Question

10. Calculate voltage drops (VR1c =?, VR2C = ?, VR3C = ?, VRAC =?, VRSC =? ) across the resistors by using superposition principle.

Answer 1

Super Position theoremi In any active I bilatrad Nw Consisting No. of Energy sources Ę Resistencer etc! The Effect produced in any elanent when all source act at a time- is equal to sum of Effects when Each source is considerd indvidualet .. how to solved while The No.of Sob ckth shook SPT - Independent pourcen

4701 slor - given dakes - only or Source - Remain Voltage short I 5602 4702 AT 11 R= (1+//2.2k) + 470)|lik) + 560 22 12V 1157.5 X1000 +560 1157.5+1000 1096.5 - 336-5+560 I = 0.01094 3/A R = 1096.5005). | URI IUR :- UR = I xsbo NR, = 6.12 8 slv 157.5 VR, IK(0.01904 11 57.5+ VR₂ = 5.87 lui/v. VR- VR3 = (0.010943 - 5.87141x10 3x uto vez = 2.3839/v VRU VRS: VR u = 5.87141 & 2.3839 only (5) Source: – – Remain voltage Short 56. 470-1 ) TOT Î o.uz vi- 2.35 + 1.034 VT 2.2(11-12)=0 13.700 vi- 2.2 V2 = 2.35 V - V) 470 + 2 --> 0.56 V-VI) + 0.263. V2 + 0.47250 0.56 v, + 1. 2932 Vza o

Tom 0 0 :- 3-704 v - 2.2 V2 = 2.35 Xoos6 - 1.232 V, = 1.316 0.56 v. + 2932 V. rox3704 4.79001 V₂ = 0 z = 0-36986 lv Submite (ve) in er : 0.4763) ..So N = o.fsululv TUR = 2 = 0.36986v TURS = vi= o.ssullv VR3 = Vi-v2 - oorsul - 0.3696 25 - Ossul VR - Colusa y IVR = 0.48 4274 Total voltages - VR, 6. 1283 + 0.3698 VR2 = Sistituilusa IVR = 6.6983 VR₂ = 10.0169 v VR3 = 2.38 39+ cour427 VRS = 5.255 3 + 0.854g VR- 2.86817 V R = 9.112uu.

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