Measurements And Instrumentation (Math Problem Sample)

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 ±10V and ammeter reads 10A with probable error of ±1A. Find the probable error in the computed value of resistance. SOLUTION R=VI Voltage, V=160±10V Current, I=10±1A Probable error, e=102x12=10.0499Ω error, e=a2xb2=101602+1102=0.1179 Resistance=160V10A=16Ω Probable error in resistance=±16x0.1179Ω=±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 0.1Ω 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 0.1Ω 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=0.04124Ω Current through the unknown resistor: I=VSRT=0.4221+1.02350.04124+0.1=10.2351A Power lost through the resistor: P=I2R=10.23512x0.04124Ω=4.3201899W QUESTION 3 A thermocouple circuit uses a chromel-alumel thermocouple which gives an e.m.f. of 33.3mV when measuring a temperature of 8000C with reference temperature of 00C. The resistance of the meter coil, Rm is 50Ω and a current of 0.1mA gives full scale deflection. The resistance of junction and leads, Re is 12Ω. Calculate: * Resistance of the series resistance if a temperature of 8000C is to give full scale deflection. V=IR emf=I(Re+Rm+Rs) Rs=33.3x10-31x10-4-12+50Ω=271Ω * The approximate error due to a rise in 100C in the copper coil of the meter, given the resistance temperature co-efficient of the coil is 0.00426/0C. △R=10Ω ∆R=50x0.00426x10Ω=2.13Ω Current in the circuit, I=emfRm+Re+Rs+∆R=33.3x10-3V50+2.13+271+12=0.09936mA Approximate error in temperature, e=0.09936-0.10.1x800=-5.12 0C QUESTION 4 A Barium Titanate pickup has the dimensions of 4mm by 4.5mm by 1.0mm. The force acting on it is 10N. The charge sensitivity of Barium Titanite is 145pC/...