops, I4 was backwards, but not I2 though. So there's no I5 and I6 for each of the voltage sources?

 

[Cyrus]

Your right, sorry I2 is fine. There is an I5 and I6, but you do not know them because you do not know the resistances. You can find the values of I5 and I6 by going back and applying KCL at each of the nodes after you find all the currents (if you want to.)

 

[Elysium]

Your right, sorry I2 is fine. There is an I5 and I6, but you do not know them because you do not know the resistances. You can find the values of I5 and I6 by going back and applying KCL at each of the nodes after you find all the currents (if you want to.)

That would make sense since I would know the current comining in and out after. Ok then, thank you for helping me out. :)

 

[Cyrus]

Tell me what values you get and I can tell you if its right or wrong.

 

[Elysium]

for I1. I get... 13000? That can't be right...

 

[Elysium]

Whoops again, I have to divide that number by the determinant.
I get:
I1 = 0.5 A

This is hard doing it manually. :s

 

[BerryBoy]

Whoa! I think this problem has been made complicated. I think I1 is the easiest current to find, as it has a potential difference of 5V across it.

I would reccomend using nodal analysis to find the other currents, as it is a linear equation with 1 unknown variable. The only node where you don't know the voltage is the dot at the bottom of your diagram.

Thoughts?

Regards,
Sam

 

[Elysium]

Whoa! I think this problem has been made complicated. I think I1 is the easiest current to find, as it has a potential difference of 5V across it.

I would reccomend using nodal analysis to find the other currents, as it is a linear equation with 1 unknown variable. The only node where you don't know the voltage is the dot at the bottom of your diagram.

Thoughts?

Regards,
Sam

I'm at a loss. How is it that the top resistor at I1 has only 5.0V? Can you explain? So I only need to use Cramer's Rule for I2, I3 and I4? (I4's polarity is wrong in the diagram)

Thanx

 

[BerryBoy]

You know that the potential at the right side of the resistor is +15V and the left side is at +10V, because the two cells are either side of it and they have their 0V at the same node (they are connected together).

Sorry, I have to go, I'll try to log on again in 15 minutes...

Sam

 

[Cyrus]

He does not know about node voltage analysis yet. Just let him do it his way for now.

I would urge caution in saying that they have their 0V at the same node. You do not know that the - is zero volts. All you know is the potential difference across the sources is 15V, you have no information what it is on each end relative to ground. For all you know it could be 10V-0v and -15v and 0 V. Then it would 0V and -15V adjacent to each other, and not both zero.

 

[BerryBoy]

All potential is relative, I didn't say it was 0 V relative to ground. Ground is the usual reference for zero, but I used it as the lowest point of potential, so there are no negative voltages. I always try to simplify things before using maths, it makes life easier.

As for Kirchoff's current loops, now there only need to be 2 loops, so that is a simulaneous equation with only 2 unknowns. I am surprized though that Loops are taught before nodal analysis.

Sam

 

[Cyrus]

Why, loops are the most basic forms. Node analysis is derived from KVL and KCL. You can't do it without first knowing loops.

 

[BerryBoy]

I'm at a loss. How is it that the top resistor at I1 has only 5.0V? Can you explain? So I only need to use Cramer's Rule for I2, I3 and I4? (I4's polarity is wrong in the diagram)

Thanx

It doesn't matter what way you have drawn your diagram, I always draw the current direction in whatever way I feel; the beauty is that if you've drawn it in the wrong direction, it doesn't matter, you just a negative answer.

 

[BerryBoy]

Why, loops are the most basic forms. Node analysis is derived from KVL and KCL. You can't do it without first knowing loops.

But loops is KVL -

The sum of al the potentials round a loop (V=IR) is zero.

Sam

 

[Elysium]

ARGH! I've been wasting my time.

I'll simply row-reduce this matrix and call it quits.

\left(\begin{array}{cccc}0&40&30&15\\20&0&30&10\\1&1&-1&0\end{array}\right)

Zat ok?

 

[Galileo]

Using Cramer's rule to solve a system of linear equation is madness. Gaussian elimination is way more efficient.

Also, using KCL at the nodes and substituting the V-I relation immediately is often easier (no loops!) and faster, but that's my opinion.

 

[Elysium]

I'm getting a negative value for I_5 (At the 10V voltage source in the direction of the diagram. I didn't label it before the scan) . Did I make a mistake?

 

[BerryBoy]

What answers are you getting, and I'll see if I agree with you.

If you get a negative answer, it doesn't mean you got the answer wrong, it could be that you drew the current in the other direction.

Sam

 

[BerryBoy]

Using Cramer's rule to solve a system of linear equation is madness. Gaussian elimination is way more efficient.

Also, using KCL at the nodes and substituting the V-I relation immediately is often easier (no loops!) and faster, but that's my opinion.

Totally agree, but cyrusabdollahi says its not taught this way.

Sam

 

[Elysium]

Using Cramer's rule to solve a system of linear equation is madness. Gaussian elimination is way more efficient.

Yea... I know. I first started with Cramer's Rule because of that stupid textbook, but then the linear algebra knowledge kicked in at the last minute...

If you get a negative answer, it doesn't mean you got the answer wrong, it could be that you drew the current in the other direction.

Yea, I know about this. I was just surprised that the current of the 10.0V voltage source was occurring in the in the opposite way of the indicated poles. I guess how the 15.0 V voltage source was placed outdid the 10.0 V one. At the end I still considered both voltage sources to 'exert' positive power, otherwise it didn't sum up with the power that the resistors absorbed and converted into thermal energy.

 

[Elysium]

What answers are you getting, and I'll see if I agree with you.

Too late, sent it out. I spent a lot more than an hour on the problem, but I did solve it (I feel extremely awful about the time it required to solve it, bu I did learned from the experience). I got an answer in the magnitude of 10^(-2) for I_2. Eh.

 

[Cyrus]

Ref the matrix, it will take you a whole 5 mins max.