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Example (2): Design an Ayrton shunt by indirect method to provide an ammeter with current ranges IA, SA, and 10A, if PMMC meter have internal resistance of 50 and full scale current of ImA. Sol: Rm-502 Ipsp=Im=ImA IA Ra Ir. Rm + R Im SA Im +o Where R-Ra+ Rb+ Rc Rb Rm 10A And r- parallel resistors branch with the meter 1- For 1A Range: Rc 1 Rm + R Im R

Example (2): Design an Ayrton shunt by indirect method to provide an ammeter with current ranges IA, SA, and 10A, if PMMC meter have internal resistance of 50 and full scale current of ImA. Sol: Rm-502 Ipsp=Im=ImA IA Ra Ir. Rm + R Im SA Im +o Where R-Ra+ Rb+ Rc Rb Rm 10A And r- parallel resistors branch with the meter 1- For 1A Range: Rc 1 Rm + R Im R

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Example (2): Design an Ayrton shunt by indirect method to provide an ammeter with current ranges 1A, SA, and 10A, if PMMC metet have internal resistance of 502 and full scale current of ImA. Sol.: Rm-502 Irsp=Im=lmA 1A Ra Ir. Rm+ R Im SA Im +o Rm Where R=Ra+ Rb+ Rc And r= parallel resistors branch with the meter Rb 10A 1- For 1A Range: Re I1 Rm + R Im R
Transcribed Image Text:Example (2): Design an Ayrton shunt by indirect method to provide an ammeter with current ranges 1A, SA, and 10A, if PMMC metet have internal resistance of 502 and full scale current of ImA. Sol.: Rm-502 Irsp=Im=lmA 1A Ra Ir. Rm+ R Im SA Im +o Rm Where R=Ra+ Rb+ Rc And r= parallel resistors branch with the meter Rb 10A 1- For 1A Range: Re I1 Rm + R Im R
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