Find the admittance Yab in the circuit seen in the figure. Take that R1 = 6 , R2 = 6 , R3 = 4 and R4 = 12.8 Part A Express Yab in rectangular form. Express your answer in complex form. The answer is NOT Yab = (4+j3)10^(5) mS Yab = 40+j30 mS 510^(5) e^(j36.86)Yab = 510^(5) e^(j36.86) mS j12.8 n Yab bo 12 j10 n Solution Z3=R3||j10=4*j10/(4+j10) Z2=R2+j12=6+j12 Z1=R1-j2=6-j2 Zequ1=Z1||Z2||Z3=2.1517 +j 0.5793 Z=R4+Zequ1-j12.8=12.8-j12.8+2.1517 + j0.5793=14.9517 -j12.2207 Y=1/Z=1/14.9517 -j12.2207 Y=0.0401 + j0.0328 convert rect to polar a=sqrt((0.0401^2)+(0.0328^2))=0.0518 b=tan-1(-0.0328/0.0401)=- 0.6856 in radins, if conver to degree=-39.2 ans: 0.051806*expj-39.2.