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Online Homework 14 Solutions

# Online Homework 14 Solutions - {a Cainulate the equivalent...

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Unformatted text preview: {a} Cainulate the equivalent resistant: ﬁfth: netwnrk 51mm in the diagram. The mistmﬂsarecﬂl =9ﬂ,Rg=5 SLR; = 129, andR4= 16 {1. {h} Whm a battery with E = '3' V and 2m internal resistance: is man-had acres; this netwm'k, What current ﬂaws through resistur RI? :9 2.11 I = 1.5%1- I. r J. - Exit-9:133 (3291’- P-IHJ'Quﬁt'E'l :w In 3:03: E -. __ _ 0.318(31 ) Rail {337.9— M Current divisiun The circuit above Danish: 5 resistor: {R}. R1. RA. Ra. Ra) wiﬂl aqua] minutiae E. Find the min £3.51. 1:3:im: @— : 5&1"; Ez+EB QR Fma Van *- Q'L-ﬂ‘ 3E” R1~B 1:, - I I I 7. EP— ﬁﬁi '3 M = 33R. +23 = €52, Egg-swim; . gum Vat: = 6'9“"2'RL'. (”HQ IHenlenps A eireuit composed et'ene battery and four resistors is ananged as shewn in the ﬁgure. E = 20 v R| ' 13“ R: = 2 ‘1 R3 = la n R. =31); (a) Calculate the eurrent I. through RI! {b} What is the potential VA at points! relative to the gtmtnﬂ Ground means Pa = D. in the ﬁgure, paint G? Three loops The resistors in the eireuit shown in. the diagram hatte the values listed below. The current through R] is measured to be I1 = 2.? A. R1=lﬂ 32:93.1 R3=14§2 R4=32I1 (a) Cnieulete the EMF: [h] Unionists the currents through the ether resistors: {Positive entree! directions are shown by arrows on the diagram.) [1:] In a time interval A. t = 5D 5, how mush charge ﬂows though the battery? (djlnstirne inmnl.Ar=SDs,howtnuehtets1hestQis generatedﬁn allresistors}? a) ﬁlial m3 2‘1 £1.11: LL; WM W ﬁn”, and C...» m M“ at? :21, FL: __ ‘25 Meters The points in and e on the diagram are the terminals are battery with EMF E and internal reaistmee r. The eireuit resistors have the Foﬂowing values: R1=3Jﬂ Rz=4l ﬂ R3=51 ﬂ Re=55§1 {a} An ideal voltmeter is connected between he and o; it reads PL - Pa. = 12 V. What is the EMF of the battery? {b} What resistance would an ohmnteter connected aeroea points a and e lead? {e} An ideal ammeter is inserted between points a and h and it reads {1.2 A, Detennine the internal reeiatanee of the battery. {d} What does the voltmeter connected across I: and c read now? {e} CaIeulate the magnitude of the current through R1. {f} At what rate ie ﬂtermai energy.r [heat] being produced inside the battery? {g} What is the power being provided by the halter to the extemal eireuit [outside the battery}? § '1de mm =3 CUW‘tﬂ ﬂing!) mm .5) W:o Ere- Vii—Vb: a J E 511-2“ J9) 9—1-1—3 3 121 1" PJJ- RE ‘1 [EFL-O” = 1M r :35“ Pm = i3“— -ﬁom, : Qo.q_a‘,ﬂ..i QZQ CL) V‘D _ E‘ '1'!" (Gr ELIZi'E'E-tl‘) Multilccp Resistors R1, R1, R3, and R4 arcarrangcd in a circuit as shown in thc figurc above. Thc direction of pcciﬁvc currents 1‘3, 1’1, I3, and I; through the rﬂistcrc arc Shawn. (112.. negative cttrrcttt values remand to cmcnts in thc oppositc dirccﬁm tc marshcm ahcvc}. Thc valucc for thc rccistcrs and thc hattcrics in thc circuit arc: R1=iﬂﬂﬂ.ﬂz=200ﬂ,ﬂg=30ﬂ.R¢-4ﬁiﬂ, Vl-4V1ttndi’z=12vr What is lhc valuc of 12, the current thrcugh R3“? {Remembcrlg may be ncgativc.) ': ‘l'q-I-ig g L’- - [6131.1 "" 7591.3 .... I ' -— tho G) 1?... +%13 I :.O @4169 LI" —Lm(i1+®3) “ {3951.3 tr- 1061: = ﬂ*_J 13 11350 L m4m© 1'14- 7WD IW IL :. D *%4neamhﬁ%ml ,. Igg am “L1- 2% ...
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