Answer:
Therefore the value of the line currents IR, IY, and IB are
[tex]I_{R}=I_{1}=83\angle 6.87^{o}A\\I_{B}=-I_{2}=-71.88\angle-23.13^{o}A\\I_{Y}=I_{2}-I_{1}\\I_{Y}=-I_{2}=41.50\angle113.3^{o}A[/tex]
Explanation:
Apply KVL for loop 1
[tex]415\angle 120^{o}=\left ( 3+j4+3+j4 \right )I_{1}-\left ( 3+j4 \right )I_{2}\\415\angle 120^{o}=\left ( 6+j8 \right )I_{1}-\left ( 3+j4 \right )I_{2}\\415\angle 120^{o}=\left ( 10\angle 53.13^{o} \right )I_{1}-\left ( 5\angle 53.13^{o}\right )I_{2} \rightarrow \left ( 1 \right )[/tex]
Apply KVL for loop 2
[tex]415\angle 0^{o}=\left ( 3+j4+3+j4 \right )I_{2}-\left ( 3+j4 \right )I_{1}\\415\angle 0^{o}=\left ( 6+j8 \right )I_{2}-\left ( 3+j4 \right )I_{1}\\415\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{2}-\left ( 5\angle 53.13^{o}\right )I_{1} \rightarrow \left ( 2 \right )[/tex]
Solving the above equations,
[tex]415\angle 120^{o}+830\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{1}-\left ( 5\angle 53.13^{o}\right )I_{2} +\left ( 20\angle 53.13^{o}\right )I_{2}- \left ( 10\angle 53.13^{o} \right )I_{1}\\415\angle 120^{o}+830\angle 0^{o}=\left ( 20\angle 53.13^{o}\right )I_{2}- \left ( 10\angle 53.13^{o} \right )I_{2}\\415\angle 120^{o}+830\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{2}\\I_{2}= 71.88\angle -23.13^{o}A[/tex]&
[tex]415\angle 0^{o}=\left ( 10\angle 53.13^{o} \right ) \times 71.88\angle -23.13^{o}-\left ( 5\angle 53.13^{o} \right )I_{1}\\415\angle 0^{o}=718.8\angle 30^{o}-\left ( 5\angle 53.13^{o} \right )I_{1}\\\left ( 5\angle 53.13^{o} \right )I_{1}=415\angle 60^{o}\\I_{1}= 83\angle 6.87^{o}A[/tex]
Hence,
[tex]I_{R}=I_{1}=83\angle 6.87^{o}A\\I_{B}=-I_{2}=-71.88\angle-23.13^{o}A\\I_{Y}=I_{2}-I_{1}\\I_{Y}=-I_{2}=41.50\angle113.3^{o}A[/tex]
