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

In summary: A. The voltage across the generator is 3 V. ehildThe generator supplies 2...2 mA. The voltage across the generator is 3 V. ehildoh ok. 5I2 + 2I2 + 1I2 = -38I2 = -38(-0.375)=-3-3=-3Is this correct? ehild-3=-3Yes, but the current is positive. The direction of I2 is clockwise. ehildYes, but the current is positive. The direction of I2 is clockwise. ehildSo it should be 8(0.375)=3I2 = 0.
  • #1
nchin
172
0
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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  • #2


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...
 
  • #3


find volt across the 2mA source
 
  • #4


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??
 
  • #5


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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  • #6


nchin said:
how many mesh loops do we have here??
A fair few - but you don't need all of them.
 
  • #7


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
 
  • #8


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
 
  • #9


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
 
  • #10


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
 
  • #11


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

I1 = 3/6+5 = 0.2727A right?
 
  • #12


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

ehild
 

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  • #13


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

ehild

applying KVL = 6I1 - 3 +5I1 + 2I2= 0?
 
Last edited:
  • #14


nchin said:
applying KVL = 6I1 - 3 +5I1 + 2I2= 0?

yes.


ehild
 
  • #15


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?
 
  • #16


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)?
 
  • #17


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

What should be 0.002?

ehild
 
  • #18


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
 
  • #19


im not sure what I am suppose to do after i got the currents though
 
  • #20


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

ehild
 
  • #21


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?
 
  • #22


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

I2 = 2 + -0.0769 = 2.0769mA?
 
  • #23


im trying to find the voltage across the 2mA
 
  • #24
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?
 
  • #25


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

ehild
 
  • #26


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?
 
  • #27


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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  • #28
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 ?
 
  • #29


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
 
  • #30
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
??
 
  • #31


The sum of voltages along the loop is zero. Include the voltage of the generator and write up KVL,

ehild
 
  • #32
ehild said:
The sum of voltages along the loop is zero. Include the voltage of the generator and write up KVL,

ehild

Whats the generator? Isn't it 2mA? How do you get 2mA in volts?
 
  • #33


The generator supplies 2 mA current, but there is some voltage across it. You need to find that voltage from KVL.The sum the voltages across the resistors + the voltage across the generator is zero.

ehild
 

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

1. What is a hard intro circuit problem?

A hard intro circuit problem is a type of problem in the field of electrical engineering that involves designing and analyzing complex circuits. These problems often require advanced knowledge of circuit theory and can be challenging to solve.

2. How do I approach solving a hard intro circuit problem?

The best way to approach solving a hard intro circuit problem is to break it down into smaller, more manageable parts. Start by identifying the different components in the circuit and their relationships, and then use circuit analysis techniques to solve for unknown values.

3. What are some common techniques used to solve hard intro circuit problems?

Some common techniques used to solve hard intro circuit problems include Kirchhoff's laws, Ohm's law, Thevenin's theorem, and nodal analysis. These techniques can help simplify complex circuits and make them easier to analyze.

4. What should I do if I am stuck on a hard intro circuit problem?

If you are stuck on a hard intro circuit problem, try taking a step back and looking at the problem from a different perspective. You can also consult with your peers or a professor for guidance and clarification. It may also be helpful to review your notes and textbook for relevant concepts and equations.

5. How can I improve my skills in solving hard intro circuit problems?

The best way to improve your skills in solving hard intro circuit problems is to practice regularly. You can also try challenging yourself with more complex problems and seeking feedback from others. Additionally, staying up-to-date with new technologies and techniques in the field can also help improve your problem-solving abilities.

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