Re: little math help on schematic
From: Rich Grise (rich_at_example.net)
Date: 02/16/05
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Date: Wed, 16 Feb 2005 05:31:23 GMT
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
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