BEGIN{ARRAY}{l}{V_{dc}} = \frac{{2{V_m}}}{\pi } = \frac{{2 \times 100 \times \sqrt 2 }}{\pi } = 90.03V\\{R_{TOTAL}} = \frac{{{V_{dc}}}}{{1mA}} = \frac{{90.03}}{{1mA}} = 90.03k{\rm{\Omega }}\\{\rm{Rs}} = 90.03k{\rm{\Omega }} - 100{\rm{\Omega }} = {\rm{}}89.93k{\rm{\Omega }}\END{array}\)

"> BEGIN{ARRAY}{l}{V_{dc}} = \frac{{2{V_m}}}{\pi } = \frac{{2 \times 100 \times \sqrt 2 }}{\pi } = 90.03V\\{R_{TOTAL}} = \frac{{{V_{dc}}}}{{1mA}} = \frac{{90.03}}{{1mA}} = 90.03k{\rm{\Omega }}\\{\rm{Rs}} = 90.03k{\rm{\Omega }} - 100{\rm{\Omega }} = {\rm{}}89.93k{\rm{\Omega }}\END{array}\)

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An ac voltmeter uses the circle shown below where the PMMC meter has an internal resistance of 100 Ω and requires a DC current of 1 mA for full scale deflection. Assuming the diodes to be ideal, the value of Rs to obtain full scale deflection with 100 V (ac rms) applied to the input terminal would be

Electrical Engineering Indicating Instruments in Electrical Engineering . 7 months ago

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\(\BEGIN{ARRAY}{l}{V_{dc}} = \frac{{2{V_m}}}{\pi } = \frac{{2 \times 100 \times \sqrt 2 }}{\pi } = 90.03V\\{R_{TOTAL}} = \frac{{{V_{dc}}}}{{1mA}} = \frac{{90.03}}{{1mA}} = 90.03k{\rm{\Omega }}\\{\rm{Rs}} = 90.03k{\rm{\Omega }} - 100{\rm{\Omega }} = {\rm{}}89.93k{\rm{\Omega }}\END{array}\)

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Posted on 15 Nov 2024, this text provides information on Electrical Engineering related to Indicating Instruments in Electrical Engineering. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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