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Unformatted text preview: the use of the cuflentdivider 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 simplifying the circuit with seriesparallel 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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This note was uploaded on 07/05/2010 for the course EE 100 taught by Professor Boser during the Spring '07 term at Berkeley.
 Spring '07
 Boser

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