Electric Circuits 8th Edition 88

Electric Circuits 8th Edition 88 - the use of the...


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64 Simpte R€sistive Circuits rigure 3.15 A Ihe drnent-divider circuit. t o,*1,' , * The Current-Divider Circuit The curent-divider cfucuit showr in Fig- 3.15 consists of two tesistors con- nected in parallel adoss a current source. The current divider is designed to divide the current i. between R1 and R2. We find t]le relationship between ttre curent i, a.nd the current in each resistor (that is, i1 and ir) by directly applying Ohm's law and Kirchhoffs current law. The voltage aooss the parallel resistom is Ffom Eq.3.26, Analyzing a Cufient-Divider Circuit ,.= R, , "' R, + R7-"' ,.= R' , "' R, + Rz-'' (3.26) 13.27) (3.28) Equations 3.27 and 3.28 show that the current divides between two resis tors in parallel such that the curent in one resistor equals the curent entering the parallel pail multiplied by the other resistance and divided by the sum of the rcsistors.
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Unformatted text preview: the use of the cuflent-divider equation. Find the power dissipated in the 6 f) resistor shown inFig.3.16. Solution First, we must find the curent in the resistor by sim-plifying the circuit with series-parallel rcductions ltus, the circuit shown ir Fig. 3.16 rcduces to the one shown in Fig. 3.17. We find the qrllent i, by using the formula for current division: 1 6 j" = _--(10) :8 A. t b + 4 Note that i, is the cunert in the 1.6() iesistor in Fig. 3.16. We now car turther divide i, between the 6 O and 4 O resistols. The current in the 6 O resistor is rd : _(8) : 3.2 A, and the power dissipated in the 6 O resistor is p=(3.2)'(6)=61.44w. 1.6 r} t ) f r r o 3 o n figure 3.16 A The ci,cuitfor Erampte l.l. 1 0 A I ) 1 6 0 4{)} Figure 3.17 A A sjmptjfication ofthe ciftuitshown in Fig.3.16....
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