Finding current using current dividers

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Discussion Overview

The discussion revolves around calculating the current i3 in a circuit involving current dividers and resistors. Participants explore various methods to find i3, including the use of Ohm's law and the principles of parallel and series resistances. The context is primarily homework-related, focusing on the application of electrical circuit theory.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes using the equations i1=i(R3/(R3+R6)) and i2=i(R6/(R3+R6)) to find i3, but expresses uncertainty about their calculations.
  • Another participant calculates i=V/R4 and finds i=3sin(t) A, then attempts to derive i1 and i2, leading to an incorrect conclusion about i3.
  • A different approach involves calculating the source current Is and using KCL to determine the distribution of current through the resistors, leading to an expression for i3 as 12/9 Sin(t) A.
  • One participant combines the 6 ohm and 3 ohm resistors in parallel to find an equivalent resistance, then calculates the total current and derives i2, questioning the correctness of their method.
  • Another participant confirms the correctness of the latter method while critiquing the initial calculations of another participant, indicating confusion in the earlier posts.

Areas of Agreement / Disagreement

Participants express differing methods and calculations to find i3, with some agreeing on certain approaches while others challenge the correctness of initial claims. The discussion remains unresolved regarding the best method to calculate i3.

Contextual Notes

There are various assumptions made regarding the distribution of current and the equivalent resistances, which may not have been fully clarified or agreed upon by all participants. Some mathematical steps and definitions are also not explicitly resolved.

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Homework Statement


find i3?


Homework Equations


i1=i(R3/(R3+R6))
i2=i(R6/(R3+R6))
R(parallel)=R6(R3)/(R6+R3)
v=iR


The Attempt at a Solution


i=V/R4=(12sin(t))/(4 ohm)= 3sin(t) A then using R parallel I get 2 ohms as the resistor then again using v=iR. I get 1.333sin(t) A. Did I do this wrong or would this be acceptable
 

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I tried this way using v=iR4 finding i=3sin(t) A. Then using the equation to find i1 I get i1=sin(t) A and using the equation to find i2=2sin(t). i3=i1+i2 so I get i3= 3sin(t) A which is incorrect. By the way i1 is the current on the leg with the 6 ohm resistor and i2 is the current on the leg with the 3 ohm resistor.
 
Ugg no answers did I make it confusing again I always do that bad grammar and what not anyways please help!
 
i=V/R4=(12sin(t))/(4 ohm)= 3sin(t)

Perhaps it's me but I don't follow that at all.

I did it this way...

Let the source current be Is and the source voltage Vs =12 Sin(t)

Then

Is = Vs/(4 + 6//3)
Is = Vs/6......(1)

By inspection 2/3rds of Is goes through the 3 Ohm and 1/3rd through the 6 Ohm resistor (they must sum to 3/3rds Is due to KCL)

so

I3 = 2/3 * Is

Using eqn 1...

I3 = 2/3 * Vs/6
I3 = Vs/9

Then substitute Vs with 12Sin(t)

I3 = 12/9 Sin(t)
I3 = 1.333 Sin(t) Amps
 
Oh yah well would this work as well what I did is combine the 6 ohm and 3 ohm resistor into a parallel and got 2 ohms. I then put it in series with the 4 ohm resisor and get 6 ohms. then using v=ir i get i=2sin(t) A then using my formula for i2= 2sin(t)(6/9)=(4/3)sin(t) A which is correct. My question is what I did correct as well?
 
DODGEVIPER13 said:
Oh yah well would this work as well what I did is combine the 6 ohm and 3 ohm resistor into a parallel and got 2 ohms. I then put it in series with the 4 ohm resisor and get 6 ohms. then using v=ir i get i=2sin(t) A then using my formula for i2= 2sin(t)(6/9)=(4/3)sin(t) A which is correct. My question is what I did correct as well?

You have it right now. But your first post wasn't correct.
i=V/R4=(12sin(t))/(4 ohm)= 3sin(t)
There isn't a current of 3 anywhere.
 

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