(g) i1= 0, and (h) i2= 0. 56. (a) The inductor prevents a fast build-up of the current through it, so immediately after the switch is closed, the current in the inductor is zero. It follows that 112100V3.33A.10.0+20.0iRRε===+ΩΩ(b) 213.33A.ii(c) After a suitably long time, the current reaches steady state. Then, the emf across the inductor is zero, and we may imagine it replaced by a wire. The current in R3is i1– i2. Kirchhoff’s loop rule gives ()112 212300.iR iRiRi i R−−=−−−=We solve these simultaneously for i1and i2, and find ( )( )2311320.030.010.020.010.030.020.030.04.55A,i+Ω++ΩΩ+ΩΩ+ΩΩ=(d) and ( )( )3230.010.020.010.030.020.030.02.73A.RiΩΩΩ+ΩΩ+ΩΩ=(e) The left-hand branch is now broken. We take the current (immediately) as zero in that
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.