2 pages/≈550 words
Measurements And Instrumentation (Math Problem Sample)
THE TASK IS AN ASSIGNMENT ON THE QUESTION: A resistance is determined by voltmeter - ammeter method. The voltmeter reads 160V with a probable error of ±10V and ammeter reads 10A with probable error of ±1A. Find the probable error in the computed value of resistance. source..
Name: Instructor: Course Title: Electrical And Electronic Engineering Unit Title: Measurements And Instrumentation Date: QUESTION 1 A resistance is determined by voltmeter - ammeter method. The voltmeter reads 160V with a probable error of Â±10V and ammeter reads 10A with probable error of Â±1A. Find the probable error in the computed value of resistance. SOLUTION R=VI Voltage, V=160Â±10V Current, I=10Â±1A Probable error, e=102x12=10.0499Î© error, e=a2xb2=101602+1102=0.1179 Resistance=160V10A=16Î© Probable error in resistance=Â±16x0.1179Î©=Â±1.88679 QUESTION 2 During the measurement of a low resistance using a potentiometer the following readings were obtained. * Voltage drop across the low resistance under test is 0.4221V * Voltage across the 0.1â„¦ standard resistance is 1.0235V Calculate the value of the unknown resistance, current through it and power lost in it. SOLUTION Let Voltage drop across the low resistance be VL=0.4221V Voltage drop across the 0.1Î© standard resistor be V0=1.0235V Let also Rs be standard resistor and Rx be the unknown resistor. VL=RSRS+RXxVin 1.0235=0.10.1+RXx(0.4221+1.0235) RX=0.14456-0.102351.0235=0.04124Î© Current through the unknown resistor: I=VSRT=0.4221+1.02350.04124+0.1=10.2351A Power lost through the resistor: P=I2R=10.23512x0.04124Î©=4.32...
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