Question Number 148825 by Zainalarifin last updated on 31/Jul/21

Answered by Zainalarifin last updated on 31/Jul/21

ΣIR+ΣE=0 i_1 +i_2 =i_3 →i_2 =i_3 −i_2 ......(1) 2i_1 +3i_3 +9−6=0 2i_1 +3i_3 =−3 ....(2) 6i_2 +3i_3 +9−12=0 6i_2 +3i_3 =3 6(i_3 −i_1 )+3i_3 =3 −6i_1 +9i_3 = 3 ....(3) (2)→(3) 2i_1 +3i_3 =−3 ×3 −6i_1 +9i_3 = 3 −−−−−−−+ 6i_1 +9i_3 =−9 −6i_1 +9i_3 = 3 −−−−−−−+ 18i_3 = −6 i_3 = −(1/3) A 2i_1 +3i_3 =−3 2i_1 −1=−3 2i_1 =−2 →i_1 =−1 A i_2 =−(1/3)+1=(2/3) A

Answered by JDamian last updated on 31/Jul/21

{: ((V_(AB) +V_(BC) = V_(AC) )),((V_(DB) +V_(BC) =V_(DC) )) } {: ((2I_x +3(I_x +I_y )+9= 6)),((6I_y +3(I_x +I_y )+9=12)) } ⇒ {: ((5I_x +3I_y = −3)),((3I_x +9I_y = 3)) } I_x =( determinant (((−3),3),(( 3),9))/ determinant ((5,3),(3,9)))=((−27−9)/(45−9))=((−36)/(36))= −1 A I_y =( determinant ((5,(−3)),(3,( 3)))/ determinant ((5,3),(3,9)))=((15+9)/(45−9))=((24)/(36))= (2/3) A