A (I, + 12) 40 -M 360 3.0 I 12 F +15V E D + 3v Now we will a apply Kinchoff's Voltage law ( LVL ) for loop ABEFA . + 15 = 41 -612- -0 We take clockwise direction as the ] for loop BCDEB : + 3 = 6 12+ 3 ( 1 1 + 12) .(2 2) 3 = 9 12 + 3 1 1 1 = 312 + I, 9 I1 = (1 - 312 putting this in @ 1'5 = 4( 1- 312) - GI2 15 = 4- 12 12-672 15 = 4 - 18 12 1812 = 4-1.5 12 = 0 .14 A I) = 1 - 312 = 0.583 A ( I,+ [2 ) = 0.723 A current in left loop - I, = 01583 A in upward direction Current in right loop ( Iit Iz ) = 0.723 A in downward direction Currents in bridge section - I , = 0.14 A in upwoand direction