It's a Visual Thing for Me!

Look up Circuit Cellar on the web.

Also, Digikey and Mouser are great places to get supplies.

You are on your way!

No slap for you! That's not a dumb question. I was actually the other way. I had trouble with fluids till I started treating fluid flow like it was electricity and then it made sense.

In my first electronics class (in the Army), there was a guy who both knew his stuff and who could do vocal effects and he could imitate an electron flowing, then straining through a resistor, oscillating in a tank circuit and so on. A real hoot, but when we were studying at the local beer hall, it was a great aid.

I don't know if you're far enough along for something like PSPICE to be effective, but that works for some people. I'll see what else I can remember (I'm at that age where it takes me as long to remember as it does to go to the bathroom).

JScott13, That looks like a good starting block! I'll take a look at that now. Cheers

JScott13 posted the one I was trying to remember. Maybe 10 years back (?), I got a couple modules for a tech I hired who had "exaggerated" his resume and I thought they were good.

This looks like the sort of thing I'm after! It looks basic from the first lesson but that's a good thing! I need to start from scratch and get it in my head!

You had mentioned being a visually oriented person, that's what made me think of them. Very reasonable price for very well animated course. Circuits are "seen" working which helps bring home the concepts.

You hit the nail on the head there Scott, cheers again

Even the word 'electronics' sends shivers down my spine! Cheers AH (don't know why just yet but cheers anyway) TP45, I'm a first step begginer looking to get off on the right step! I'll take a look at that PSPICE and see if I can gleen some info from it, cheers!

Would you please take a look here. Electronics Workbench is so good software for education.

There's some thoroughly readable book(s) called, "The Art of Electronics". Others have enjoyed them, and recommend them.

What do you want to do with your electronics? If you want to learn a language you can either start at the bottom and learn the vocabulary, rules of grammar, correct sentence order, collocation and so on. Or you can use the phrase book approach, where you learn useful phrases to communicate in particular situations.

If you consider your intended application(s) you could play with standard circuits that do a particular part of that application, build them and see what happens, a slightly different application a different circuit, eventually you spot a pattern. Even the best designers cut and paste circuits, why re-invent the wheel? If you haven't already invest in a breadboard for experimenting. I apologize if you know this already http://en.wikipedia.org/wiki/Breadboard Digital logic is the easiest for immediate results and works immediately, but LEDs going on and off is about as visual as it gets, analogue electronics is where it gets a bit tricky, again consider what you want to do, and good luck

I'd suggest playing with logic chips first, these will be easy to visualise using the 'gate' analogy or seeing them as valves/water or relays or a mixture of all 3. The great advantage is you can use switches as inputs and leds as outputs and you can use a simple battery supply and a multimeter.

The next project I'd suggest is multivibrators using transistors or a 555 timer, however you really need a 'scope at this point....
If you have a scope then move onto single transistor audio amplifiers.

Then design anti garvity, perpetual motion and time travel... (PS I show you how to build that last one in 10years time )

There was a thread a while back askingf about good electronics text books.
'The Art of Electronics' was recomended by several including me.

Have fun.... if you need a hand we will be gentle with you.

Del

But Del, you showed me the time travel circuit 10 years ago using the over unity ball up a slope trick method!

Are you looking for a useful analogy to visualize electricity? This could help:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir.html

Good point about the books...

the 'Ladybird' books are still seen as a very good introduction to electronics/computers/'levers,pulleys and engines' is agood one
There is a good one on how a car works, very good for new learner drivers...it's amazing how many people drive the damn things without the faitest idea of how it works.

Del

Cheers Chris, Some good visual stuff there! My version of a variable resistor was a lock gate that you could raise or lower, and the diode was a like a pea valve but I guess they are what makes sense to us the best! I'm going to keep on pushing on with the web course and do some experimenting too so I'm keeping my fingers crossed I don't blow something up in the process! Then again don't ya just love the smell of burning resistors in the morning!

Good post Chris!

I'd like to offer a hydraulic analogy for the field-effect transistor...

My old teacher told us that you could think of it as a flow controller (hydraulic analogy). Instead of you controlling the flow with a knob, the flow (current) is controlled by the water pressure (voltage) from somewhere else.

The junction transistor is a bit more difficult to find an analogy for since the output current is determined by the input current, not voltage.

I forgot to say thanks Tom! I hope you stay around to tell us more stories

I like to envisage electronic circuits as a playground with children running about. The boys are the electrons and the girls the positive counterparts, called 'holes', which move in the opposite direction. The battery or power supply is the entrance to the park, with queues of kids (full of energy!)

Resistors are narrow passages that the kids have to squeeze through.

Inductors are turnstiles or roundabouts that don't change speed or direction easily.

Capacitors are footpaths along opposite sides of a canal that the boys and girls line up opposite but can't cross.

Transformers are roundabouts that are geared together.

Diodes are one-way gates.

Transistors are special gates that allow large groups of children through when one child appears at another entrance.

I hope this helps to envisage some of what's happening in some circuits. As you've hinted, there's analogies in the world of plumbing but some things are harder to visualise with water.

I had the same problem. Back in the day, (WAY back), I attempted to study electronics, but once it moved beyond my ability to also liken it to the flow of water, I was lost.

So I went for aptitude testing, and discovered, to my surprise, that I have a strong mechanical aptitude and exceptional ability to visualize in 3D.

So I got into mechanical design, and haven't looked back since. Except I do think I could make use of a stronger electromechanical knowledge base.

Perhaps this would be a field where you could transfer and apply your skills and knowledge.

But when I find myself wanting in this regard, I call in a few reps from the right companies and pick their brains! They fill in the blanks for me, and in return they get the sale.

Sometimes you have to play to your strengths.

The trouble with electrons is that you cant see them but you can if they are turned into photons and all you need is an LED to do that. Further trouble though because an LED needs a certain forward voltage before it starts to conduct, think of it like a concrete step into a building, the step must be overcome with a potential equivilent to the band gap in electron volts of the material thats makes up the LED.

Once the step is overcome current floods through the LED and if not limited by a resistor will PoP the LED.

I have developed a device called INDI-LINK that solves most of these problems and so to your own in visualising the dynamic operation of an electric circuits. Sadly I cant find anyone prepared to exploit this development in the UK and I find myself £60,000 poorer with the cost of Patents.

Im just about exhausted trying to recoup my savings but it was born of the same problem you have posted.

It has a current range of 3mA to 3-10 Amps AC or DC while requiring the minimum of forward volts (0.35 V) to produce a light signal. Its enough for the potential difference between a small Copper electode and similar Zinc to energise the device, yet you can place it in series with a 220 volt AC 10 amp electric appliance whereby it produces more light in proportion to the current level.

Anyone over the pond like to help me commercialise this product? because the UK community seems to be dead from the neck up as far as embracing invention.

INDLINK's in the photo are 0.003 to 3Amp types but this can be easily increased, A CT is also usfull if currents in the 100's of Amps are of interest.

"Anyone over the pond like to help me commercialise this product? because the UK community seems to be dead from the neck up as far as embracing invention."

Maybe KrisDel™ could help.

docrobar,

Great idea! This would be a Godsend for the field electrical Instrument & Electrical technicians troubleshooting tools. It would provide the much needed "at-a-glance" visual that could/would cut troubleshooting downtime by a large margin. There are several companies that might be interested.

Try contacting: Phoenix Contact Ltd.

P.O.Box 4100

Harrisburg, PA 17111-0100

www.phoenixcon.com

Ph: 717-944-1300

Fax: 717-944-1625

or

TURCK INC.

3000 Campus Drive

Plymouth, MN 55441-USA

Ph: 763-553-7300

Fax: 763-553-0708

www.turck_usa.com

Good luck!

Thanks ever so much everybody, I have a good understanding of electrical circuitry when it comes to industrial or domestic stuff ie 415V three phase and 230V single phase, even down to 12V control circuits because I can follow the wires to there destination, so all that lot is mostly ingrained now! It's the stuff I can't follow around that foxes me!

I started the course yesterday on the net as suggested and up to now, it's just been stuff I did years ago at college! I'm going to go through it all if I can! My other problem is I live in Spain so Spanish colleges are out for me because I only know basic Spanish!

I'm going to see if my sister in England can send me over the Ladybird book of basic electronics to see if that can help also!

I'll let you all know how I get on and Cheers one again

and don't forget Truman, you ALWAYS have the electronicers here when things don't make sense.

Bill

Bill, it's a great comfort to know that if I get stuck I can always ask you guys for a bit of help without getting slapped about too much!

"I'm going to see if my sister in England can send me over the Ladybird book of basic electronics to see if that can help also!"

Also have your sister get in touch with Del for an introductory copy of "Deltronics".

Incidentally, the link to MIT open courseware by JasBond is a good one.

-John

LOL I'm going to look at the MIT site while I wait for a reply from Twisted Sister Pair and of course the reply from my sister about the ladybird book! Do you refer to the Deltronic book of easy circuits for the use in Over unity and Time travel systems 'as seen on TV' ! I've read the reviews, Mrs. Jones from Cumbridge Barton said and I quote, 'Now I've discovered the secrets to unlocking the awesome power of the Over unity space time warp drive, I have so much more time in the day to do the things I never had time to do before. A must have for any modern girl'. I think it's a god send that Del wanted to share this research with us, don't you?

On the subject of student help...

My student efforts are not what I had hoped. It seems as though much of the math I took for granted has flown the coop. I am faced with simple problems in this beginners course and yet can not find a good source for remedial help.

I need to be able to deduce a formula given a specific circuit analysis problem. Any circuits that require straight forward applications of well known formulas don't seem to be a problem, but when I need to manipulate the formula to suit the specifics I seem lost.

Any suggestions? (thanks in advance)

I've just finished the first course and got 100%! I took 3 attempts repeating the course from start to finish until it sunk in! The first time, I was a bit baffled so I did it again and again until it finally clicked! Put it down on paper, ie draw the circuits and fill in the missing info using ohms law until you have the info you need! Be careful with the units. I'm so pleased with myself, that I'm going to buy the rest!

At the risk of getting castigated...

Electronics requires very few formulae unless maybe you are into RF. All the Thevenins teorems and the like are fine in theory...but I've never used 'em.
You seldom have fancy networks, and when you have, they can generally be simplified, rationalised, approximated, tested empirically or simulated.

Ok you need to understand the stuff so that you recognise how to rationalise to,.. say a constant current source or a voltage source and a resistance...but it's more about understanding than maths. Also most circuits are designed around standard values such that ratios rather than absolute calculated values are more important.

This maybe viewed as heresy..I'd be interested to see what the other electronics guys think.

Del

Maybe I should qualify my previous post..

You need Ohms Law, and the ability to work out a potential divider.
A lot of the rest is knowing stuff....like,
'I can ignore the base current if I make the current through the potential 50 times greater than the base current'... it's that sort of approximation that makes it work...
and the understanding that the emiter resistor will provide the DC stability at some where near the calculated point.

If you can bias a single transistor common emiter circuit and understand it you are a looong way towards knowing your stuff .

(Ducks rapidly in anticipation of floods of derision)

Del

Though I fully appreciate what you're saying about likelihood of using this math, at this point I just have to know.

In brief my problem is :

I am trying to solve for the resistance of a resistor in series. I have been given the source voltage and the resistance of two of the three resistors. I have also been given the voltage dropped over the mystery resistor. I need to solve for total circuit current, resistance of the unknown resistor and power consumed by the circuit.

Given that I can't calculate current without total resistance I was stumped. Knowing the voltage over the mystery resistor was 4.5 volts and that the applied current was 20 v I thought that:

4.5=20(Rx/Rt) solve for Rx

However that provided an incorrect answer.

Isthis a real problem or a paper problem?

If it is a real problem...you measure the voltage drop (V) across one of the known resistors (R)... this gives you the current (I) through the chain of resistors I= V/R
You then measure the voltage (u) drop across your unknown resistor (x).

x= u / (V/R)

If it is a paper problem...we need to see the whole problem...

(I'm assuming it's a DC circuit)

Dunno if this is of any help?

Del

For the paper problem...
If the prob is 3 resistors in series with just one unknown.
Say 100ohm, 220ohm and unknown R. With a 20volt supply.

Current (I) down the chain of all 3 resistors is I= 20/ (100+220+R)
(from re-arranged ohms law I=V/R)

If you know the voltage drop across unknown R...

R=V/I ......you know the volt drop V....we've just worked out an expession for I (highlighted)

So plug the numbers into the expression and solve for R...

If you need help doing that, I'm sure we can show you step buy step.... but it will be easier if you give us the relevant figures.

Del

Information provided regarding this series circuit (and yes dc is assumed)

R1=4.7k

R2=1.5k

Voltage over R3= 4.5V

Total applied voltage = 20V

Using what I know about ohms law I deduced-

Voltage over R1=11.75V

Voltage over R2=3.75V

My confusion arises from the fact I only have voltages. I have no resistance to work ohms law for R3 and I can't figure total current with total resistance (an assumption I'm beginning to understand must not be true).

I first see a major contradiction "the applied current was 20v". Current is not measured in volts. Now if it should be "the applied voltage is 20v", this we can work with.

So we have 3 resistors in series... Lets call them R₁, R₂, and R_x. Lets assume they are in that sequence with R_x being on the ground side.

Lets first lump the two known resistors into one and call it R_A. R_A=R₁+R₂. The voltage across R_A would be 20v-4.5v=15.5v. The current through the string would be I=15.5v/R_A. This same current flows through R_X so R_X=4.5v/I. Total power would be 20v x I.

Ha... you sound like you know what you are doing... but this is begining to sound familiar...

A guy I work with (we shall call him Mick...'cos that's his name) has been doing a correspondence course in electronics....the exercises were full of missprints, missing information and barking mad questions which bore little resemblance to real electronics.
To add to the problem ..his 'tutor' was an imbecilic Scotsman who appeared to be illiterate, drunk or both...

I helped Mick by reviewing his work before he sent it off for marking and nudging him in the right direction...one lot came back marked 'D' despite being near perfect... he sent it back with a nasty letter...it came back an 'A' after re-marking.!
Be carefull you may find questions with silly units and missing information. Questions with resistors in ohms and currents in amps are generally unrepresentative of real electronics where kΩ and mA are the norm

Any how...good luck with the course... feel free to drop me a Private Message if you wish..

Del

Thank you! and yes, I meant applied voltage (sorry bout that)

I did add the two known resistances to find the current. And I did know that current over the two knows would be the same current over the unknown. But for some reason this all didn't come together until just now.

Thank you all for being so patient with me as I reacquire long lost knowledge!

Man I feel a lot better now!

Now that we have numbers, I=2.5 mA

R_X=1.8KΩ

P_T= 50mW

I must admire whomever came up with the problem because they use real world component values. Even R₁ + R₂ is another standard value (6.2K).

WANT SOME MORE FUN??

I have a 10 volt power supply and 3 10K resistors. I need a resistor network such that I get 3.3 volts out. There are actually 2 solutions to this problem. Can you find both of them?

Bill

well the first solution would be for the three resistors to be in series

VRx=Va(Rx\Rt)

for the second...

Oh, this will require some thought... "I'll be back"

And am I right with the math here, all three in parallel will yield 3V? Not an exact answer, but within the tolerances I'm accustomed to. Forgive me but I only just began learning the math respecting parallel circuits.

I can see you getting hooked on this like me! I've just managed to download the 27 video courses on the MIT site. I've been doing a whole load of examples, and it's clicked into place now. That first lesson was only about series resistors, so we've got a long way to go before we're making 'remote controlled, with web cam vision, automatic pest shooting peashooters connected via internet to the webpage of your choice'! circuits!

Yes, it is bit of an addiction. And as you mentioned, the more I satiate the need for information the more understanding I accumulate. Or more succinctly... studying works! The MIT link is fantastic! Hard to believe all of that is openly available to the public. I will have to go over what's there once I get some more free time.

Your right on the first one. Topologically it is the same as your problem... and with 20K to the power supply and 10K to ground, you would have 1/3 the supply voltage or 3.33 volts.

The second solution is a wee bit tougher. Two equal value resistors in parallel have an equivalent resistance of 1/2 the value of a single resistor so in our case, 2 parallel resistors = 5K. So we run 2 parallel 10K resistors to ground... and the 3rd 10K resistor to the power supply, and we still get 3.33 volts out. The second solution does draw twice the power from the supply though.

I could go on and on with this, but I think I had better be good and quit for now.

Have fun

Bill

You only have 15.5 V! The resistors split the voltage so the other two resistors must share 15.5 V left, so now you can use the V/R=I 15.5/(R1+R2)=I to give you the circuit current! Then you can work out the VR1 and VR2 and the resistance of R3 and also the power used by all!

I wrote this reply before I noticed that there were several other replies. Apologies if I've covered the same ground.

--------SNIP end of edit--------------

That looks right to me.

4.5=20(Rx/Rt)

i.e. 4.5Rt = 20Rx

so 4.5(Rx+R1+R2)=20Rx

OR 4.5Rx + 4.5(R1+R2)=20Rx

So 15.5Rx = 4.5(R1+R2)

Or Rx = 9(R1+R2)/31

Did you mean like this?

Or perhaps it's easier if you just see that Rx/(R1+R2) = 4.5/15.5

Yes, that's the circuit. And no problem on the late entry. Every one of these explanations reinforces the concept and bolsters my understanding. I can not thank you all enough.

Resistors in series all see the same current flow, and their individual voltage drops add up to the total voltage across the series.

Your problem gives you the total voltage applied to the series, and the voltage drop across the mystery resistor - so by subtraction you can get the voltage drop across the pair of known resistors. Using Ohm's law you can get the current flowing through them.

The current in the mystery resistor is the same as in the others, and you have the voltage across the mystery resistor. Use Ohm's law again to calculate its resistance.

That's what I said in post #55(ish). I'm glad you agree, I've just started the same course and I feel very at ease with resistors in series! Can't wait to do something a little more complex like resistors and a light bulb but as yet, I have had no reply from Twisted Pair or the recently published Deltronic book of easy circuits for the easily distracted so I am in limbo. What I don't want to do is go rushing in like a nutter and miss stuff out! I tried to get to grips with Electronics twenty odd years ago but couldn't get my head round it, this time I intend nailing the beast!

Hi Odessey2001,

-John

Excellent post Andy.

-John

Fascinating, Andy. I studied Electronics (component repair and troubleshooting, designing and building with discretes) in the US Navy, and have used the plumbing analogy many times in explaining electronics. Now I have a new model. Plumbing as a model never developed to the level of detailed logic yours reaches. I'll use it (with your permission, of course. But then, this IS a public forum, so I guess the idea isn't well protected, either, is it?) whenever I am teaching physics and general science students about electron flow, and electronics theory. Thanks for a great, and detailed, analytical tool.

Micah

Micah

No problem to use the model; I'm glad you find it helpful. As I say, water is great for getting a qualitative feel for what is happening but is harder for quantitative analysis unless you are very familiar with working with flow rates and pressure drops. The marbles idea mentioned below also works on the same principle. It's fun thinking through the analogies; effectively it's looking at what is happening on the electron level and you can analyse even very complex systems and components with the tool.

Enjoy

One old teacher of mine used marbles instead of people but the learning was the same.

In his lecture, the speed at which the marbles moved was constant but it was the number of marbles that varied. Packing in the marbles more tightly represented more current. If you over-pack the marbles (exceed the current rating of the wire), you run the risk of bursting the pipe (burning the wire).

It was a nice analogy which really helped us understand things.

Oh Jeez!!!

I believe this requires a technique called superposition (which fits in between series/parallel circuits and Thevenin). I studied that stuff way back in 1966 and would have to dig out the old textbook to solve it. I can see the textbook on the shelf but I would need a ladder to get to it so I will pass for now.

Bill

This is probably not the most elegant solution to the problem, but it should work.

Lets assume that we apply a voltage E at point A. We don't care what E is as we will get rid of it later. This will result in 3 currents in the network... I1, I2, and I3.

Using Ohms law (I = E/R)

I1 = E/200 (200 = R1+R2)

I2 = E/300 (300 = R3+R4)

I3 = E/700 (700 = R1+R5+R4) {I noted that current flow would be in this direction because voltage at the junction of R1& R2 is greater than at the junction of R3&R4}

Total current I = I1+I2+I3 = E/200 + E/300 + E/700

the least common denominator in the above equation is 4200 so

I = 21E/4200 + 14E/4200 + 6E/4200 = (21E + 14E + 6E)/4200 = 41E/4200

I/E = 41/4200 thus E/I = 4200/41

Of course E/I = R so R = 4200/41 or R = 102.43 ohms.

The above utilized Kirchoffs current law which you probably havent gotten to yet.

What say John... am I right??

Bill

I was not correct...

I3 = E/600 not E/700

Total current = I1 + I2 + I3 = E/200 + E/300 + E/600

LCD is 600 so

I = 3E/600 + 2E/600 + E/600 = 6E/600 = E/100

I/E = 1/100 and E/I = R = 100 ohms

Bill

Bill

The reason why this doesn't give you quite the right answer is that I3 is also flowing through R1 and R4, increasing the voltage across those resistors and upsetting the balance of things. To do the full, mathematical analysis, I suggest the following:

Let the current through R2 be I1; the current through R3 be I2; and the current through R5 be I3.

This gives the currents though R1 to be I1+I3 and that through R4 to be I2+I3.

The voltages across the resistors are therefore:

R1: 100 (I1+I3)

R2: 100 I1

R3: 200 I2

R4: 100 (I2+I3)

R5: 400 I3

We can then pull out a couple of simultaneous equations about the voltages at the top of R1 & R3 and that at the junction of R1, R2 & R5:

100 (I1+I3) + 100 I1 = 200 I2 + 100 (I2 + I3)

100 I1 = 100 (I2+I3) + 400 I3

Using standard mathematical methods to simplify these, you get:

I1 = 15 I3 and I2 = 10 I3

The total voltage is therefore 3100 I3 and the total current is 26 I3. The overall resistance is therefore 3100 I3 / 26 I3 = 119.23076923Ω

Apologies to those for whom algebra makes them queasy.

Andy

Note this is a reply to 68 not 69 which you posted while I was composing.

----------------SNIP end of edit ----------------------------

I don't think so:

Look at it from the point of view of R5 initially. Looking to the left it sees R1 and R2 in parallel to E/2; looking to the right it see R3 and R4 in parallel to E/3.

Altogether that gives you 50 + 400 + 67 (well 66 and 2/3) Ohms across E/6.

You now have the current (and hence voltage) across R5 (which are different to the values you got I think). You gave me the clue I needed to get going (Thevenin!).

If I said that the total resistance was between 100 ohms and 1000 ohms, would I be close? (I didn't see this problem in the ladybird book!)

Okay class, try it with a slight modification as shown here:

-John

This one can be done solely by observation. Given voltage E at point A, voltage at R1,R2 is .5E. Similarly voltage at R3,R4 is also .5E. With no voltage difference between the ends of R5, there is NEVER any current flow through R5, thus it is not necessary to the circuit. We thus have (R1+R2) in parallel with (R3+R4), or 200 in parallel with 200, and R= 100. R5 is open.

Now to take the opposite extreme, lets make R5 = 0. We will short that sucker out.

R1 and R3 are in parallel as are R2 and R4... each with an equivalent resistance of 50 ohms. Because R1||R3 is in series with R2||R4 (|| indicates parallel), the total resistance is still 100 ohms.

Believe it or not, the above circuit does have application in the real world (ignoring R5 for the moment). Lets say that I need a 100 ohm resistor that can dissipate 1 watt of power. All I have is a handful of 100 ohm 1/4 watt. I can plug them in to R1 through R4 and I now have my 100 ohm 1 watt resistor.

Let me take this one step further, then I will quit. Lets use 1 watt resistors for R1 through R4, and build an identical circuit next to it. Now, if we connect the two circuits in parallel, we have a 50 ohm resistor which will dissipate 8 watts. If we use carbon comp resistors (low inductance properties), we now have an excellent dummy load for a hand-held transciever. Transcievers typically put out no more than 5 watts. My Dad built this in a spam can many many years ago, and I still use it today.

Sincerely

Bill

Try this one: construct a cube using twelve 1 Ohm resistors as the edges. What's the resistance between (any) two diagonally opposite corners.

I decided to work through this problem to a result (mainly because Andy's answer was so close to the total resistance if R5 wasn't there i.e. 120 Ω). I've changed the values of the resistors (temporarily) and put 186V at point A to make the arithmetic (especially in this editor) easy. In each case, for example: R₁ is the "name" V₁ is the voltage across the resistor and I₁ is the current through it.

As I said before, look at it first from the point of view of R5:

To the left it sees 93 Volts through 3Ω, and,

to the right it sees 62 Volts through 4Ω

Total resistance 31Ω, and, total Voltage 31V

So V₅ = 24V, and, I₅ = 1A

Now V₂ = V₄ + 24

V₁ = 186 - V₂ = 162 - V₄

V₃ = 186 - V₄

I_Total = I₁ + I₃ = (162-V₄)/6 + (186 - V₄)/12 = (510-3V₄)/12

Also I_Total = I₂ + I₄ = (V₄ + 24)/6 +V₄/6 = (4V₄ + 48)/12

Clearly these two are the same so V₄ = 66V

So V₁ = 96V, V₂ = 90V, and V₃ = 120V

And now I_Total (either of the two above) = 26A

So R_Total = 186/26 = 7.154Ω

Divide by 6 and multiply by 100 to get back to the original problem gives 119.231Ω

I'm going to have a bash at this one myself! I think I'll need a few attempts at it but nothing ventured, nothing gained! What I really don't want to do is get beyond myself! This happened last time and as a result, I lost focus, got dis-heartened and couldn't find my way back! I'm going to start part 2 of the course in the next few days (work permitting) I think it's about resistors in parallel (easy for you guys I know) but I'm after a good base to work from so bear with me for a while.

I looked at the other circuits and tried to visualize what was going on, and to a degree, I could follow but not in a quantitative manor!

Thanks to all for your input and wish me luck, I'm going in!!

Keep the faith Truman...

I concede that alot of our playing here is a bit beyond series and parallel circuits so no one expects you to understand it all yet. If it is confusing now, that will change... the muddy waters will clear. Especially with all our handsome features (good looks).

Sincerely

Bill

"Especially with all our handsome features (good looks)."

And someday all of us will be parallel to ground.

I found a novel way to solve this circuit using delta-wye conversion. First I will redraw the circuit emphasizing the delta portion.

Rx, Ry, and Rz are the designators in the delta Wye conversion equations which are:

RA = Ry*Rz/(Rx+Ry+Rz)

RB = Rx*Rz/(Rx+Ry+Rz)

RC = Rx*Ry/(Rx+Ry+Rz)

The conversion results in the following circuit:

RB+ R3 = 157.143 ohms

RC+R4 = 128.571 ohms

(RB+R3)||(RC+R4) = 70.714 ohms (|| means in parallel)

This result added to RA = 114.286 + 70.714 => RT = 185 ohms

My other attempt at this particular problem was definitely FUBAR (bad).

Bill

Bill

An excellent approach to this problem (even if it is dependant on somewhat esoteric knowledge). There's also a confusion of RA and RC in the scheme as you've labelled them. This is where visualising the circuit can be really helpful in checking that the maths is in order. Of the three resistors in the delta configuration, the smallest value is R1 or Rx. In the star (or wye?) configuration, the smallest two resistors (RB & RC) should be between the same two legs. If you look at the equations, there should be a symmetry there.

If you run through the rest of the calculations with that correction made {R = 28.571 + (57.143 + 100) || (114.286 + 100)}, you will come to the same result of 119Ω derived above.

Well done for identifying a good approach to this problem.

Andy

THANK YOU ANDY!!

The print in the book I got this from has shrunk horribly since I bought the book in 1966. Maybe I managed to wash it in hot water some time over the years. After getting a set of glasses, I see that I transposed Rx and Rz. It is 02:30 am here so I am going to get some sleep and I will recrunch the numbers in the morning.

Sincerely

Bill

As I mentioned to Andy, I accidentally transposed Rx and Rz. Here is the corrected version.

Rx, Ry, and Rz are the designators in the delta Wye conversion equations which are:

RA = Ry*Rz/(Rx+Ry+Rz)

RB = Rx*Rz/(Rx+Ry+Rz)

RC = Rx*Ry/(Rx+Ry+Rz)

The conversion results in the following circuit:

RB+R3 = 157.143 ohms

RC+R4 = 214.286

(RB+R3)||(RC+R4) = 90.659 ohms

This result added to RA = 90.659+28.571 => R_Total = 119.230 ohms

This agrees with Andy's results.

Sincerely

Bill

Slap, slap, slap, slap...

Hey Truman! Everyone's given good answers here and I can't offer much in the way of additional help. So, I just gave you a massage.

'Dya like it?

Just ask, chief. I'll be here to help you if you need it.

You got me there Vulcan, I was looking the other way! It's going to be a long road so get your boots shined up and ready mate!

Another fine option I've just discovered:

Electronics and Computer Math

Seventh Edition

By Bill Deem
and Tony Zannini

I got the text for $7 (delivered) from Amazon. It contains a "study wizard" on CDROM that is worth five times the price. However, there is also a website dedicated to further study and it's available to all (buy the book or not).

http://wps.prenhall.com/chet_deem_electronic_7/2/644/164962.cw/index.html