Find unknown current in circuit, voltage division law

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Discussion Overview

The discussion revolves around determining the unknown current (I L) in a circuit using voltage division law and Kirchhoff's rules. Participants explore various methods to analyze the circuit, including applying Millman's theorem and writing Kirchhoff's Current Law (KCL) equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes the lack of a given potential difference and questions how to find the current I L, suggesting that voltage may be equal across the circuit.
  • Another participant proposes applying Millman's theorem to simplify the circuit analysis.
  • A different participant suggests using current and voltage division laws after simplifying the circuit to two parallel wires, expressing uncertainty about how to determine the equivalent current (I eq).
  • One participant recommends writing a KCL node equation and defining the top node's potential (V1) to express unknown currents in terms of V1 and resistances.
  • Another participant attempts to derive expressions for the currents through various resistors and arrives at a potential (v) of 1.2 V for the node, leading to a calculation of I L.
  • There is a correction noted regarding a potential mathematical error in the calculation of I L, although the method is acknowledged as correct.

Areas of Agreement / Disagreement

Participants express various methods and approaches to solve the problem, but there is no consensus on a single solution or method. Uncertainty remains regarding the application of certain theorems and the correctness of calculations.

Contextual Notes

Some participants express confusion about the application of theorems and the determination of equivalent currents, indicating potential limitations in their understanding of circuit analysis.

Color_of_Cyan
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Homework Statement



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

http://imageshack.us/a/img31/1530/circuits215.jpg

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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Try to apply millman's theorem
 
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?
 
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.
 
gneill said:
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

?
 
Color_of_Cyan said:
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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