.MCAD 310000000 [  docDocument LmcObjectZ  dim_formatSmasslengthtimecharge temperature luminosity substanceNumericalFormatPdii  shpRectU8^rmcDocumentObjectState[ mcPageModelJ????mcHeaderFooterH@H |P CHeaderFooterI@{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fnil\fprq15 Arial;}{\f3\fswiss\fprq15 Arial;}} {\colortbl\red0\green0\blue0;} \deflang1033\pard\plain\f3\fs18 \par } @{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fnil\fprq15 Arial;}{\f3\fswiss\fprq15 Arial;}} {\colortbl\red0\green0\blue0;} \deflang1033\pard\qc\plain\f3\fs18 \par } @{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fnil\fprq15 Arial;}{\f3\fswiss\fprq15 Arial;}} {\colortbl\red0\green0\blue0;} \deflang1033\pard\qr\plain\f3\fs18 \par } @I@{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fnil\fprq15 Arial;}{\f3\fswiss Arial;}{\f4\fswiss\fprq15 Arial;}} {\colortbl\red0\green0\blue0;} \deflang1033\pard\plain\f4\fs18 \par } @{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fnil\fprq15 Arial;}{\f3\fswiss Arial;}{\f4\fswiss\fprq15 Arial;}} {\colortbl\red0\green0\blue0;} \deflang1033\pard\qc\plain\f3\fs20 \par } @{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fnil\fprq15 Arial;}{\f3\fswiss Arial;}{\f4\fswiss\fprq15 Arial;}} {\colortbl\red0\green0\blue0;} \deflang1033\pard\qr\plain\f4\fs18 \par } @I@I MbP?MbP? TextState> TextStyle=@ ArialSerial_ParPropDefaultVNormal=@Arial@V Heading 1=@ Arial@V Heading 2=@ Arial@V Heading 3 =@ Arial@V Paragraph =@ Arial@VList =@ Arial@VIndent =@Times New Roman@V Title =@Times New Roman@V Subtitle font_style_listN font_styleO  VariablesTimes New Roman@O  ConstantsTimes New Roman@O TextArial@O Greek VariablesSymbol@O User 1Arial@O User 2 Courier New@O User 3Arial@O User 4Times New Roman@O User 5Times New Roman@O User 6Arial@O User 7Times New Roman@O SymbolsSymbol@O Current Selection FontArial@O Undefined Font@O HeaderArial@O FooterArial@O Rotated Math FontTimes New Romanw TextRegion* docRegionFshpBoxT JBBB CharacterMap-RangeMap:BProblem 1-6 A ferromagnetic core with an relative permeability of 1500 is shown in Figure P1-3. The dimensions are as shown in the diagram and the depth of the core is 7cm. The air gaps on the left and right sides of the core are 0.070 and 0.050 cm respectively. Because of fringing effects, the effective area of the air gaps is 5 percent larger than their physical size. If there are 300 turns (ignore this...use 400 turns as shown in the diagram) in the coil wrapped around the center leg of the core and if the current in the coil is 1.0 A, what is the flux in each of the left, center, and right legs of the core? What is the flux density in each air gap? Also, what is the terminal voltage of the coil for a 60Hz alternating current?  ChrPropMap6 ParPropMap8 RangeElem; ParPropData9 RangeData<@VEmbedMap0;LinkMap.;LinkData/@NormalArial eqRegionA@T"tree? p? ?dturns?1@A@T\ ? p? ?dN.c??t400?turns @A@T !? p"? !#?d"I.c$?"%?t$1.0&?$A'@A@Tx(? p)? (*?d)\m.r+?)1500,@A@T2-? p.? -/?d.\m.00?.1?@02?@13?t244?2\p5?16?t5107?K58?779?0:?d9H;?9m<*@T'B-(Dimensions give path lengths as follows:6(8(=;(>9@V0?;.(@@;(@A/@NormalArial @B@A@T ]"4@C? p@D? @C@E?d@D l.center@F?@D@G?t@F37@H?@Fcm@I@A@T "@J? p@K? @J@L?d@Kl.left@M?@K@N?t@M111@O?@Mcm@P@A@T S"$@Q? p@R? @Q@S?d@Rl.right@T?@R@U?t@T111@V?@Tcm@W@A@Tx " @X? p@Y? @X@Z?d@YA.c@[?@Y@\?p@@[@]?@\@^?t@]7@_?@]cm@`?p@[@a?@`@b?t@a7@c?@acm@d@A@T;" @e? p@f?@e@g?d@fA.c@h?@f@i?+@@hSerial_DisplayNodeW@j?@h@k?d@jcm@l?@j2@m@A@TDZZ(P@n? p@o? @n@p?d@og.left@q?@o@r?t@q0.07@s?@qcm@t@A@TDZP@u? p@v? @u@w?d@vg.right@x?@v@y?t@x0.05@z?@xcm@{*@Ts( B-Reluctances are as follows:68@|;@}9@V0@~;.@;@/@NormalArial @@A@Ty9 @? p@? @@?d@ R.center@?@@?d@ l.center@?@@?@@@?d@\m.r@?@\m.0@?@A.c@@A@TC@? p@? @@?d@R.left@?@@?d@l.left@?p@@?@@?@@@?d@\m.r@?@\m.0@?@A.c@@A@T@? p@? @@?d@R.right@?@@?d@l.right@?p@@?@@?@@@?d@\m.r@?@\m.0@?@A.c@@A@T8@? p@?@@?d@ R.center@?@@?+@@@W@?@@?t@1@?@H@@A@TL@? p@?@@?d@R.left@?@@?+@@@W@?@@?t@1@?@H@@A@T@? p@?@@?d@R.right@?@@?+@@@W@?@@?t@1@?@H@@A@TFD0@? p@? @@?d@ R.gap_left@?@@?d@g.left@?p@@?@@?@@@?d@\m.0@?@A.c@?@1.05@@A@TwF$0@? p@? @@?d@ R.gap_right@?@@?d@g.right@?p@@?@@?@@@?d@\m.0@?@A.c@?@1.05@@A@T[Cp@? p@?@@?d@ R.gap_left@?@@?+@@@W@?@@?t@1@?@H@@A@T[w#p@? p@?@@?d@ R.gap_right@?@@?+@@@W@?@@?t@1@?@H@*@T<:zz-Magnetomotifve force68@;@9@V0@;.@;@/@NormalArial @@A@TM%@? p@? @@?d@F.c@?@@?d@N.c@?@I.c@@A@T9#@? p@?@@?d@F.c@?@@?+@@@W@?@turns@*@T[ikh%:YY-Find the fluxes.68@;@9@V0A;.A;A/@NormalArial A@A@T{>(A? pA?,AA?p@AA?0AA?0AAA ?@AA ?AF.cA ?KAA ?A F.cA ?AA?p@A A?0AA?0AAA?0AAA?0AAA?@AA?AA?@AA?dAR.rightA?A R.gap_rightA?A R.centerA?KAA?A R.centerA?KAA?A R.centerA?AA?@AA?dAR.leftA ?A R.gap_leftA!?A R.centerA"?pA A#?0A"A$?0AA#A%?@A$A&?A$ \f.rightA'?A#\f.leftA(@A@TJ*A)? pA*? A)A+?p@A*A,?0A+A-?0AA,A.?@A-A/?A- \f.rightA0?A,\f.leftA1?A*A2?@A1A3?p@A2A4?0A3A5?0AA4A6?0AA5A7?0AA6A8?@A7A9?A7A:?@A9A;?dA:R.rightA?KA6A??A> R.centerA@?KA5AA?A@ R.centerAB?A4AC?@ABAD?dACR.leftAE?AC R.gap_leftAF?AB R.centerAG?KA2AH?AG1AI?pA1AJ?0AIAK?0AAJAL?@AKAM?AKF.cAN?KAJAO?ANF.cAP@A@T!XQ@+AQ? pAR?AQAS?p@ARAT?0ASAU?0AATAV?@AUAW?AU \f.rightAX?AT\f.leftAY?ARAZ?+@AY@WA[?AYA\*@Ts=:-%Sum fluxes to get the center leg flux6%8%A];%A^9@V0A_;.%A`;%Aa/@NormalArial Ab@A@T O,Ac? pAd? AcAe?dAd \f.centerAf?AdAg?dAf \f.rightAh?Af\f.leftAi@A@T}-Aj? pAk?AjAl?dAk \f.centerAm?AkAn?+@Am@WAo?AmAp*@T<?:,,-9Find the flux densities. Include the effect of fringing.6989Aq;9Ar9@V0As;.9At;9Au/@NormalArial Av@A@T 5\ .Aw? pAx? AwAy?dAx B.gap_leftAz?AxA{?dAz\f.leftA|?AzA}?dA|A.cA~?A|1.05A@A@TX* /A? pA?AA?dA B.gap_leftA?AA?+@A@WA?AA@A@T Eud`2A? pA? AA?dA B.gap_rightA?AA?dA \f.rightA?AA?dAA.cA?A1.05A@A@TTYj#`3A? pA?AA?dA B.gap_rightA?AA?+@A@WA?AA*@TA:-@QTo find the voltage, first find the flux linkages at the coil, in the center leg.6Q8QA;QA9@V0A;.QA;QA/@NormalArial A@A@T P6A? pA? AA?dA \l.centerA?AA?dA \f.centerA?AN.cA@A@Tg7A? pA?AA?dA \l.centerA?AA?+@A@WA?AturnsA*@T+C2-@Ivoltage is the time derivative of flux linkage. For a sinusoidal input, 6I8IA;IA9@V0A;.IA;IA/@NormalArial A@A@T v@10:A? pA? AA?dA\wA?AA?@AA?@AA?tA2A?A\pA?A60A?AA?dAradA?AsecA@A@T [rQh9A? pA? AA?dAVoltageA?AA?dA \l.centerA?A\wA@A@T\Rmh;A? pA?AA?dAVoltageA?AA?+@A@WA?AA*@T>B6P6P-AProblem 1-7 A two legged core is shown in Figure P1-4. The winding on the left leg has 600 turns and the winding on the right leg has 200 turns. (The diagram shows 400 turns and 300 turns, respectively, so we will use that.) The coils are wound in the directions shown in the figure. (This will be important for algebraic signs in our calculations.) If the dimensions are as shown, what flux would be produced by the currents i1=0.5A and i2=0.75A? Assume ur=1000 and constant. 68A;A9@V0A;.A;A/@NormalArial A@A@T$A? pA? AA?dAturnsA?A1A@A@T a"%A? pA? AA?dAN.lA?AA?tA400A?AturnsA@A@T "A? pA? AA?dAN.rA?AA?tA300A?AturnsA@A@T0 n"AA? pA? AA?dAi.1A?AA?tA0.50A?AAA@A@T "A? pA? AA?dAi.2A?AA?tA0.75A?AAA@A@T "A? pA? AA?dA\m.rA?A1000A@A@T(;^R<HA? pA? AA?dA\m.rA?A1000A@A@T3XHA? pA? AA?dA\m.0B?AB?@BB?@BB?tB4B?B\pB?BB?tB10B?KBB?B7B ?BB ?dB HB ?B mB *@T(s2B-@HThe dimensions give the following length of path and cross section area.6H8HB ;HB9@V0B;.HB;HB/@NormalArial B@A@T0AB? pB? BB?dBl.cB?BB?@BB?tB4B?pBB?BB?tB15B?B50B?BcmB@A@TB? pB ?BB!?dB l.cB"?B B#?+@B"@WB$?B"B%@A@T8PB&? pB'? B&B(?dB'A.cB)?B'B*?p@B)B+?B*B,?tB+15B-?B+cmB.?pB)B/?B.B0?tB/15B1?B/cmB2@A@TB3? pB4?B3B5?dB4A.cB6?B4B7?+@B6@WB8?B6B9?dB8cmB:?B82B;*@T0::-The path reluctance is68B<;B=9@V0B>;.B?;B@/@NormalArial BA@A@T0FBB? pBC? BBBD?dBCR.cBE?BCBF?dBEl.cBG?BEBH?@BGBI?dBH\m.rBJ?BH\m.0BK?BGA.cBL@A@Ty%BM? pBN?BMBO?dBNR.cBP?BNBQ?+@BP@WBR?BPBS?tBR1BT?BRHBU*@T0K[:X:-@XThe currents are drawn to give fluxes that add. Use the right hand rule to verify this.6X8XBV;XBW9@V0BX;.XBY;XBZ/@NormalArial B[@A@T0tEB\? pB]? B\B^?dB]F.cB_?B]B`?@B_Ba?dB`N.lBb?B`i.1Bc?B_Bd?dBcN.rBe?Bci.2Bf@A@Tt9Bg? pBh?BgBi?dBhF.cBj?BhBk?+@Bj@WBl?BjturnsBm@A@T8hLBn? pBo? BnBp?dBo\f.cBq?BoBr?dBqF.cBs?BqR.cBt@A@TvBu? pBv?BuBw?dBv\f.cBx?BvBy?+@Bx@WBz?BxturnsB{*@T8 > 2-At 60 Hz, find the voltages68B|;B}9@V0B~;.B;B/@NormalArial B@A@T8 8 I( B? pB? BB?dB\wB?BB?@BB?@BB?tB2B?B\pB?B60B?BB?dBradB?BsecB@A@T 8 ( B? pB?BB?dB\wB?BB?+@B@WB?BB?dBradB?BsecB@A@T8K qb JX B? pB? BB?dB\l.lB?BB?dBN.lB?B\f.cB@A@TK 9b X B? pB?BB?dB\l.lB?BB?+@B@WB?BturnsB@A@TpK b X B? pB? BB?dB\l.2B?BB?dBN.rB?B\f.cB@A@TK <b X B? pB?BB?dB\l.2B?BB?+@B@WB?BturnsB@A@T8 k J B? pB? BB?dBv.lB?BB?dB\wB?B\l.lB@A@T # B? pB?BB?dBv.lB?BB?+@B@WB?BB@A@T   B? pB? BB?dBv.rB?BB?dB\wB?B\l.2B@A@T ,  B? pB?BB?dBv.rB?BB?+@B@WB?BB*@T B * ** * -@vOpen circuit voltage: set high voltage-side current to zero. Recalculate the terminal voltage on the secondary side:6v8vB;vB9@V0B;.vB;vB/@NormalArial B@A@T( `* 9 B? pB? BB?dBi.1B?BB?tB0.0B?BAB@A@T * B? pB? BB?dBi.2B?BB?tB0.75B?BAB@A@T(< R =H B? pB? BB?dBF.cB?BB?@BB?dBN.lB?Bi.1B?BB?dBN.rB?Bi.2B@A@T< 1R H B? pB?BB?dBF.cB?BB?+@B@WB?BturnsB@A@T0f ` D B? pB? BB?dB\f.cB?BB?dBF.cB?BR.cB@A@Tl n B? pB?BB?dB\f.cC?BC?+@C@WC?CturnsC@A@T0 i B C? pC? CC?dC\l.lC?CC?dCN.lC ?C\f.cC @A@T 1 C ? pC ?C C ?dC \l.lC?C C?+@C@WC?CturnsC@A@Tx   C? pC? CC?dC\l.2C?CC?dCN.rC?C\f.cC@A@T <  C? pC?CC?dC\l.2C?CC?+@C@WC?CturnsC@A@T0 c B C ? pC!? C C"?dC!v.lC#?C!C$?dC#\wC%?C#\l.lC&@A@T  C'? pC(?C'C)?dC(v.lC*?C(C+?+@C*@WC,?C*C-@A@Tx   C.? pC/? C.C0?dC/v.rC1?C/C2?dC1\wC3?C1\l.2C4@A@T ,  C5? pC6?C5C7?dC6v.rC8?C6C9?+@C8@WC:?C8C;*@T B3  :: : -@xThe voltages are in a 4:3 ratio, which is what we would expect for an open circuited transformer with a 4:3 turns ratio.6x8xC<;xC=9@V0C>;.xC?;xC@/@NormalArial CA*@T C 7c AP *  -@yMagnetizing inductance is found from the open circuit test. We have no losses, so core loss resistance is insignificant.6y8yCB;yCC9@V0CD;.yCE;yCF/@NormalArial CG@A@T(~ _ F CH? pCI? CHCJ?dCIX.MCK?CICL?dCKv.rCM?CKi.2CN@A@T  CO? pCP?COCQ?dCPX.MCR?CPCS?+@CR@WCT?CRCU*@T : " **0*0-@Open circuit voltage: in the event that we worked from the other side (unconventional, but for the sake of completeness, we will do the calculations). Set secondary current to zero. Recalculate the terminal voltage on the primary side:68CV;CW9@V0CX;.CY;CZ/@NormalArial C[@A@T( `" 9 C\? pC]? C\C^?dC]i.1C_?C]C`?tC_0.5Ca?C_ACb@A@T "  Cc? pCd? CcCe?dCdi.2Cf?CdCg?tCf0Ch?CfACi@A@T(4 J =@ Cj? pCk? CjCl?dCkF.cCm?CkCn?@CmCo?dCnN.lCp?Cni.1Cq?CmCr?dCqN.rCs?Cqi.2Ct@A@T4 1J @ Cu? pCv?CuCw?dCvF.cCx?CvCy?+@Cx@WCz?CxturnsC{@A@T0^ ` Dx C|? pC}? C|C~?dC}\f.cC?C}C?dCF.cC?CR.cC@A@Td n x C? pC?CC?dC\f.cC?CC?+@C@WC?CturnsC@A@T0 i B C? pC? CC?dC\l.lC?CC?dCN.lC?C\f.cC@A@T + C? pC?CC?dC\l.lC?CC?+@C@WC?CturnsC@A@Tx   C? pC? CC?dC\l.2C?CC?dCN.rC?C\f.cC@A@T <  C? pC?CC?dC\l.2C?CC?+@C@WC?CturnsC@A@T0 c B C? pC? CC?dCv.lC?CC?dC\wC?C\l.lC@A@T  C? pC?CC?dCv.lC?CC?+@C@WC?CC@A@Tx   C? pC? CC?dCv.rC?CC?dC\wC?C\l.2C@A@T ,  C? pC?CC?dCv.rC?CC?+@C@WC?CC*@T >+ "( :&&-@aThe voltages are in a 4:3 ratio, which is what we would expect for an open circuited transformer.6a8aC;aC9@V0C;.aC;aC/@NormalArial C*@T ; 7[ AH *  -@yMagnetizing inductance is found from the open circuit test. We have no losses, so core loss resistance is insignificant.6y8yC;yC9@V0C;.yC;yC/@NormalArial C@A@T(v ^ F C? pC? CC?dCX.MC?CC?dCv.lC?Ci.1C@A@T  C? pC?CC?dCX.MC?CC?+@C@WC?CC*@T ? " :' ' -@The ratio between magnetizing inductances, measured from opposite sides, should be the turns ratio squared. Checking this, we find a ratio of 16:9, which is the square of the 4:3 turns ratio. 68C;C9@V0C;.C;C/@NormalArial