Question

A balanced star-connected three-phase load is shown in Figure 4. Determine the value of the line currents IR,IY and IB using mesh-current analysis.

Answer 1

Therefore the value of the line currents IR, IY, and IB are

`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`

Step-by-step explanation:

Apply KVL for loop 1

`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 )`

Apply KVL for loop 2

`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 )`

Solving the above equations,

`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`&

`415\angle 0^(o)=\left ( 10\angle 53.13^(o) \right ) * 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`

Hence,

`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`