# Homework Help: Finding current using current dividers

1. Feb 1, 2013

### DODGEVIPER13

1. The problem statement, all variables and given/known data
find i3?

2. Relevant equations
i1=i(R3/(R3+R6))
i2=i(R6/(R3+R6))
R(parallel)=R6(R3)/(R6+R3)
v=iR

3. 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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2. Feb 1, 2013

### DODGEVIPER13

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.

3. Feb 1, 2013

4. Feb 3, 2013

### CWatters

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

5. Feb 3, 2013

### DODGEVIPER13

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?

6. Feb 4, 2013

### Staff: Mentor

You have it right now. But your first post wasn't correct.
There isn't a current of 3 anywhere.