sol-w2-1

sol-w2-1 - Problem 2.43 30;” “U”: EWIFHN...

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Unformatted text preview: Problem 2.43 30;” “U”: EWIFHN Hafiw-H‘E-Irpnfl l'rr finnan nun-udlmmmum 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) plum-e: 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 The-refme, r' = 9.5 A . c} P=Tf=533|15 =5 'LE' :1] Applying film's law: v1 —IS:I'=3 1" and v] ---'l:'=-2 1". e} FEED: :33] = 11V. Cum-11- I 11- nun-n- -I :n—wu: I-n. 'nnw-crn-uu 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. Hateful-e. 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:- -:.-o-u- - IHIII Problem 2.45 Sch: lion: Kn cm H q uantities: Circuits of Figure 1-15. 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 Prat-lam 2.55 SOIL-lion: 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 dim-ugh each IESiEIIII. AnalysiE: F'W“ Fl“ Applying K‘fl. m meahlal_rnesh|b}a.11d mesh {n}: Iru-GRD + {re 4-331: 0 + tea -!'a:i*= '1 '[5a —-'b,'|'R1—EIER2+[1} "5113-33 =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 =1r-ii, =29. [pusifiveinflledimcbimufz'rj Prfl blem Cw.th Thr' HOE-rm? 'IICflflm'lfli. rIr Fern-55 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 PESiET-IJJIDE' 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_; ::= agar-4'3}? m: ...
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This note was uploaded on 02/29/2012 for the course 332 373 taught by Professor Shoane during the Fall '09 term at Rutgers.

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sol-w2-1 - Problem 2.43 30;” “U”: EWIFHN...

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