Hard intro circuit problem help please im stuck on this one for hrs

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nchin
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http://imageshack.us/photo/my-images/585/circuitzzz.png/

i thought maybe i should ignore everything that is on the right side of the 3v and just focus on the left side. but this one is not looking too good for me.

any ideas on this one??
 

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It looks like it should respond to mesh analysis.
What's the problem?

One way to sort your head out about this sort of problem is to redraw the circuit in a way that makes more sense to you. Since there is an explicit ground in this diagram, it may help to draw a ground rail across the bottom of the page and draw the other components in above that ... i.e. the ground rail has three resistors connected to it...
 


find volt across the 2mA source
 


Simon Bridge said:
It looks like it should respond to mesh analysis.
What's the problem?

One way to sort your head out about this sort of problem is to redraw the circuit in a way that makes more sense to you. Since there is an explicit ground in this diagram, it may help to draw a ground rail across the bottom of the page and draw the other components in above that ... i.e. the ground rail has three resistors connected to it...

how many mesh loops do we have here??
 


nchin said:
http://imageshack.us/photo/my-images/585/circuitzzz.png/

i thought maybe i should ignore everything that is on the right side of the 3v and just focus on the left side. but this one is not looking too good for me.

any ideas on this one??

Yes, it is a good idea. See picture.

ehild
 

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ehild said:
Yes, it is a good idea. See picture.

ehild

I1 = V/R
Resistors add up right?
I1 = 3/3+3+2+3+2
 


ehild said:
Yes, it is a good idea. See picture.

ehild

i found i1 = 0.6409 A
i2 = 0.2307
total Resistance = 13.588Ω

so i1 - i2 x total R = volt across the 2mA?

is that correct
 


nchin said:
I1 = V/R
Resistors add up right?
I1 = 3/3+3+2+3+2
No.
Only resistors in series add up. There is I2 current flowing through the left 2kΩ resistor. It is not in series with the other ones.

ehild
 


ehild said:
No.
Only resistors in series add up. There is I2 current flowing through the left 2kΩ resistor. It is not in series with the other ones.

ehild

i see, so what do i do? I am not sure how to solve I1, i think i did I2 correctly though
 


i think i got it, since one of the nodes are grounded out

I1 = 3/6+5 = 0.2727A right?
 


ehild said:
Write KVL for the loop in the figure, and substitute I2 = I1+2

ehild

applying KVL = 6I1 - 3 +5I1 + 2I2= 0?
 
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ehild said:
yes.ehild

it should be .002 right? because 2 mA = .002 A. after i got the currents, what would the total resistor be to plug in the equation V = i x R?
 


Also I got I1 = -0.2727A and I2 = 1.727A
do i add them together to get the the current i need to solve the problem?

1.727 + (-.2727)?
 


ehild said:
You substituted the resistance in kΩ-s, then the currents must be in mA.

What should be 0.002?

ehild


oh i see. i thought it was KVL = 6I1 - 3 +5I1 + .002I2= 0
 


im not sure what I am suppose to do after i got the currents though
 


ehild said:
You multiply the current with the resistance to get voltage.
Have you got the currents? What are they?

ehild

I1 = -0.0769 mA
I2 = 2.079 mA

correct?
 


6I1 - 3 + 5I1 +2(I1 + 2 ) = 0
13I1 = -1
I1 = -0.0769mA

I2 = 2 + -0.0769 = 2.0769mA?
 


im trying to find the voltage across the 2mA
 
ehild said:
6I1 - 3 +5I1 + 2I2= 0
and I2=2+I1. (I are in mA)
What are I1 and I2?

Solve.

ehild


6I1 - 3 + 5I1 +2(I1 + 2 ) = 0
13I1 = -1
I1 = -0.0769mA

I2 = 2 + -0.0769 = 2.0769mA

Correct?

What do I do next?
 


ehild said:
Look at the loop with the current source. What are the voltage across the resistors?

ehild

V = IxR

VR5KΩ = 2.0769 x 5 = 10.2845 V
VR2KΩ = 2.0769 x 2 = 4.1538 V
vR1KΩ = 2.0769 x 1 = 2.0769 V

correct?

would the final answer be just adding up all the volts?
 


nchin said:
V = IxR

VR5KΩ = 2.0769 x 5 = 10.2845 V
VR2KΩ = 2.0769 x 2 = 4.1538 V
vR1KΩ = 2.0769 x 1 = 2.0769 V

correct?

would the final answer be just adding up all the volts?

No, I2 flows through the vertical 2kΩ resistor only. The others have the generator current.And your I2 is not correct. Write up KVL again for the loop in the figure. The voltage drop across the generator is Ug. Determine Ug.

ehild
 

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ehild said:
No, I2 flows through the vertical 2kΩ resistor only. The others have the generator current. Write up KVL again for the loop in the figure. The voltage drop across the generator is Ug. Determine Ug.

ehild

5I2 + 2I2 + 1I2 = -2
8I2 = -2
8(2.08)=-2?

Why did we need to find the voltage across each resistor at first ?
 
ehild said:
It is wrong. Write the voltages. What current flows through the 1 kΩ and 5 kΩ resistors? What is the current through the 2 kΩ resistor?


ehild

V = I x r
I2= 1.9231

V2 kohm = 1.9231 x 2 = 3.85v
V5kohm = 2mA x 5 = 10.0v
V1kohm= 2mA x 1 = 2.00v
??