Have to find Ix.


Attempt:

Solving for current source:

I'x= 3.5 v / 7+3 + 15 ll 5 ?

the solution manual did not consider 5 ohms at all , instead

I'x= 3.5 v / 7+3+15

I don't understand why.


Solving for voltage source:

I''x= 2 x ( 10 / 10 + 15)

According to the manual 5 ohms is zero since it is parallel to the short circuit ( 3.5 v).

Isn't 5 ohms also parallel to 10 ohms ( 7+3) ?


Any help will be appreciated.

 

the solution manual did not consider 5 ohms at all

Because 5Ω is in parallel to a perfect voltage source. You can have many resistors in parallel to the voltage source and you can do the trick as I did in the image.

 

Thanks for your help.

 

First of all just a little note. You called the current source a voltage source and vice versa. Proper naming speeds up the comprehension process.

As your book and you did, I will solve this using superposition.

Considering the voltage source and open circuiting the current source. Ix belongs to the branch that contains the resistances of 3, 7 and 15 Ω in series. You know that I=V/R, and the voltage across this branch is the whole 3.5V of the source. We do not have here a voltage divider as the formula you wrote implies.

Considering the current source. When a resistance is put in parallel with a short circuit, the total resitance of the branch falls to zero. This can be found either by the formula \(R_t=\frac{R \cdot 0}{R + 0}=0\) or intuitively, by thinking that the current will choose the "easiest" path to go, the one that will have no resistance. This analysis leaves us with two parallel branches, one with 10Ω and one with 15Ω, led by a source of 2Amps. You can work the rest of the math yourself.