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

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Chapter8: Electromagnetism And Em Waves
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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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