Answer:
The values are:
![I_(o) =0.06+j0.045A](https://img.qammunity.org/2021/formulas/physics/college/txtkt4br0cd49p65s6fkmtoootxda3ky09.png)
![I_(1) =0.3+j0.225A](https://img.qammunity.org/2021/formulas/physics/college/jnj5iy8mwzst72ath4a0p1a0dnw775fjw2.png)
![V_(o) =12+j9V](https://img.qammunity.org/2021/formulas/physics/college/f2v8ugvc1n9vllzb50ppwmlaighoo2pfu7.png)
Step-by-step explanation:
the attached figure shows the diagram. The equation is the following:
(eq. 1)
Voltage across the 40 Ω is equal:
(eq. 2)
Applying the Kirchhoff´s voltage law:
(eq. 3)
Equation 1 and 2:
(eq. 4)
Substituting in eq. 3
![75=I_(o) (600+j150)+5I_(o) (40-j150)\\75=I_(o) (800-j600)\\I_(o) =(75)/(800-j600) =0.06+j0.045A](https://img.qammunity.org/2021/formulas/physics/college/d8tyjnayi3mkw0yekd625095ba4djkwgg5.png)
Substituting in eq. 4
![I_(1) =0.3+j0.225A](https://img.qammunity.org/2021/formulas/physics/college/jnj5iy8mwzst72ath4a0p1a0dnw775fjw2.png)
Substituting in eq. 2:
![V_(o) =12+j9V](https://img.qammunity.org/2021/formulas/physics/college/f2v8ugvc1n9vllzb50ppwmlaighoo2pfu7.png)