little math help on schematic

--
Since the 12K, the 6K, and the 4K resistors are all in parallel, their
total resistance becomes:


                            1
	Rt = -------------------------------- = 2000 ohms
                  1         1         1
               ------- + ------- + ------
               12000R     6000R     4000R


Now, if we redraw your schematic with that in mind, we'll have:


+8V>--[1K]--+---E2
            |
           [2K]
            |
0V>---------+
    

and we can say:

              8V * 2K
	E2 = --------- = 5.333V
              1K + 2K




If we now look at the original schematic:


+8V>--[1K]---+------+------+----5.333V
             |      |      | 
           [12K]   [6K]   [4K]
             |      |      |
0V>----------+------+------+

we can see that 5.333V will be across all of the parallel resistors,
so the current in any one of them will be:

             E
	I = ---
             R

Thus, for the 12000 ohm case,:

             5.333V 
        I = -------- ~ 0.000444A ~ 444µA 
             12000R

for the 6000 ohm case,:

             5.333V 
        I = -------- ~= 0.000888A ~ 888µA 
             6000R


and for the 4000 ohm case:

             5.333V 
        I = -------- ~ 0.00133A ~ 1.33mA 
             4000R

His sincerity was not in doubt and I did not impugn it.

Hmmm, so "not going to do your homework" becomes chiding. Makes me wonder what some real criticism or an honest-to-god flame would be called.

The problem was not likely a real world problem and, if it had been, those terms would be well known to the OP. Since it looked like a homework problem, and because people assigned such problems are exposed to both of those concepts, and because learning to use them is an important part of learning circuit theory, I merely set out to remind him that those methods were applicable. The OP was, of course, welcome to ask what the terms meant, or do a web search for them, if he was not already familiar with them or lacked a textbook in which to look them up. There is certainly no reason for me to explain them ahead of time.

I don't grok 'seb', but having tutuored several electronics students with good results, I can say that the biggest obstacle (at least for those who have any business in such a curriculum) is coming to the realization that they can apply the abstract analytical tools they have (or should have) learned. Naming applicable tools was, I thought, the best contribution under the circumstances.

I categorically reject the contention that my reply was "beneath" my other contributions. Where practical advice is requested, just answering the question is a beneficial contribution. Not so for homework requests. The OP wanted correction of a formula presented without any hint of how it was derived. That would be tantamount to doing his homework, if it was homework. As everyone understands, or can readily figure out, helping any student pass courses without doing their homework is a bad idea, even for the ostensible beneficiary.

If the OP was solving a real circuit, and has not yet been exposed to superposition and nodal analysis, he would clearly benefit from learning those techniques. This formum is not the place for tutorials on those subjects.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

I would like to see the math involved in calculating the current in the 4k resistor and the current in the 12k resistor. I have tried to do a node current at the top -right of 1k resistor and top of all parallel items) which I have called 'node A' (Va) but I keep getting the wrong answer. There is 8mA in the 4K and 2.66mA in the 12k (this means the voltage at 'node A' must be 32V. (I hand drew the 50mA and 90mA current sources so that's what they are supposed to be -- if there was a symbol I missed it). I took the current at node A (top) to be:

((Va-8)/1k)-50mA - (Va / 12k) - (Va/6k) + 90mA - (Va/4k) = 0 but that does not yield 32v. Can someone point an error I've made (as I assume I've set something up wrong) or correctly indicate the math for me? Thanks.

1K ___ |-|___|-----------------------------------|-------- | | | | | | | | | | | | | | .-. .-. |90ma .-. /+\ | | | | | | | | 8v ( ) 50ma (\|/) 12K | | | | 6K (/|\) | | 4K \-/ | '-' '-' | '-' | | | | | | --------------------------------------------------- (created by AACircuit v1.28.4 beta 13/12/04

Not counting the 1K resistor, the other resistors are equivalent to a single 2K resistor. The nodal impedance at Va is therefor

1K || 2K == (2/3)K. The voltage divider from the 8V source to Va is 2K/(1K+2K) == 2/3. Applying superposition from the 3 sources to get 3 terms: Va == 8V * (2/3) + 90mA * (2/3)K - 50mA * (2/3)K Va == 16/3 V + 80/3 V == 96/3 V == 32 V.

You can get any of the currents in the parallel collection by applying Ohm's law using 32V and the resistance.

If you are trying to learn this stuff on your own, I have to applaud the effort. My father did that and, after a long haul and much difficulty, became a professional engineer. We still argue about which way current is best considered to flow. He keeps talking about electrons.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

--- Poor, sincere dude's downloaded Andy's program and very quickly learned to use it just so he can post a schematic of what he needs help with and you chide him because it's homework, and then you hit him with nodal analysis and superposition as _the_ solution for his problem without even explaining what you mean? From your earlier contributions, that seems to be beneath you, so why don't you cut the guy some slack and help him out? _seb_, OK?

-- John Fields

OK, but now you have to add in the current sources. 90-50 = net 40 ma flowing into the 5.333 volt node. The node impedance is 1k||2k = 666.6 ohms, so the additional 40 ma pulls the voltage up from 5.333 to 32.00 volts.

John

[big snip]

You've gone off the deep end here. My language was simply a way to refer to the collection of resistors that act in a shunt role for the divider fed by the 8V source. I was not trying to make the 1K series resistor go away and in fact used its value in the next two sentences.

I urge you to settle down and consider more than the first interpretation that occurs to you before your next post. Such precaution may keep your foot from any further oral insertion.

You get to 32V, as I did, by summing the contribution of the 8V source, as divided by the 1K,2K divider, with the product of each current source and the nodal impedance. There is no "starting off with 8V" in the circuit originally posted. There were current sources and a voltage source. Neither implied or was stated to have a limit in their effect.

I hope you are either kidding or having a really bad day today because I have found more intelligence in most of your other on-topic posts.

Would a smiley have helped you understand? Or do I need to elaborate a trivial aside comment?

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

How about the two current sources?

John

4k

There

A'

what

does

Thanks for your help, John. I am actually not in school, but this problem is from a book. The current in the 4k resistor is 8mA not 1.33mA (the solution is given). Also, the current in the 12k is 2.66mA (not given but power is given at 85.3mW -- works backwards to 2.66mA at 32v which also establishes the 8mA in the 4k). In both cases, node A , where you have derived 5.33V is actually 32V to establish those voltages. Any ideas?

4k

There

A'

what

does

Thanks for your help, John. I am actually not in school, but this problem is from a book. The current in the 4k resistor is 8mA not 1.33mA (the solution is given). Also, the current in the 12k is 2.66mA (not given but power is given at 85.3mW -- works backwards to 2.66mA at 32v which also establishes the 8mA in the 4k). In both cases, node A , where you have derived 5.33V is actually 32V to establish those voltages. Any ideas?

There

does

Thanks to a Mr. Don Taylor for pointing out my math error (sign on the 1K resistor contribution).----- Original Message ----- From: "Don Taylor"

Sent: Sunday, February 13, 2005 7:23 PM Subject: Re: little math help on schematic

does

Maybe I'm off base her... by why is ((Va-8)/1k positive when -(Va/12k) and -(Va/6k) and -(Va/4k) are all negative?

Yes, Don. Thanks, that was my error. With the sign correction made the proper voltage of 32v at node A is given and the correct resultant currents are achieved. Thank you for pointing it out.

--
_Not_ counting the 1k resistor?  It's in there isn't it?, so how do
you propose to do away with it?  Oh, I get it... it's inconvenient for
you to deal with,  so you want to want to make it go away so you don't
have to deal with it.  Never mind that the load isn't reactive so you
can't realize a voltage greater than the DC source voltage feeding it,
you still want to assume that, somehow, 32VDC is there.

--
Yup. (extricates foot from mouth) For some reason, I completely
glossed over them. 

Sorry, guys...

--
It was the wrath of grapes...

It's Valentine's Day!

John

WHOOPS, I sort of wondered.

DNA

4k

There

A'

what

the

does

set

That's quite sweet.

DNA

There

does

Taking + currents into node A you should have (8-Va)/1k) rather than (Va-8)/1k).

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

Quite true.

However, ((8-Va)/1k) -50mA - (Va / 12k) - (Va/6k) + 90mA - (Va/4k) = 0

*does* yield 32V for Va

Look at the first term: Currents are positive if they flow *into* a node. Hence (8-Va)/1K is the correct current with reference to node "A"

I leave it to you to verify the result.

--
Then there's duct tape ... 
              (Garrison Keillor)