| Author |
Message |
John Fields
Guest
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 Posted:
Tue Feb 15, 2005 4:20 am Post subject:
Re: little math help on schematic |
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On Mon, 14 Feb 2005 21:38:05 GMT, "Genome" <ilike_spam@yahoo.co.uk>
wrote:
| Quote: |
"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
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Genome
Guest
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| Posted:
Tue Feb 15, 2005 5:51 am Post subject:
Re: little math help on schematic |
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"John Larkin" <jjSNIPlarkin@highTHISlandPLEASEtechnology.XXX> wrote in
message news:pa60115as2sbdi3ic7n1k26tgki39je94o@4ax.com...
| Quote: | 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 |
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John Larkin
Guest
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| Posted:
Tue Feb 15, 2005 6:10 am Post subject:
Re: little math help on schematic |
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On Mon, 14 Feb 2005 23:51:42 GMT, "Genome" <ilike_spam@yahoo.co.uk>
wrote:
| Quote: |
That's quite sweet.
|
It's Valentine's Day!
John
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Rich Grise
Guest
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| Posted:
Wed Feb 16, 2005 6:10 am Post subject:
Re: little math help on schematic |
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On Mon, 14 Feb 2005 16:20:47 -0600, John Fields wrote:
| Quote: | 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 |
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Rich Grise
Guest
|
| Posted:
Wed Feb 16, 2005 6:10 am Post subject:
Re: little math help on schematic |
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On Sun, 13 Feb 2005 21:36:00 -0500, R.Spinks wrote:
| Quote: | "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_.
| Quote: | 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 |
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Rich Grise
Guest
|
| Posted:
Wed Feb 16, 2005 6:10 am Post subject:
Re: little math help on schematic |
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On Sun, 13 Feb 2005 21:59:35 -0600, John Fields wrote:
| Quote: | 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 |
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Don Kelly
Guest
|
| Posted:
Wed Feb 16, 2005 6:10 am Post subject:
Re: little math help on schematic |
|
|
"R.Spinks" <rspinks1@wowway.com> wrote in message
news:tsCdna2X3v4dQ5LfRVn-sw@wideopenwest.com...
| Quote: | 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)
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 |
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Fred Abse
Guest
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| Posted:
Fri Feb 18, 2005 2:41 am Post subject:
Re: little math help on schematic |
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On Sun, 13 Feb 2005 18:21:24 -0500, R.Spinks wrote:
| Quote: | ((Va-8)/1k)-50mA - (Va / 12k) - (Va/6k) + 90mA - (Va/4k) = 0 but that
does not yield 32v
|
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) |
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