Current in a Resistor network ( 2 parts of part b)

[hitman0097]

Homework Statement

Consider the resistor network shown in the figure below, where R1= 5$$\Omega$$ and R2= 7$$\Omega$$ .

(a) Find the equivalent resistance between points a and b
R_eq=([1/6 +1/5]+7)+12+6=(9.73)^-1+18^-1=6.32$$\Omega$$

(b) If the potential drop between a and b is 12 V, find the current in each resistor.
I_12$$\Omega$$=I_{6$$\Omega$$upper}=12/18=2/3A
I_{6$$\Omega$$lower}=.56A
I_5$$\Omega$$=?
I_7$$\Omega$$=?

Homework Equations

R_eq=V/(I_net)
I=V/R; resistor in parallel I/2

The Attempt at a Solution

For I_{6$$\Omega$$lower} I think I just did V/11$$\Omega$$ and rounded.
Now for I_5$$\Omega$$ shouldn't it be .53A or the same .56A they both got marked wrong though...
And for I_7$$\Omega$$ shouldn't it be .56A+I_5$$\Omega$$
wrong answers for I_7$$\Omega$$:.92A,.615A, 1.09A,.67A I have one more attempt. I want to make it count.
wrong answers for I_5$$\Omega$$:.56A, 1.23A,.53A 2 more tries.
which I think is more important.

Also the total I would be 1.899A?

 

[rock.freak667]

If the voltage across ab is 12 V, then isn't the voltage in 6 and R1 resistors also 12V?

 

[Delphi51]

Check the current for the lower branch again. The 6 and 5 in parallel make 2.727 ohms, right? So the lower branch has resistance 7 + 2.727 and the current should be 12/9.727 = 1.23 A. That's the current through R2. You haven't said what I5 or I7 mean so I don't know what else you are having trouble with.

 

[hitman0097]

I5 would mean the current threw the 5 ohms resistor and I7 the current threw the 7ohms resistor

 

[hitman0097]

and for the 6ohms resistor the answer was .56A

 

[hitman0097]

Ah I got it thanks the current threw the bottom wire was the key. So that would make I5=.67A and I7=I6lower+I5= 1.23A