Electric Circuits 8th Edition 101

Electric Circuits - D = 2R(R2 2R 2R2Rb(3.56 R Figure 3.37 a A sjmptified modeL ofthe Rt Rz R" RA Ra R(3.54(3.57 and therefore It foLtows

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PmcticaLP€rspectjve 77 Fig.3.37. Note furtherthat after finding R1, R2, R3, Ra, and R, in the circuit in Fig. 3.37, we have alio found the vaLues for the remaining rcsjstors, since R1 Begin analysis of the simpLified grid circuit in Fig. 3.37 by writing expressions for the currents i1, i2, i?, and ib. To find ir, descrjbe the equiva- lent resistance in paratlel with R3: - ^^ xr(R1 +2&) _ (& + 2R.)(R2 + 2R) + 2R2Rb (Rt I R, 2RJ For convenience, define the numerator
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Unformatted text preview: D = (\ + 2R)(R2 + 2R) + 2R2Rb, (3.56) , R . Figure 3.37 a A sjmptified modeL ofthe Rt - Rz, R" : RA, Ra - R,. (3.54) (3.57) and therefore It foLtows directty that and ( R r + R , + 2 R " ) . v,t '' R3 Expressjons for i1 and 12 can be found djrectly from ib usjng current division. Hence ' . = &-' R. Y d c ( R r + R r + 2 & ) D (1 58) iiR1 %,R' ' R - - R , r ' R . , D . ib(R] + 2R4) vd.(Rr + 2&) ( R t ' R , 2 R D The expression for i3 js simpLy (3.61)...
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This note was uploaded on 07/05/2010 for the course EE 100 taught by Professor Boser during the Spring '07 term at University of California, Berkeley.

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