
#1
Feb413, 08:13 AM

P: 289

1. The problem statement, all variables and given/known data
Find the voltage V_{0} in the network. 2. Relevant equations V = IR Voltage Division: (Voltage across series resistor) = [(resistance) / total series resistance)](total input V) Current Division (for 2 parallel resistors): (current across parallel resistor) = [(other resistor) / (sum of parallel resistors)](total incoming current)] 3. The attempt at a solution I can simplify the 1k and 3k resistors to 4k no matter what. I'm not sure if I can simplify the 6k and 12k on the bottom as parallel though, but they will be 4k. 4k + 4k + 2k in series becomes 10k, assuming you can do that with the current source. Then 10k + 6k makes R total = 16k ohm, and then I tot = 24 mA V = IR means Vtot = 384V. Then voltage division would make V_{0} = [ (6k)/(6k+10k) ](384V) = 144V would this be correct? 



#2
Feb413, 08:34 AM

Mentor
P: 11,446





#3
Feb413, 08:35 AM

HW Helper
P: 4,716

Why don't you draw a simpler schematic?
You said the 12 and 6 make 4. The 3 and 1 make 4. The 2 and 6 make 8. So draw it with those 3 resistances and see what you can make of it. BTW, it's not a dependent current source, it's a constant current source. 



#4
Feb413, 09:09 AM

P: 2,861

Dependent current source, find VoIf you redraw the circuit at this point you should see that a further simplification can be made. Should be possible to treat it as a current divider at that point. 



#5
Feb413, 09:30 AM

P: 289

I got confused and thought you could simplify the resistors in series like that since it was a constant current source, heh. Never mind then. 384V seemed way too high too.
So then: : and then doing current division: Current through right branch = (4k/16k)(24 mA) = 6 mA and then V knot is just V = IR, since current is the same through parallel elements V knot = (6 mA)(6 kΩ) = 36V Thanks. 



#6
Feb413, 05:57 PM

P: 2,861

That's what I make it.



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