2016/4/6 6.3 example 3-continued 14 2 R L3 (positive as shown)Rs PU23cR,=0:FU235--R5 1k U202 EP.=0:(F.1.)3/5=R. 0.833 0.417 1.L.FU2Lg (w LV243·-53R -0.417 FU243·+53R1 -0.417 米十米 See discussion in Example Problem 6.3 6.3 example 3-continued 32 2M,=0:32R,+12,w ,,=-R 1 U TR132 EML30:32R1+12FU3Ua0,u."-3212 +0.667 +0.667 I.L.FUgUs (WN ku-3212R9u,u-212R 10 62016/4/6 5 9 6.3 example 3 – continued 10 6.3 example 3 – continued