# Some basic circuit questions....

Discussion in 'Homework Help' started by RyanKim, Oct 6, 2011.

1. ### RyanKim Thread Starter Member

Sep 18, 2011
37
1
Hi again....once again I'm kind of stuck here and I come here for help! So I have this question I sort of understand....at least to some degree but what I'm having difficulty understanding is how to get I(source). Just the way its drawn throws me off but if you give me a second to explain to what extent i understand I hope you can fill in my gaps of knowledge.

So....I assume at the left most dot it reads 24V...I initially mistook it as a source voltage........so along that red line i drew about the middle black line is a potential of 24V all across. Now I know across the 10K and 2K resistors since they are to ground the Voltage across has to be 24V. Conversely the 4k has an 8V reading meaning the V across that must be 24-8 = 16V. Thats all fine and dandy but what about I(source)?. The asnwer is actually 1.84mA but my question is How can I have a source current smaller than a value of I? KCL states that what goes in must come out....if my I value is 4mA how can my Is be smaller?

Any help is much much appreciated...

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2. ### justtrying Active Member

Mar 9, 2011
329
351
I am pretty sure it is a mistake... I get 18.4 mA - a bit too suspicous 18.4 vs 1.84? I think your logic is pretty solid on solving it.

3. ### RyanKim Thread Starter Member

Sep 18, 2011
37
1
Ahh must be it. Though I still am wondering why its 18.4....I know that the 2k resistor is drawing 16 mA and the 10k is drawing 2.4....thus thats 18.4...but what about that 4K? is it because its just hanging loose off there not connected to ground? If theres potential across it there must be current being drawn no?

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Well Is = 24V/10K + 24V/2K + (24V - 8V)/4K = 2.4mA + 12mA + 4mA = 18.4mA

5. ### RyanKim Thread Starter Member

Sep 18, 2011
37
1
Ahhh....I see what i did there. Sorry bout that been a long day. Thanks for the help though I get it now.