1 ) |Step One : Find current. VE IR R, : V= IR R2 : V= IR Rq: V= IR RS : V=IR 5 8 V = I ( 120 12 ) 58 V = I 18252) 3V= I ( 25 12 ) 3V= I( 110-12) R = 120 I = 0. 483 A I= 707 A I = 0. 12 A I = 0.027 A Ra = 82 R2= 64 Loop I Voltage Law : Loop 2 Voltage Law: Ry= 25 I RI + I2 R2 + 13 R3 = 58 V Ig Ry + I3 R3 + Is R5 = 3. 0 V RS = 110 Loop 3 Current Law: at the first split I1 = I3+ 15 >using current law, substitute I, in Voltage Law eq. D ( I 3 + Is ) R, + 12 R 2 + I3 R3 = SBV 2 Iy Ry + Ig R3 + Is RS = 3.0V plug In R, R2 , R3 , Ry, RS O 120 12 + 120 Is + 82 12 + 6413 = 58 V -> 18413 + 12018 + 82 12 = 68V 2 25 14+ 6413 + 11015 = 3.0V