{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Electric Circuits 8th Edition 82

Electric Circuits 8th Edition 82 - rr grves-ol i"Rr...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
58 Simpte Resistjve Circuih \t'i 'ii i )" R, RO Rr n;c-ir-i" figurc 3.1 A Resjstors connected jn sedes. R, RO RS 3.1 Resistors in Series In Chapter 2, we said that when just two elements connect at a single node.lhey are said to be in series. Seri€\-connected circuit eternents canr lhe same cuffenr.The resi.rors in rhe circuil shown iD frg.3.t are con- necled in series. \ e can show lhal lhe.e fesistorc cafry lh; .ame currerr b\ applling Kirchhoffs currenl law to cacb node in rh; circuir.The scrics interconnection in Fig. 3.1 requires that is=4= i2=ir=ia: is= -i6= i. , (3.1) which states that if we know any one ofthe seven currents, we know them all. Thus we can redraw Fig.3.1 as shown in Fig.3.2, retainhg rhe idenrity of the single
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: rr grves-ol + i"Rr + irn2 + irR3 + j.Rj + lrR5 + trn6 + irRT:0, (3.2) z ' . = t " ( R 1 + R 2 + & + & + R s + R 6 + R ? ) . (3.3) h g f e Figure 3.2 A Series resistors with a sjngte unkno\i/n The significance of Eq.3.3 for calculating ir is that the seven resistors can be replaced by a single resistor whose numeiical value is the sum of the individual resistors. that is. R e q = R 1 + R 2 + R 3 + R 4 + R 5 + R 6 + R 7 \3.4) and h Figure 3.3 A A simplified version of the cjrcujt shown in Fig.3.2. (3.5) Thus we can redraw Fig.3.2 as sho$n in Fig.3.3. In general, if tr resistors a.re connected in series, the equivalenl sircle re.islor has a resrstance equal lo lhe sum ot lhe & re,i.rancir. or (3.6) Note that the resistanc€ of the equivalent resistor is always larset than !har ot lhe largcsl resi.tor in rhe series conneclion. Combining resistoc in series >...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online