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# Measurements And Instrumentation (Math Problem Sample)

Instructions:

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..

Content:

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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