Assignment 1 Solutions.doc

Assignment1Solutions1)Usingmeshanalysis,findthecurrentthroughthe5Ohmresistor:Meshequations:10I1-5I2-4I3=18-5I1+10I2-3I3=12-4I1-3I2+9I3=0Solvetogive:I1===,Ix=I1-I2=)UsingKVLandKCLorMeshanalysis,:I1=2AMesh2:-4I1+6I2=10VI2=IX=3A2)UsingaTheveninequivalent,hecurrentthroughthe2ohmresistorCalculateRTH:Remove2ohm,short10Vsourceandopen2Asourcethenfindresistance:RTH=4ohmCalculateVTH:Remove(open)the2ohm,thenVTH=10+2*4=18VNowwehaveVTH=18VinserieswithRTH==18/(4+2)=3A3)UsingaNortonequivalent,hecurrentthroughthe2ohmresistorRN=RTH=4ohmCalculateIN:shortthe2ohm,thenusingKCL,2A=I1+IShortI1=-10/4=-=Ishort=:IX=IN*RN/(RN+2)=3A1)UsingKVLandKCLorMeshanalysis,:I1=6AMesh2:-3I1+(3+6)I2-6I3=18Mesh3:I3=-3APut(1)and(3)into(2)toget9I2=18,I2=2A2)UsingaTheveninequivalent,:shortvoltagesource,opencurrentsources,remove6ohmr