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LIMIT ERROR

LIMIT ERROR. BILA : q 1 = kuantitas sesungguhnya q = kuantitas terukur q e = limit error MAKA : q = q 1 ± q e q = q 1 ( 1 ± % q e /100 ) q = q 1 ( 1 ± e ) dimana e = q e / q 1. SATUAN LIMIT ERROR:

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LIMIT ERROR

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  1. LIMIT ERROR BILA : q1 = kuantitas sesungguhnya q = kuantitas terukur qe = limit error MAKA : q = q1 ± qe q = q1 ( 1 ± %qe /100 ) q = q1 ( 1 ± e ) dimana e = qe / q1

  2. SATUAN LIMIT ERROR: • Besaran yang digunakan misalnya tegangan dlm Volt  V = 220V ± 22V. • Persentase %  V = 220V ± 10% • Frictional (pecahan) e = qe / q1 = 22/22 = 0,10 V = 220V ( 1 ± 0,10 )

  3. OPERASI HITUNG LIMIT ERROR • PENJUMLAHAN: bila: qa = q1a± qea qa = q1a ( 1 ± ea ) qb = q1b± qeb qb = q1b ( 1 ± eb ) --------------------------+ qa + qb = (q1a + q1b)( 1 ± ea + b ) qa + b = (q1a + q1b){ 1 ± (qea + qeb) / (q1a + q1b)} atau qa + b = (q1a + q1b) ± (qea + qeb)

  4. 2. PENGURANGAN. qa – qb = (q1a – q 1b)( 1 ± ea – b ) atau qa – b = (q1a – q1b ) ± (qea + qeb) 3. PERKALIAN. qa.b = (q1a . q1b ) ± (%qea + %qeb) 4. BAGI qa/b = (q1a/q1b ) ± {%(qea) – %(qeb)}

  5. Latihan: 1.Tentukan nilai limit error Rx dari rangkaian Jembatan Wheatston yang mempunyai lengan pembanding R1= 250 Ω± 10%; dan R2 = 500Ω ± 25Ω ; R3= 1500 Ω ± 30Ω . 2. Hitung tahanan pengganti bila R1 = 750Ω ± 75Ω dan R2= 500 Ω ±10% terhubung: a.seri b.paralel. 3.Tentukan rugi daya dan efisiensi trafo yang memp. input 5000w ± 1% dan output 4800w ± 120w. 4.Motor listrik berkekuatan 1HP ± 15 % memp. data sbb: 220 V ± 22 V , 50 Hz. Tentukan arus yang dibutuhkan bila Cos φ = 0,8 ± 4% agar dapat memikul beban sebesar kemampuan maksimumnya.

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