P31_053 - . 0 )(30 . 0 ) = 2 . 73 A . (c) The left-hand...

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53. (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 i 1 = i 2 = E R 1 + R 2 = 100 V 10 . 0Ω+20 . 0Ω =3 . 33 A . (b) 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 R 3 is i 1 i 2 . KirchhoF’s loop rule gives E− i 1 R 1 i 2 R 2 = 0a n d E− i 1 R 1 ( i 1 i 2 ) R 3 =0 . We solve these simultaneously for i 1 and i 2 . The results are i 1 = E ( R 2 + R 3 ) R 1 R 2 + R 1 R 3 + R 2 R 3 = (100 V)(20 . 0Ω+30 . 0Ω) (10 . 0 Ω)(20 . 0Ω)+(10 . 0 Ω)(30 . 0Ω)+(20 . 0 Ω)(30 . 0Ω) =4 . 55 A , and i 2 = E R 3 R 1 R 2 + R 1 R 3 + R 2 R 3 = (100 V)(30 . 0Ω) (10 . 0 Ω)(20 . 0Ω)+(10 . 0 Ω)(30 .
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Unformatted text preview: . 0 )(30 . 0 ) = 2 . 73 A . (c) The left-hand branch is now broken. We take the current (immediately) as zero in that branch when the switch is opened (that is, i 1 = 0). The current in R 3 changes less rapidly because there is an inductor in its branch. In fact, immediately after the switch is opened it has the same value that it had before the switch was opened. That value is 4 . 55 A 2 . 73 A = 1 . 82 A. The current in R 2 is the same as that in R 3 (1 . 82 A). (d) There are no longer any sources of emf in the circuit, so all currents eventually drop to zero....
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