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Unformatted text preview: Problem 2.43 30;” “U”: EWIFHN HaﬁwH‘EIrpnﬂ l'rr ﬁnnan nunudlmmmum mu
351 l'I 5.! Kn ow n q uantities: Circuit Dfl‘ig'ure 2.4.3. Fin :1:
3) equivalent resistaiwe .—
11] current I
c) plume: delivered
:1] voltages e) minimum IIU'WEE ﬁlling I'm R]
Analysis: a} The equivalent resistance seen 115' Lhe some is
R = 3 +6 +l =12}? 11] Applying K‘v‘L: Ei +121: = CI
Therefme, r' = 9.5 A .
c} P=Tf=53315 =5 'LE'
:1] Applying ﬁlm's law: v1 —IS:I'=3 1" and v] 'l:'=2 1".
e} FEED: :33] = 11V. Cum11 I 11 nunn I :n—wu: In. 'nnwcrnuu u:u.1.h:<:wu_ .11] SOIL:th n:
Kn ow n q uant ities: Circuit: of Figure 1.14.
Find: Equivalent resistanm andi_a'_: 1' . AnalysiE:
35.3 = 2—19 7:]:13 52. Hatefule. By me C'ment diﬁder rule:
1": 7
i = '_ ll].9:=—_IEI.9!=E.EA.
' T2 —9 31 9.150. :mm 2111! "3 [land 1': ['1 resistors are :41 parallel. we :11: cunnlude that
Y = 91'] = 13 ‘5" '.W$W'" r. w:..a¢<. c I'\.:: :.v:u u: :.ou  IHIII Problem 2.45 Sch: lion:
Kn cm H q uantities: Circuits of Figure 115.
Find: Equivalent resistance and CW f KL! Analysis:
Ssepl: :J, +.:—:2=1451
Seep 2: 24 E = I5 1'! Thereime. the equivalent circuir is as shown in the ﬁgure: 3] 0 " 6 I] Further. [4 + 6] 90 = 9 w The new equit'elm circuit is shown heluw.
Thus Emm = 109. We can new film] I'l'Le merit i by ﬁle current divider rule as fuﬂnws: a I. g D 51w "" Hm Pratlam 2.55 SOILlion: Known quantities:
SEhEl'Ialic of Lhe chnuit Shawn in Figure P255 wiﬂ: resistant:
R0 = 21131 =Lﬂlﬂi=ifﬂﬂlﬂ3 =51": alumna? me vs =12v.
Find: a} The mesh current: in. ii“ I': h] The Elma dimugh each IESiEIIII.
AnalysiE: F'W“ Fl“ Applying K‘ﬂ. m meahlal_rneshb}a.11d mesh {n}: IruGRD + {re 4331: 0 + tea !'a:i*= '1
'[5a —'b,''R1—EIER2+[1} "511333 =ﬂ 3‘ '[l'c—'a:}—i1='e ‘5 7': ‘13)“:
l E = ("c ‘1} JR: 5436 — .5}: 1;
Solving the 55:5[211 we obtain:
(IR. =1'E = 2 A {pusiﬁve in [he directim‘L ufz'a :l
'2‘ = 2 '3‘ .13! = r5 — .'_, = .1. A {jusiﬁve in me diiactim ufz'b] j '13: =1}, = E A (positive in [he directim ufz'g :l
ZIP.I =1rii, =29. [pusiﬁveinﬂledimcbimufz'rj Prﬂ blem Cw.th Thr' HOErm? 'IICﬂﬂm'lﬂi. rIr Fern55 onmqllm rtr'r'utmmunn mum. R Sofuliﬁn: ."
H [T_1.I1L:II.H:3 . .. Known quantities:
Schematic of due citauit aha.111 in Figure P162. Find: The equix'nJmt PESiETIJJIDE' Rd: Elf the infinite :I'LETWka ufi'esistm. Analysia: “’9 can see the Maire nEIE'WIJﬂC of IESiEID'IS as due equivﬂmt TO the circuit
in the picture: Ti'LEfEfD'IE_ RR, .
REE=R+[R Rcéj— =2R——R_; ::= agar4'3}?
m: ...
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 Fall '09
 SHOANE

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