How to setup KVL equations with element laws?

[(*steve*)]

Your reasoning looks correct but I would still solve it using KVL and KCL. You have 4 unknowns so you will need 4 equations.

It's worth the practice.

I'm pretty sure I also spent around 15 hours per week on this course. I took shortcuts at the beginning and it came back to bite me later and probably cost me the final question

 

[chopnhack]

Your reasoning looks correct but I would still solve it using KVL and KCL. You have 4 unknowns so you will need 4 equations.

It's worth the practice.

I'm pretty sure I also spent around 15 hours per week on this course. I took shortcuts at the beginning and it came back to bite me later and probably cost me the final question

I can tell that after week 1 I am in big trouble... Unless I can become more proficient at being able to recognize and label a network I will wash out for sure! Nodal analysis is a bit vague currently, that will be the next challenge as soon as I shore up the other deficiencies...

 

[(*steve*)]

I can tell that after week 1 I am in big trouble... Unless I can become more proficient at being able to recognize and label a network I will wash out for sure!

I wouldn't worry too much about that The beauty of these courses is that you can do them multiple times at the same cost ($0). The other thing to remember is that this course is a *real* engineering unit. It's not fluff put out on the internet to give people entertainment. Learn what you can. Move through the weeks at your own pace and try to become proficient. Being able to understand this stuff is more important than being able to get the right answers one time on a test.

As I was doing this stuff there were times that I had to sit and think for quite some time to do the problem in the way they wanted it done. In many cases I could almost solve the problem by inspection (knowing many shortcuts from years of experience) but I had never done this using the basic laws.

Nodal analysis is a bit vague currently, that will be the next challenge as soon as I shore up the other deficiencies...

Several pointers:

1) Take notes during the lectures and keep them by your side.
2) have another book of rules and identities for various solved problems which look significant. For example the equation for a voltage divider, or resistors in parallel. Have you covered conductance vs resistance yet? Converting from one to the other can make some things far easier.
3) Dilligently do all the exercises, not just the assessed ones.
4) Find some web sites which can find symbolic answers to equations. Later on, you'll find the equations you generate are *horrible* and solving them is a PITA These will help with the mechanical work. (I have some of these bookmarked on another computer. I'll try to remember to post the links later.)
5) make sure you understand how the MIT software handles formula input. It can be dicky and the last thing you want is to use up your three tries just getting the right answer into a format the software understands.
6) Use the forums in the MIT software. I found them to be somewhat useful. They were very useful when I did this course, but were buggerized about for other courses which made them much harder to use. Beware that people do occasionally post the complete method for answering a question. Great for getting the right answer, not so great to learn from.

When you come to the exams in this course, beware that they are as hard as any exam I have done. They have questions that range from easy to really hard -- they are designed to produce a result which distinguishes skilled from non-skilled people, NOT to allow everyone to pass. The points in steps 1 and 2 above will be a great help.

 

[chopnhack]

I wouldn't worry too much about that The beauty of these courses is that you can do them multiple times at the same cost ($0). The other thing to remember is that this course is a *real* engineering unit. It's not fluff put out on the internet to give people entertainment. Learn what you can. Move through the weeks at your own pace and try to become proficient. Being able to understand this stuff is more important than being able to get the right answers one time on a test.

As I was doing this stuff there were times that I had to sit and think for quite some time to do the problem in the way they wanted it done. In many cases I could almost solve the problem by inspection (knowing many shortcuts from years of experience) but I had never done this using the basic laws.



Several pointers:

1) Take notes during the lectures and keep them by your side.
2) have another book of rules and identities for various solved problems which look significant. For example the equation for a voltage divider, or resistors in parallel. Have you covered conductance vs resistance yet? Converting from one to the other can make some things far easier.
3) Dilligently do all the exercises, not just the assessed ones.
4) Find some web sites which can find symbolic answers to equations. Later on, you'll find the equations you generate are *horrible* and solving them is a PITA These will help with the mechanical work. (I have some of these bookmarked on another computer. I'll try to remember to post the links later.)
5) make sure you understand how the MIT software handles formula input. It can be dicky and the last thing you want is to use up your three tries just getting the right answer into a format the software understands.
6) Use the forums in the MIT software. I found them to be somewhat useful. They were very useful when I did this course, but were buggerized about for other courses which made them much harder to use. Beware that people do occasionally post the complete method for answering a question. Great for getting the right answer, not so great to learn from.

When you come to the exams in this course, beware that they are as hard as any exam I have done. They have questions that range from easy to really hard -- they are designed to produce a result which distinguishes skilled from non-skilled people, NOT to allow everyone to pass. The points in steps 1 and 2 above will be a great help.

You're awesome Steve!! Thank you. I have a notebook that I have been using to jot down questions as I watched the videos, great being able to pause a lecture to formulate a questions when something pops into your mind I realize that the MIT course was probably a step or two above me, but I figured that if I could work through it, I might be able to understand what the hell I am doing!! Lovely, I can see the light sometimes and basic circuits are becoming less of a mystery (conceptually at least ).

I am currently reviewing KVL/KCL at allaboutcircuits.com, they have a video series that has already helped me tremendously. There are some points that the MIT course either didn't cover or assumed that have helped me with starting the analysis of a loop.

 

[chopnhack]

R1=1ohm
R2=2ohm
R3=6ohm
I=11A

Currently working on this problem, one of the first steps I wanted to take was to simplify the parallel combination of r1 and r2 which yields 2/3Ω.
But what becomes of the currents i1 and i2?

So far,

KCL
-i+i2-i5=0
i1-i2-i4=0
-i3=0

KVL
+v5=v1+v3____________________a: 6V=i1+6i3___________> these two can be equated_____:i1=2i2 <----interestingly this confirms the parallel resistor ratio
+v5=-v2+v3____________________b: 6V=2i2+6i3
+v5=v1+v4____________________c: 6V=i1+v4?
+v5=-v2+v4___________________d: 6V=2i2+v4?

Element Laws
v5=6V
v1=(i1)1Ω
v2=(-i2)2Ω
v3=(i3)6Ω
v4=?? how do you write a current source in terms of v? I assume that I would have to solve for voltage across the top rail first somehow, but when solving neglecting the current source, strictly a series resistance equation across the top, the result is incorrect.

i1=v1/1Ω or i1=v1
i2=v2/2Ω
i3=v3/6Ω

further substitution of KVL with above equations:

a: 6v=v1+6(v3/6) or 6v=v1+v3
b: 6v=2(v2/2)+6(v3/6) or 6v=v2+v3
c: v1+v4
d: 2(v2/2)+v4 or 6v=v2+v4

So now the proofs show that I should be able to use a: to get my v1 and v3, but that leads me to think of treating the top rail as a simple series equation that yields 0.122A for i1 and 0.734 for i3 which I know from a sim is incorrect...

I feel like I made some progress here, but the current source is still confusing me. I read a bit about series aiding and opposing, but haven't found much on parallel current sources other than they can be added together so long as they are at the same potential regardless of where they are in a circuit. Let me know if I made a good showing or if I should pack it in It's been close to 12 hours of studying today, my mind is addled - I'm off to watch the doctor...

 

[(*steve*)]

Have you been reading the recommended text?

I was perhaps less dedicated to doing this than I should have been, but when I did, it proved useful.

 

[chopnhack]

Have you been reading the recommended text?

I was perhaps less dedicated to doing this than I should have been, but when I did, it proved useful.

Yes, I have read the recommended reading for this past week. Some sections I have reread 3 or 4 times! It's a slow process... not realizing certain items are important until you see them in questions!

 

[(*steve*)]

Yeah, parallel current sources add to the current the same way that series batteries do.

Batteries in parallel make no sense if they're a different voltage, and thus current sources in series make equally no sense unless they have the same current. In terms of theoretical devices, you will never do it.

You can kinda think of a current source as the reciprocal of a voltage source. Everything is backwards!

Voltage sources can sink or source whatever current is required to maintain the voltage. Current sources can have whatever voltage is required to allow the configured current to flow.

 

[(*steve*)]

R1=1ohm
R2=2ohm
R3=6ohm
I=11A

Currently working on this problem, one of the first steps I wanted to take was to simplify the parallel combination of r1 and r2 which yields 2/3Ω.
But what becomes of the currents i1 and i2?

Well this is drawn is a way to make that deliberately difficult. I wouldn't do it. It will fall out.

So far,

KCL
-i+i2-i5=0
i1-i2-i4=0
-i3=0

Let's fist say that

I2 - I1 - I5 = 0

so I5 = I2 - I1

Then looking at the next node

I1 - I2 - I4 - I3 = 0

i4 = -I, and I1 - I2 - I5, so

I5 + I - I3 = 0

KVL

+v5=v1+v3____________________a: 6V=i1+6i3___________> these two can be equated_____:i1=2i2 <----interestingly this confirms the parallel resistor ratio
+v5=-v2+v3____________________b: 6V=2i2+6i3
+v5=v1+v4____________________c: 6V=i1+v4?
+v5=-v2+v4___________________d: 6V=2i2+v4?

Vr1 + Vr2 = 0, so Vr1 = -Vr2

Vr1 = I1.1
Vr2 = I2.2

so I1 = -I2.2

and I5 = I2 - (-I2.2) since I5 = I2 - I1
so I5 = -3.I2 (that's i2 not 12) (now we can calculate I2 later)

Element Laws
v5=6V
v1=(i1)1Ω
v2=(-i2)2Ω
v3=(i3)6Ω
v4=?? how do you write a current source in terms of v? I assume that I would have to solve for voltage across the top rail first somehow, but when solving neglecting the current source, strictly a series resistance equation across the top, the result is incorrect.

There's a loop there.

V3 - V4 = 0, so V3 - V4

and another loop V1 + V3 - V = 0

i1=v1/1Ω or i1=v1
i2=v2/2Ω
i3=v3/6Ω

yeah, but remove the ohm symbol

further substitution of KVL with above equations:

See if I have shed any light on where you may have gone wrong, and see if you can figure it out.

 

[chopnhack]

Well this is drawn is a way to make that deliberately difficult. I wouldn't do it. It will fall out.



Let's fist say that

I2 - I1 - I5 = 0

so I5 = I2 - I1

Then looking at the next node

I1 - I2 - I4 - I3 = 0

i4 = -I, and I1 - I2 - I5, so

I5 + I - I3 = 0



Vr1 + Vr2 = 0, so Vr1 = -Vr2

Vr1 = I1.1
Vr2 = I2.2

so I1 = -I2.2

and I5 = I2 - (-I2.2) since I5 = I2 - I1
so I5 = -3.I2 (that's i2 not 12) (now we can calculate I2 later)



There's a loop there.

V3 - V4 = 0, so V3 - V4

and another loop V1 + V3 - V = 0



yeah, but remove the ohm symbol



See if I have shed any light on where you may have gone wrong, and see if you can figure it out.

Thank you Steve, I can already see the correct direction and I thank you for it - I passed on the idea that a node could have more than 3 inputs because I had never seen an example of such... should trusted my instinct! I will review and try again tomorrow, thanks again!

 

[(*steve*)]

I passed on the idea that a node could have more than 3 inputs because I had never seen an example of such...

I guess that it is worth pointing out that a node is a fairly complex notion. Typically, anything connected along an unbroken conductor could be assumed to be connected to a node.

 

[chopnhack]

I guess that it is worth pointing out that a node is a fairly complex notion. Typically, anything connected along an unbroken conductor could be assumed to be connected to a node.

Indeed!

Some preliminary questions:

Let's fist say that

I2 - I1 - I5 = 0

so I5 = I2 - I1

Then looking at the next node

I1 - I2 - I4 - I3 = 0

i4 = -I, and I1 - I2 - I5, so

I5 + I - I3 = 0

When you write "then looking at the next...."

Where did you come up with i1-i2-i5? Is it a typo for the first line?

I found that I had the wrong polarity for i2 - I wasn't following the loop correctly... Aside from fixing that I found that v4 = 6i3 !!!! So the potential across the 11A source is 6x i3.

 

[(*steve*)]

yes typo. - and = are too close on the keyboard.

 

[chopnhack]

yes typo. - and = are too close on the keyboard.

Hey, at least I could figure that out ;-)

I have a number of equations at this point, some I have been able to equate and reduce. I seem to need more data to compute - if I had a current I could progress. At this point could I apply ohm's law and split the current supplied from i4 to i2 and i3? Would that be a logical next step?

EDIT:
I just realized while looking at the schematic for a way to start that voltage in parallel is equivalent everywhere, so the entire top rail should have +6v!!! That should give me a place to start calculating values

 

[(*steve*)]

It's only 6v until those two resistors.

I've taken some opiates for pain and my wife calls me "wasted"so I better not offer too much assistance at the moment

 

[chopnhack]

It's only 6v until those two resistors.

I've taken some opiates for pain and my wife calls me "wasted"so I better not offer too much assistance at the moment

ooh... feel better Steve. Not to worry, I remember there is an additive effect caused by the current source. If I can figure out how to apportion it, I may be able to do more ;-)

Rest up and feel better. Thanks again for all your input and guidance.

Some progress:

with 6v distributed to the parallel resistors I got:

i1=6/1 or 6A
i2=-6/2 or -3A
i1=i2-i5
6A=-3A-i5
9A=-i5 or i5=-9A

Sum of charges doesn't add up...

 

[(*steve*)]

Those figures aren't right. It is quite unlikely that 6v is dropped across those 2 parallel resistors. For that to be true the current through the other resistor would have to be zero and the current through the voltage source would have to equal that through the current source meaning the voltage across the current source would likewise be zero.

perhaps you could look again at the voltage divider solution and modify it for an unequal current through both legs.

If I have my head together later on I'll take another look at the equations. I'll see if I can point you one more step along the way

In the absence of my analytical skills I recommend you look for a single value, say the voltage across the resistor parallel to the current source if you can find two ways of expressing that then you may be able to solve for that. Once you have any single value the test will drop out.

The big hint is that it will usually require the solving of a series equations. If you find yourself getting an answer just by simplifying a single equation then you're either very lucky or you've made a mistake.

Actually... No I'm not going to talk specifics, I'm not confident I'd get them right.

Laplace, where are you?

 

[chopnhack]

Those figures aren't right. It is quite unlikely that 6v is dropped across those 2 parallel resistors. For that to be true the current through the other resistor would have to be zero and the current through the voltage source would have to equal that through the current source meaning the voltage across the current source would likewise be zero. Perhaps you Chou

Rest Steve! Tomorrow is another day

 

[Laplace]

Watching this exercise of trying to apply Kirchoff's Laws directly has been painful. Maybe I need some of those opiates, too.

This circuit has one node plus ground so a single node equation is required. But I never fuss with writing down node currents since the purpose of KCL is to find voltages, so I sum all the currents at the node in terms of the node voltage. To keep things simple I assume that all currents are positive (flowing away from the node) except for current sources flowing into the node which are negative. Then I can type the node equation directly from the schematic into MathCad. V3=12V. 2A flows through R3, 9A flows in reverse through the 6V source.

 

[chopnhack]

View attachment 14973

Watching this exercise of trying to apply Kirchoff's Laws directly has been painful. Maybe I need some of those opiates, too.

This circuit has one node plus ground so a single node equation is required. But I never fuss with writing down node currents since the purpose of KCL is to find voltages, so I sum all the currents at the node in terms of the node voltage. To keep things simple I assume that all currents are positive (flowing away from the node) except for current sources flowing into the node which are negative. Then I can type the node equation directly from the schematic into MathCad. V3=12V. 2A flows through R3, 9A flows in reverse through the 6V source.

Glad I could entertain you, pass the bottle when you get it... I have pages of equations but not much clarity, when they do a remake of A Beautiful Mind, they can have my desk and papers to toss out the window!

The shame is the concept is elusively simple, but I can't get the simple straight path you came up with...

BTW, there are actually three nodes in this circuit.