9/29/2005 Kirchoffs Current Law.doc 1/5 Jim Stiles The Univ. of Kansas Dept. of EECS Kirchoff’s Current Law So, we now know that: ()SdQIrdsdt== ⋅∫∫J Consider now the case where S is a closed surface: ()SdQIrdsdt== ⋅∫∫Jw The current I thus describes the rate at which net charge is leaving some volume V that is surrounded by surface S. We will find that often this rate is I=0 ! Q: Yikes! Why would this value be zero?? A: Because charge can be neither created nor destroyed! Think about it. If there was some endless flow of charge crossing closed surface S—exiting volume V—then there would have to be some “fountain” of charge creating this endless outward flow.9/29/2005 Kirchoffs Current Law.doc 2/5 Jim Stiles The Univ. of Kansas Dept. of EECS Alternatively, if there was some endless flow of charge crossing closed surface S—entering volume V—then there would have to be some charge “drain” that disposed of this endless inward flow. * But, we cannot create or destroy charge—endless charge fountains or charge drains cannot exist! * Instead, charge exiting volume V through surface S must have likewise entered volume V through surface S (and vice versa). * As a result, the rate of net charge flow (i.e., current) across a closed surface is very often zero! In other “words”, we can state: ()0Srds⋅=∫∫Jw For example, consider a closed surface S that surrounds a “node” at which 3 conducting wires converge: 1 2 3 S9/29/2005 Kirchoffs Current Law.doc 3/5 Jim Stiles The Univ. of Kansas Dept. of EECS Since current is flowing only in these wires, the surface integral reduces to a surface integration over the cross section of each of the three wires: () ()() ()31 21 32sSSss sS Srdr rd ss d dss r⋅⋅++⋅⋅=∫∫∫∫∫∫∫∫J JJJwwww The result of each integration is simply the current flowing in each wire! ()321sSrs Id II⋅=++∫∫Jw 1 2 3 S2 S2 S1 1 2 3 S I1 I2 I39/29/2005 Kirchoffs Current Law.doc 4/5 Jim Stiles The Univ. of Kansas Dept. of EECS But remember, since we know that charge cannot be created or destroyed, we have concluded that: ()0sSrds⋅=∫∫Jw Meaning: 1 320I II=++ More generally, if this node had n wires, we could state that: 0nnI=∑ Hopefully you recognize this statement—it’s Kirchoff’s Current Law! Therefore, a more general, electromagnetic expression of Kirchoff’s Current Law is: ()0Srds⋅=∫∫Jw Note that this result means that the current density ()rJ (for this case) is solenoidal! In other words, the above integral likewise means that ()0r∇⋅ =J .9/29/2005 Kirchoffs Current Law.doc 5/5 Jim Stiles The Univ. of Kansas Dept. of EECS Gustav Robert Kirchhoff (1824-1887), German physicist, announced the laws that allow calculation of the currents, voltages, and resistances of electrical networks in 1845, when he was only twenty-one (sowhat have you been doing)! His other work established the technique of spectrum analysis that he applied to determine the composition of the
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