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Pages:
2 pages/≈550 words
Sources:
2 Sources
Level:
APA
Subject:
Engineering
Type:
Math Problem
Language:
English (U.S.)
Document:
MS Word
Date:
Total cost:
$ 10.8
Topic:

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 ±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=4.32...
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