little math help on schematic

[R.Spinks]

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 www.tech-chat.de)

 

[Larry Brasfield]

"R.Spinks" <rspinks1@wowway.com> wrote in message
news:tsCdna2X3v4dQ5LfRVn-sw@wideopenwest.com...

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.

I'm not going to do your homework for you. But I can
indicate a better approach than you have shown.

Write a single node equation for the right side of the 1K
resistor, treating the bottom-most node as 0 V. Use
superposition to separately calculate the contribution
of each of the 3 sources and get the net effect by
adding those independent contributions.

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

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

 

[John Fields]

On Sun, 13 Feb 2005 18:21:24 -0500, "R.Spinks" <rspinks1@wowway.com>
wrote:

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 www.tech-chat.de)

---
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 ~ 444ľA
12000R

for the 6000 ohm case,:

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


and for the 4000 ohm case:

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


--
John Fields

 

[John Fields]

On Sun, 13 Feb 2005 15:24:38 -0800, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

I'm not going to do your homework for you. But I can
indicate a better approach than you have shown.

Write a single node equation for the right side of the 1K
resistor, treating the bottom-most node as 0 V. Use
superposition to separately calculate the contribution
of each of the 3 sources and get the net effect by
adding those independent contributions.

---
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

 

[Larry Brasfield]

"John Fields" <jfields@austininstruments.com> wrote in message
news:4esv01d0n7bgbk2l3b12gqhectfnn9t5hp@4ax.com...

On Sun, 13 Feb 2005 15:24:38 -0800, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:


I'm not going to do your homework for you. But I can
indicate a better approach than you have shown.

Write a single node equation for the right side of the 1K
resistor, treating the bottom-most node as 0 V. Use
superposition to separately calculate the contribution
of each of the 3 sources and get the net effect by
adding those independent contributions.

---
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

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

and you chide him because it's homework,

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.

and then you hit
him with nodal analysis and superposition as _the_ solution for his
problem without even explaining what you mean?

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.

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?

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.

 

[R.Spinks]

"John Fields" <jfields@austininstruments.com> wrote in message
news:qqov01lvrh95ek6kddf6q0c17d8fb33l61@4ax.com...

On Sun, 13 Feb 2005 18:21:24 -0500, "R.Spinks" <rspinks1@wowway.com
wrote:

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 www.tech-chat.de)

---
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 ~ 444ľA
12000R

for the 6000 ohm case,:

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


and for the 4000 ohm case:

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


--
John Fields

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?

 

[R.Spinks]

"John Fields" <jfields@austininstruments.com> wrote in message
news:qqov01lvrh95ek6kddf6q0c17d8fb33l61@4ax.com...

On Sun, 13 Feb 2005 18:21:24 -0500, "R.Spinks" <rspinks1@wowway.com
wrote:

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 www.tech-chat.de)

---
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 ~ 444ľA
12000R

for the 6000 ohm case,:

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


and for the 4000 ohm case:

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


--
John Fields

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?

 

[R.Spinks]

"R.Spinks" <rspinks1@wowway.com> wrote in message
news:tsCdna2X3v4dQ5LfRVn-sw@wideopenwest.com...

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 www.tech-chat.de)

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

... 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.

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?

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

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.

 

[John Larkin]

On Sun, 13 Feb 2005 18:05:04 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Sun, 13 Feb 2005 18:21:24 -0500, "R.Spinks" <rspinks1@wowway.com
wrote:

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 www.tech-chat.de)

---
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>----------+------+------+

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

 

[John Fields]

On Sun, 13 Feb 2005 18:29:37 -0800, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"R.Spinks" <rspinks1@wowway.com> wrote in message
news:18KdnQUSkNmul43fRVn-qQ@wideopenwest.com...

"John Fields" <jfields@austininstruments.com> wrote in message
news:qqov01lvrh95ek6kddf6q0c17d8fb33l61@4ax.com...
On Sun, 13 Feb 2005 18:21:24 -0500, "R.Spinks" <rspinks1@wowway.com
wrote:

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 www.tech-chat.de)

---
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 ~ 444ľA
12000R

for the 6000 ohm case,:

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


and for the 4000 ohm case:

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


--
John Fields

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?


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.

---
_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.
---

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

---
The point is, you can't get to 32V if you're starting off with 8V and
you don't have something which can pump it up to 32V at the current
your cicuit requires.
---

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.

---
And _you_ have what to offer?

Confusion about what constitutes electron flow?

--
John Fields

 

[Larry Brasfield]

"John Fields" <jfields@austininstruments.com> wrote in message
news:le60111a3stp4cof9jrtbia1j3cmoc2674@4ax.com...

On Sun, 13 Feb 2005 18:29:37 -0800, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:
[big snip]
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.

---
_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.

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

---
The point is, you can't get to 32V if you're starting off with 8V and
you don't have something which can pump it up to 32V at the current
your cicuit requires.

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.

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.

---
And _you_ have what to offer?

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.

Confusion about what constitutes electron flow?

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.

 

[John Larkin]

On Sun, 13 Feb 2005 21:59:35 -0600, John Fields
<jfields@austininstruments.com> wrote:

1K
___
|-|___|-----------------------------------|--------
| | | | | |
| | | | | |
| | .-. .-. |90ma .-.
/+\ | | | | | | | |
8v ( ) 50ma (\|/) 12K | | | | 6K (/|\) | | 4K
\-/ | '-' '-' | '-'
| | | | | |
---------------------------------------------------

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

---
The point is, you can't get to 32V if you're starting off with 8V and
you don't have something which can pump it up to 32V at the current
your cicuit requires.

How about the two current sources?

John

 

[John Fields]

On Sun, 13 Feb 2005 20:29:33 -0800, John Larkin
<jjSNIPlarkin@highTHISlandPLEASEtechnology.XXX> wrote:

On Sun, 13 Feb 2005 21:59:35 -0600, John Fields
jfields@austininstruments.com> wrote:



1K
___
|-|___|-----------------------------------|--------
| | | | | |
| | | | | |
| | .-. .-. |90ma .-.
/+\ | | | | | | | |
8v ( ) 50ma (\|/) 12K | | | | 6K (/|\) | | 4K
\-/ | '-' '-' | '-'
| | | | | |
---------------------------------------------------



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

---
The point is, you can't get to 32V if you're starting off with 8V and
you don't have something which can pump it up to 32V at the current
your cicuit requires.

How about the two current sources?

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

Sorry, guys...

--
John Fields

 

[Genome]

"John Fields" <jfields@austininstruments.com> wrote in message
news:elc111t35g3rskeo1ohfn7p6doabqdoom1@4ax.com...

On Sun, 13 Feb 2005 20:29:33 -0800, John Larkin
jjSNIPlarkin@highTHISlandPLEASEtechnology.XXX> wrote:

On Sun, 13 Feb 2005 21:59:35 -0600, John Fields
jfields@austininstruments.com> wrote:



1K
___
|-|___|-----------------------------------|--------
| | | | | |
| | | | | |
| | .-. .-. |90ma .-.
/+\ | | | | | | | |
8v ( ) 50ma (\|/) 12K | | | | 6K (/|\) | | 4K
\-/ | '-' '-' | '-'
| | | | | |
---------------------------------------------------



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

---
The point is, you can't get to 32V if you're starting off with 8V and
you don't have something which can pump it up to 32V at the current
your cicuit requires.

How about the two current sources?

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

Sorry, guys...

--
John Fields

WHOOPS, I sort of wondered.

DNA

 

[John Fields]

On Mon, 14 Feb 2005 21:38:05 GMT, "Genome" <ilike_spam@yahoo.co.uk>
wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:elc111t35g3rskeo1ohfn7p6doabqdoom1@4ax.com...
On Sun, 13 Feb 2005 20:29:33 -0800, John Larkin
jjSNIPlarkin@highTHISlandPLEASEtechnology.XXX> wrote:

On Sun, 13 Feb 2005 21:59:35 -0600, John Fields
jfields@austininstruments.com> wrote:



1K
___
|-|___|-----------------------------------|--------
| | | | | |
| | | | | |
| | .-. .-. |90ma .-.
/+\ | | | | | | | |
8v ( ) 50ma (\|/) 12K | | | | 6K (/|\) | | 4K
\-/ | '-' '-' | '-'
| | | | | |
---------------------------------------------------



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

---
The point is, you can't get to 32V if you're starting off with 8V and
you don't have something which can pump it up to 32V at the current
your cicuit requires.

How about the two current sources?

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

Sorry, guys...

--
John Fields

WHOOPS, I sort of wondered.

DNA

---
It was the wrath of grapes...

--
John Fields

 

[Genome]

"John Larkin" <jjSNIPlarkin@highTHISlandPLEASEtechnology.XXX> wrote in
message newsa60115as2sbdi3ic7n1k26tgki39je94o@4ax.com...

On Sun, 13 Feb 2005 18:05:04 -0600, John Fields
jfields@austininstruments.com> wrote:

On Sun, 13 Feb 2005 18:21:24 -0500, "R.Spinks" <rspinks1@wowway.com
wrote:

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 www.tech-chat.de)

---
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>----------+------+------+


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

That's quite sweet.

DNA

 

[Rich Grise]

On Sun, 13 Feb 2005 21:36:00 -0500, R.Spinks wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
On Sun, 13 Feb 2005 18:21:24 -0500, "R.Spinks" <rspinks1@wowway.com
I would like to see the math involved in calculating the current in the
1K
___
|-|___|-----------------------------------|--------
| | | | | |
| | | | | |
| | .-. .-. |90ma .-.
/+\ | | | | | | | |
8v ( ) 50ma (\|/) 12K | | | | 6K (/|\) | | 4K
\-/ | '-' '-' | '-'
| | | | | |
---------------------------------------------------
(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)
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 if we restore the current sources, we have:
+8V>--[1K]--+-----+-------E2
| |
[2K] (/|\) 40 mA
| |
0V>---------+-----'

Solve _this_.

Thanks for your help, John.
....
In both cases, node A , where you have derived 5.33V is
actually 32V to establish those voltages. Any ideas?

Yeah, solve the actual circuit, rather than the erroneous simplification.

Good Luck!
Rich

 

[Rich Grise]

On Sun, 13 Feb 2005 21:59:35 -0600, John Fields wrote:

On Sun, 13 Feb 2005 18:29:37 -0800, "Larry Brasfield"

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.

_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.

Well, have you accounted for the possible voltage compliance of the two
current sources?

Cheers!
Rich

 

[Rich Grise]

On Mon, 14 Feb 2005 16:20:47 -0600, John Fields wrote:

On Mon, 14 Feb 2005 21:38:05 GMT, "Genome" <ilike_spam@yahoo.co.uk
wrote:


"John Fields" <jfields@austininstruments.com> wrote in message
news:elc111t35g3rskeo1ohfn7p6doabqdoom1@4ax.com...
On Sun, 13 Feb 2005 20:29:33 -0800, John Larkin
jjSNIPlarkin@highTHISlandPLEASEtechnology.XXX> wrote:

On Sun, 13 Feb 2005 21:59:35 -0600, John Fields
jfields@austininstruments.com> wrote:



1K
___
|-|___|-----------------------------------|--------
| | | | | |
| | | | | |
| | .-. .-. |90ma .-.
/+\ | | | | | | | |
8v ( ) 50ma (\|/) 12K | | | | 6K (/|\) | | 4K
\-/ | '-' '-' | '-'
| | | | | |
---------------------------------------------------



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

---
The point is, you can't get to 32V if you're starting off with 8V and
you don't have something which can pump it up to 32V at the current
your cicuit requires.

How about the two current sources?

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

Sorry, guys...

--
John Fields

WHOOPS, I sort of wondered.

DNA

---
It was the wrath of grapes...

Well, I've jumped on you prematurely a few branches back.

And yes, I did so with a needling attitude. Glad to see that there hasn't
been any permanent damage so far.

Cheers!
Rich