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dDp&d íeÿÿd ôÊú ô tfd;d ‰d ±m.004dÔmAA PIECE OF WIRE HAS A RESISTANCE OF .028ê. WHAT IS THE RESISTANCE"d‚mDOF AN EQUAL LENTGH WIRE WHICH HAS TWICE THE DIAMETER OF THE FIRST ? d ‚m"USING RESISTIVITY EQUATION, R = ? dèd ;d ‰d ±m.06"dÔm?A PIECE OF WIRE HAS A RESISTANCE OF .12ê. IF THE WIRE IS CUT IN"d‚m#HALF, WHAT IS THE NEW RESISTANCE ? "d ‚m"USING RESISTIVITY EQUATION, R = ? dèd
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‰d
±m10.4dÔm16380"d Ôm.128d
Ôm?THE COPPER CABLE OF AN AUTOMOBILE STARTER MOTOR HAS A LENTGH OF"d‚mI3.15 FT. TESTING OF THE SYSTEM INDICATES THAT THE CABLE RESISTANCE SHOULD"d ‚m<BE NO MORE THAN .002ê. WHAT MINIMUM DIAMETER IS NECESSARY ? d
‚m8FIND AREA A = KL / R 16380 CM =(?)(3.15) / .002 dèm
D(MILS) = eû m
A m 128 MILS = "eû m? d èmDIAMETER IN INCHES = ? "d
èd;d jd ±m92dÔm9.057"d Ôm9"d
ÔmY"dÔmFIND THE LOAD VOLTAGE ELd‚m230V omLOAD um9ê {m #8 CU 46FT —m:FROM TABLE 2 R(WIRE) = .057ê = (.628ê / 1000FT) x ? dèm'USING OHNS LAW I = 25.4A = 230 / ? "d èmEL = 228.6V = 25.4 x ? "d
èm)DO YOU THINK THE WIRE SIZE IS OK (Y/N) ? "dèd;d
‰d
±m.4dÔm250"d Ôm12d
Ôm;IT IS DESIRED TO DRIVE A 40ê HEATER THROUGH A 125 FT COPPER"d‚mFEXTENSION CORD. TO MAKE THE CORD VOLTAGE DROP NEGLIBLE, ITS RESISTANCEd ‚mESHOULD BE LESS 1% OF THE LOAD RESISTANCE. WHAT GAUGE WIRE IS NEEDED ?"d
‚m2FIND WIRE RESISTANCE R(WIRE) = .01 x 40 = ? dèm6FIND AREA A = KL / R 6500 CM = (10.4)(?) / .4 d èmFROM TABLE 2 GAUGE # = ? d
èd;d
‰d
±m100"dÔm2080d Ôm16d
ÔmBA 50FT COPPER EXTENSION CORD IS REQUIRED TO OPERATE A DRILL MOTOR.d‚mIBASED UPON THE CURRENT NEEDED TO RUN THE DRILL MOTOR, THE CORD RESISTANCE"d ‚m7CANNOT NOT EXCEED .5ê. WHAT GAUGE WIRE SHOULD BE USED ?"d
‚m6FIND AREA A = KL / R 2080 CM = (10.4)(?) / .5 dèm
D(MILS) = eû m
A m
46 MILS = eû m? d èmFROM TABLE 2, GAUGE = ? d
èd;d ‰d ±m16.1dÔm8WHAT LENTGH OF #22 COPPER WIRE HAS A RESISTANCE OF 1ê ? d‚m4FROM TABLE 2, LENTGH = (1000 FT / ?) x 1ê = 62.1 FT dèd!;d ‰d ±m100"dÔmCWHAT LENTGH OF #30 ALUMINUM WIRE IS NEEDED TO MAKE A 30ê RESISTOR ?"d‚m.L = RA / K 176.4 FT = (30)(?) / 17 dèd%;d ‰d ±mNICKELdÔm@CONSIDERING THE TEMPERATURE COEFFECIENT OF THE MATERIALS LISTED,d‚m;WHICH MATERIAL WOULD MAKE THE BEST RESISTANCE THERMOMETER ?"d ‚m!FROM TABLE 1, BEST MATERIAL IS ? "dèd);d ‰d ±d jm220V omMOTOR" um 198V, 13A" {m #14 CU COPPER —m11dÔm10.4d ÔmCFIND MAXIMUM DISTANCE THE MOTOR CAN BE PLACED FROM THE 220V SOUCRE."d‚m9FIND RESISTANCE OF 1 LINE R = E / I .85ê = ? / 13 "dèm,L = RA / K 334 FT = (.85)(4096) / ? d èe ;d ‰d ±m3"dÔm6WHAT IS THE RESISTANCE OF 36 INCHES OF #22 GOLD WIRE ?d‚m,R = KL / A .07ê = (14.7)(?) / 625 dèe ;d
‰d
±m114"dÔm6"d Ôm1000d
Ôm@A 2-WIRE COPPER FEEDER LINE IS TO SUPPLY A 1500 WATT LOAD 500 FTd‚m@AWAY FROM A 120V SOURCE. WHAT MINIMUM WIRE AREA IS REQUIRED IF Ad ‚m.5% VOLTAGE DROP CAN BE TOLERATED AT THE LOAD ?d
‚m0FIND CURRENT I = P / E 13.3A = 1500 / ? dèm-FIND R(WIRE) R = E / I .54ê = ? / 13.3 "d èm-A = KL / R 19259 CM = (10.4)(?) / .54 "d
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;d ‰d ±mNICHROMEd Ôm1600dÔm:A 50 FT PIECE OF WIRE HAS A DIAMETER OF .04 INCHES. IF ITSd‚m7RESISTANCE MEASURES 20.25ê, WHAT IS THE WIRE MATERIAL ?"d ‚m'K = RA / L 648 = (20.25)(?) / 50 "dèmFROM TABLE 1, MATERIAL IS ? d èRse¤sesd
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seÁmRESISTANCE DETERMING FACTORS–sdseÁe"eÍS–se¤sdsdÁmITHIS PROGRAM DEALS WITH THE VARIOUS PARAMETERS THAT EFFECT THE RESISTANCE"–sd!sdÁmDOF A MATERIAL AND IN PARTICULAR WIRE RESISTANCE. THESE FACTORS ARE :–sd)seÁm1) LENTGH OF MATERIAL"–se seÁm2) CROSS SECTIONAL AREA"–seseÁm%3) RESISTIVITY OF A SPECIFIC MATERIAL"–sesdÁmEPROBLEM SOLUTIONS REQUIRE USE OF THE RESISTIVITY EQUATION AND THE AWG"–sesdÁmEWIRE TABLES. THE UNKNOWS FOR EACH PROBLEM ARE INDICATED BY A QUESTION"–sesdÁm>MARK. AT THE PROMPT, TYPE IN YOUR ANSWER AND THEN PRESS ENTER.–sd¤seseÁm,TO TOGGLE SOUND OFF\ON AT ANY TIME, PRESS F1–sd)¤seseÁmPRESS ANY KEY TO BEGIN–d" Vd%deveƒd¼sd¤sd se ÁmTABLE 1 - RESISTIVITY"–sd se1ÁmTABLE 2 - COPPER WIRE"–se¤sdsdÁmMATERIAL–sdseÁmK"–sdseÁm TEMP. COEFF.–sdse-ÁmB & S"–sdse6ÁmDIAMETER–sdseÁmê/MIL-FT–sdseÁm
OF RESISTANCE"–sdse-ÁmGAUGE"–sdse8ÁmMILS–sdseBÁm ê / 1000 FT"–m#C1R250D10L30D15R60U15L30BD15D10L250" ‡mFALUMINUM 17 .004 0 325 .098 dfmEBRASS 44 .002 8 128 .628"d fmECOPPER 10.4 .004 12 81 1.588"d
fmENICKEL 42.1 .006 14 64 2.525"dfmENICHROME 648 .00017 16 51 4.016"dfmESILVER 9.77 .0038 18 40 6.390"dfmEGOLD 14.7 .0034 22 25 16.100"dfmEBRONZE 108 .0005 30 10 103.200"d!fd u kdeVÿÿX s ksdÁ uf– ud ueÿÿm5C15BM13,19R300D100L300U100BR87D100BR86U100BM13,43R300"ªm7C15BM327,19R300D100L300U100BR87D100BR86U100BM327,43R300"ªsd)¤sese ÁmR = –seseÁmK x L"–seseÁmA"–ekeŒešeŒƒd)¼seseÁmL = LENTGH IN FT–sese2ÁmA(CIR MILS) = D(MILS)"eý –se¤o ´ ÿÿÿÿÌ