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Find unknown current in circuit, voltage division law

  • Engineering
  • Thread starter Color_of_Cyan
  • Start date
  • #1

Homework Statement



Determine ( I L ) in the following circuit (the current "above" the 3 kΩ resistor).

http://imageshack.us/a/img31/1530/circuits215.jpg [Broken]

Homework Equations


Voltage division law

V (for series resistor) = (V in) * [ (resistor)(total series resistance) ]

Kirchoff's rules: I in = I out

Total potential difference in a loop = 0


The Attempt at a Solution


Okay, a lot of this is still new to me, so forgive me if I miss something.

(The "I L" should also be "I l," as it is just a listed current )

But, there's no potential difference given so how are you able to find the current asked? Would you say voltage is equal for everything in the circuit?

I try to sum all current out from left to right on the bottom as:

I tot = I l + 4(I x) - 3mA but still unknowns everywhere when all I want is I l.

Where should I start?
 
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Answers and Replies

  • #2
17
0
Try to apply millman's theorem
 
  • #3
Not sure how to apply that theorem but I found something else that might be closer and does help.


From what I think I understand you simplify everything in the circuit as much as possible and to just like 2 parallel wires except for the same node / wire for the unknown stuff / current / elements you're trying to solve?


Then you can use current / voltage division laws.

I get R eq = 1500 ohms and basically need the formula

I L = I eq [ R eq / (R eq + resistance on wire with unknown current you're trying to solve for) ]



The problem I think now is, what is (I eq)? I can simplify the independent (circles) currents to just 3 mA going up but what do I do with the dependent (diamond) current?

Or all still wrong?
 
  • #4
gneill
Mentor
20,795
2,773
Write a KCL node equation for the circuit.

Suppose the top node is at some potential V1. You should then be able to write terms for all the "unknown" currents in terms V1 and resistances. Solve for V1.
 
  • #5
Write a KCL node equation for the circuit.

Suppose the top node is at some potential V1. You should then be able to write terms for all the "unknown" currents in terms V1 and resistances. Solve for V1.

Which is the current going "out" ?


But ok I'm going to guess that it is the current going through the left 6k resistor):

(Current for 6 ohm resistor) = v / 6000 ohms


I think this is the current going out and then:


3(I X) = 3v / 2000 ohms

I X = v / 2000 ohms

I L = V / 3000 ohms


so


( 3v / 2000 ) + (v / 2000) + (v / 3000) - (v / 6000) - 0.003A = 0

v [ (4 / 2000) + (1/3000) + (1/6000) ] = 0.003A

v [ (1 / 400) ohms ] = 0.003A

v = 1.2 V for the top / bottom wire / node then?


V = IR, I = V / R

So that must mean I l = 1.2V / 3000 ohms

I l = 1.6 x 10^-4 A

?
 
  • #6
gneill
Mentor
20,795
2,773
Which is the current going "out" ?


But ok I'm going to guess that it is the current going through the left 6k resistor):

(Current for 6 ohm resistor) = v / 6000 ohms


I think this is the current going out and then:


3(I X) = 3v / 2000 ohms

I X = v / 2000 ohms

I L = V / 3000 ohms


so


( 3v / 2000 ) + (v / 2000) + (v / 3000) - (v / 6000) - 0.003A = 0

v [ (4 / 2000) + (1/3000) + (1/6000) ] = 0.003A

v [ (1 / 400) ohms ] = 0.003A

v = 1.2 V for the top / bottom wire / node then?
Yes, that looks good. Excellent.
V = IR, I = V / R

So that must mean I l = 1.2V / 3000 ohms

I l = 1.6 x 10^-4 A
?
Whoops. The method is correct but there seems to have been a slip-up in the math.
 

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