The Electrician makes really good points. I didn't say anything since I took your work to be scratch work in a notebook and not what you were planning to turn in. But the more I think about it, history shows that this isn't necessarily a good assumption.

You want to show your work in a neat and orderly progression, with explanatory labels or notes were useful. Don't make the grader have to guess where something came from, so make the steps from one line to the next incremental so that they can quickly determine what you did to get from one line to the next. If you plugged in equations into a linear equation solver to get numbers, then state that. Otherwise it looks like your values just magically appear (and, trust me, you do NOT want the grader's thoughts to go there).

I've attached the work I did so that you can see one possible model to use as a starting point.

Taking the time to really make your work stand out as polished, thoughtful effort really pays off. Even if they don't realize it, a grader is going to be biased in favor of such works. In sloppy work that is a pain to work through, a grader will unconsciously look for things to take off for and will be less inclined to give the benefit of the doubt. But in extremely well-presented work, the opposite is true; they will look for every bit of partial credit they can find.

When I was taking statics, I made a minor mistake on a problem (didn't multiply a given dimension by two even though I knew the dimension was from the midpoint of the beam to the end and I needed the total length). It was a simple good, but the rules of the course said that wrong answers get zero credit. So, on that problem, I should have received only 4/10 points (IRRC). But when I got it back I lost no points and there was a note from the grader that said something very close to, "You should have lost 60% on this problem, but this work is so beautiful that I just couldn't bring myself to deduct points for such a minor mistake."

I was a bit puzzled by this, but a semester or two later I became a grader for Physics II and after seeing the first set of homework turned in, I immediately understood where that grader was coming from.

Another thing to keep in mind -- grading is super-tedious and is usually very poorly compensated. For a large class, it's not uncommon for a grader to only be able to justify five minutes per student submission (not problem, but for the entire submission). That's generally impossible to achieve, and so the grader is basically having to spend there own time to do a halfway decent job -- and they are aware of this. Now consider that well-presented work is much easier and quicker to grade than sloppy work. Guess who gets on the grader's good side really fast.

 

We just needed to submit our answers, so this all is just rough calculation. I am not turning in this but yes I would keep in my mind to them a bit more properly.

 

The Electrician makes really good points. I didn't say anything since I took your work to be scratch work in a notebook and not what you were planning to turn in. But the more I think about it, history shows that this isn't necessarily a good assumption.

You want to show your work in a neat and orderly progression, with explanatory labels or notes were useful. Don't make the grader have to guess where something came from, so make the steps from one line to the next incremental so that they can quickly determine what you did to get from one line to the next. If you plugged in equations into a linear equation solver to get numbers, then state that. Otherwise it looks like your values just magically appear (and, trust me, you do NOT want the grader's thoughts to go there).

I've attached the work I did so that you can see one possible model to use as a starting point.

Taking the time to really make your work stand out as polished, thoughtful effort really pays off. Even if they don't realize it, a grader is going to be biased in favor of such works. In sloppy work that is a pain to work through, a grader will unconsciously look for things to take off for and will be less inclined to give the benefit of the doubt. But in extremely well-presented work, the opposite is true; they will look for every bit of partial credit they can find.

When I was taking statics, I made a minor mistake on a problem (didn't multiply a given dimension by two even though I knew the dimension was from the midpoint of the beam to the end and I needed the total length). It was a simple good, but the rules of the course said that wrong answers get zero credit. So, on that problem, I should have received only 4/10 points (IRRC). But when I got it back I lost no points and there was a note from the grader that said something very close to, "You should have lost 60% on this problem, but this work is so beautiful that I just couldn't bring myself to deduct points for such a minor mistake."

I was a bit puzzled by this, but a semester or two later I became a grader for Physics II and after seeing the first set of homework turned in, I immediately understood where that grader was coming from.

Another thing to keep in mind -- grading is super-tedious and is usually very poorly compensated. For a large class, it's not uncommon for a grader to only be able to justify five minutes per student submission (not problem, but for the entire submission). That's generally impossible to achieve, and so the grader is basically having to spend there own time to do a halfway decent job -- and they are aware of this. Now consider that well-presented work is much easier and quicker to grade than sloppy work. Guess who gets on the grader's good side really fast.

I do understand your concern and yes I feel that my work is messy. But just to inform I am not turning this in, we just have to type our answers in the questionnaire. Surely I will now show he process of making the equations from scratch, make the work presentable and as you said always keep a track of the units.. Thank you for your suggestions again.

 

Prakhargr8,

Since you did so well solving this circuit, and all you have to do is give answers, how about a little extra credit problem?:

What is the Thevenin resistance of your circuit at node A?

 

I am not sure ab

Prakhargr8,

Since you did so well solving this circuit, and all you have to do is give answers, how about a little extra credit problem?:

What is the Thevenin resistance of your circuit at node A?

I was not quite sure about what you meant so I found the Thevenin's resistance and voltage across A and D. Do let me know me if this is wrong or the question expected something else

 

I am not sure ab


I was not quite sure about what you meant so I found the Thevenin's resistance and voltage across A and D. Do let me know me if this is wrong or the question expected something else

You have done the case where the 10Ω resistor is taken to be a load to a Thevenin equivalent.

What I had in mind was to treat your entire circuit as a Thevenin equivalent which would power another external resistor load. For the case I had in mind, Vth was what you had already calculated, namely the voltage at node A. All you needed to do was to calculate the value of Rth, which is the resistance from node A to ground; it happens to be the same as what you calculated for the case you worked out.

You did calculate the mesh currents correctly, but you made a mistake at the very end where you obtained Vth as the series connection of the 12V source, the current source and the 7Ω resistor. How do you know what the voltage across the current source is?

If you calculate Vth by using the upper left hand mesh you'll get the correct result. You would have 3.4V in series with the voltage across the 4Ω resistor. What do you get if you do that?

 

I'm impressed by how you have jumped right in to do some extra calculations. Lots of practice is what makes you prepared for the mid-term exam!

If you want a little more how about this. As you know, you could have solved the original problem using superposition, but it would be lot of algebra. Try a couple of the parts. What is the voltage at node A if only the current source is active; the three voltage sources are set to zero volts. And then, what is the voltage at node A if only the 12 volt source is active; the current source is open circuited and the other voltage sources are set to zero

 

As you know, you could have solved the original problem using superposition, but it would be lot of algebra.

Not that much at all, actually. In two cases the circuit immediately reduces to a simple voltage divider. In the case of the current source, it's just that current into an equivalent resistance. As is often the case with superposition, many of the resistor combinations appear in multiple circuits, so intermediate results can be reused. In the case of the 12 V source, you immediately have a very simple two-mesh circuit in which the voltage sought is merely one of the mesh currents multiplied by one of the resistors. So each of the four circuits is immensely simpler that the full circuit with correspondingly simple and limited algebra involved.

 

Not that much at all, actually. In two cases the circuit immediately reduces to a simple voltage divider. In the case of the current source, it's just that current into an equivalent resistance. As is often the case with superposition, many of the resistor combinations appear in multiple circuits, so intermediate results can be reused. In the case of the 12 V source, you immediately have a very simple two-mesh circuit in which the voltage sought is merely one of the mesh currents multiplied by one of the resistors. So each of the four circuits is immensely simpler that the full circuit with correspondingly simple and limited algebra involved.

To calculate the equivalent resistance that the current source will be driving takes 11 add/multiply/divide floating point operations (FLOPS). To calculate the contribution made by the 3.4 volt source takes 9 flops, and the contribution of the 10 volt source takes 9 flops. I haven't evaluated the number required for the two-mesh circuit, but the contribution of the 12 volt source can be calculated with a simple source transformation with 14 flops. The two-mesh solution probably requires a similar amount of arithmetic. I won't bet my life that these flop counts are exactly correct, but they're close enough.

So the full superposition solution requires around 40 flops worth of algebra/arithmetic, plus or minus. And this doesn't count the time required to study each case to form a plan of action, which must be done 4 times. It's a good thing we don't have to do it with log tables or a slide rule.

I wouldn't describe this as "Not that much at all"

 

Not all operations having the same number of flops are equivalent. Simplifying series/parallel combinations of three to five resistors is something that can be done (after a minimal amount of experience) one the fly without needed to do a bunch of algebraic manipulations on paper. The same for calculating the output of a voltage divider. But manipulating a set of arbitrary linear equations usually involves more deliberate thinking and some amount of manipulation on paper (or elsewhere other than in your head) even if the number of operations turns out to be the same.

But how many flops are needed to formulate the node equations, put them into a form that is readily solvable, and then perform the steps needed to actually solve them (as opposed to throwing them at some program to solve them)?

Out of curiosity, let's see what my actual work entailed.

First I tackled the 10 V source. That took 10 flops. Setting it up, including redrawing the circuit with the other sources turned off, literally took twenty seconds (I just did it from scratch). Seeing the plan of attack was trivial since is was clearly a simple voltage divider, so that occurred while I was drawing the circuit. In fact, it was obvious this would be the case for both supplies that had their negative terminals tied to the signal common, which is what suggested using superposition to begin with. Doing all the computations to come up with Va, as well as two intermediate results that proved useful later on, took just under 40 seconds. So in less than a minute I had the solution for one source.

Second I tackled the 3.4 V source. That took the same amount of time to set up. Using one of the intermediate results from the prior step got the answer in less than half the time using only 6 flops. So well under a minute for that one.

Next was the 2.5 A source. That one also took the same amount of time to sketch (each of these involved one source and five resistors, so that's not surprising). Using an intermediate result from the 3.4 V case, the answer was had in 4 flops.

The 12 V source was, indeed, the trickiest. It also took the same amount of time to draw, but perhaps an additional two minutes to decide on using mesh analysis (even though I'm sure there's a better way). Setting up the equations and solving them took 5 flops and less than two minutes. Arguably it was 8 flops, since I did 10 Ω + 2 Ω + 2.857 Ω and 1.5 Ω + 2 Ω in my head.

Combining all four results then consumed another 3 flops.

So, all told, it was about seven to eight minutes and 28 flops That doesn't strike me as excessive. Plus, by working on small, independent subproblems, most people are less likely to make a mistake and more likely to catch ones that they do since the computations are directly tied to things with physical meaning compared to manipulating some linear equations in which mistakes are easily missed because the intermediate results are too abstract.

I would be a bit surprised if solving it via nodal analysis can be done in fewer flops.

I also fail to see how having to use log tables or slide rules would be any more of a burden for doing superposition than for for doing nodal analysis.

 

more how about this. As you know, you could have solved the original problem using superposition, but it would be lot of algebra. Try a couple of the parts. What is the voltage

Ye

You have done the case where the 10Ω resistor is taken to be a load to a Thevenin equivalent.

What I had in mind was to treat your entire circuit as a Thevenin equivalent which would power another external resistor load. For the case I had in mind, Vth was what you had already calculated, namely the voltage at node A. All you needed to do was to calculate the value of Rth, which is the resistance from node A to ground; it happens to be the same as what you calculated for the case you worked out.

You did calculate the mesh currents correctly, but you made a mistake at the very end where you obtained Vth as the series connection of the 12V source, the current source and the 7Ω resistor. How do you know what the voltage across the current source is?

If you calculate Vth by using the upper left hand mesh you'll get the correct result. You would have 3.4V in series with the voltage across the 4Ω resistor. What do you get if you do that?

Yes you are right. Here is the answer using the other mesh.

 

I'm assuming that actual arithmetic operations of (+ - * /) are done on a calculator by hand. What I'm calling a FLOP is a single push of a button on a calculator to carry out one of those operations. The operations I'm considering are those four, and each one is a single FLOP to the calculator user.

So when I say that it takes 11 FLOPs to calculate the equivalent resistance that the current source will be driving, I'm only counting the number of those 4 fundamental arithmetic operations that will be needed--how many button pushes.

I'm not counting the button pushes needed to type in all the numbers for the component values. And if the calculator only has one auxiliary memory some intermediate values may need to be written down on paper and later re-entered, using more button pushes. I didn't use any intermediate results because I'm using a simple calculator that only has one memory; more than one intermediate result requires writing them down and re-entering them. This particular circuit is such that intermediate results are available. This won't necessarily always be true. If done on an HP calculator which has a stack to hold intermediate values, thing would go much faster with many fewer button pushes.

Your own nodal solution in post #21 uses about 30 or so button pushes. Fewer are needed if Cramer's rule is not used. That's less than the 40 or so I count for superposition.

I think it's a lot of algebra/arithmetic to do this with a calculator by superposition, quite a bit more than "Not much at all"; you may disagree.

You misconstrue my comment about log tables. I'm not comparing superposition vs. nodal with log tables. I'm saying that even though it takes a lot of button pushes to solve this circuit with a calculator, we're lucky we aren't using log tables or a slide rule to solve it by whatever method.

 

Ye


Yes you are right. Here is the answer using the other mesh.

The 3V source should actually be 3.4V. Consider the polarity of the voltage across the 4 ohm resistor; the bottom of the resistor is positive, so you must subtract the 3.4V from the voltage across the resistor.

 

Prakhargr8,

How are you doing your calculations? Are you using a calculator, or a math app like Matlab or Mathcad? Or some app on the web?

 

I'm assuming that actual arithmetic operations of (+ - * /) are done on a calculator by hand. What I'm calling a FLOP is a single push of a button on a calculator to carry out one of those operations. The operations I'm considering are those four, and each one is a single FLOP to the calculator user.

So when I say that it takes 11 FLOPs to calculate the equivalent resistance that the current source will be driving, I'm only counting the number of those 4 fundamental arithmetic operations that will be needed--how many button pushes.

That's what I took your FLOP to mean.

I'm not counting the button pushes needed to type in all the numbers for the component values. And if the calculator only has one auxiliary memory some intermediate values may need to be written down on paper and later re-entered, using more button pushes. I didn't use any intermediate results because I'm using a simple calculator that only has one memory; more than one intermediate result requires writing them down and re-entering them. This particular circuit is such that intermediate results are available. This won't necessarily always be true. If done on an HP calculator which has a stack to hold intermediate values, thing would go much faster with many fewer button pushes.

I used the Windows calculator and didn't use its memory feature at all. I love RPN, but the Windows calculator is just too damn convenient to pull up compared to digging out my HP.

To highlight your point about FLOPs not capturing the entire picture, consider finding the equivalent resistance of two parallel resistors.

R1 = 1/(1/R1 + 1/R2) 4 FLOPs
R1 = (R1·R2)/(R1+R2) 3 FLOPs

But the number of button pushes is likely considerably less for the first approach compared to the second because of the need to enter each value twice, plus, the parens would need to be used on an algebraic calculator (and avoiding the parens would likely involve an additional multiplication taking it to 4 FLOPs, too).

Useful intermediate results are usually available when doing superposition because segments of the circuit with the same source configuration appear in multiple analyses -- and this increases as the number of sources increase.

Your own nodal solution in post #21 uses about 30 or so button pushes. Fewer are needed if Cramer's rule is not used. That's less than the 40 or so I count for superposition.

I count 32 FLOPs for my nodal solution compared to 28 for superposition. So they are quite comparable.

I think it's a lot of algebra/arithmetic to do this with a calculator by superposition, quite a bit more than "Not much at all"; you may disagree.

The "not much at all" is relative to the effort required for the alternatives (nodal analysis in particular).

You misconstrue my comment about log tables. I'm not comparing superposition vs. nodal with log tables. I'm saying that even though it takes a lot of button pushes to solve this circuit with a calculator, we're lucky we aren't using log tables or a slide rule to solve it by whatever method.

Thanks for the clarification.

 

Prakhargr8,

How are you doing your calculations? Are you using a calculator, or a math app like Matlab or Mathcad? Or some app on the web?

Oh I am using a scientific Casio calculator, we will be doing MATLAB I guess towards the end of our semester

 

YE

The 3V source should actually be 3.4V. Consider the polarity of the voltage across the 4 ohm resistor; the bottom of the resistor is positive, so you must subtract the 3.4V from the voltage across the resistor.

Yes I took 3V instead of 3.4 V in a rush. That's very silly of me. I think not all my answers in the assignment I submitted were correct but doing some many calculations and cross verifying them by myself has given me the much needed confidence. Thank you for sharing all your suggestions.