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Tha scale of a galvanometer is divided into `150` equal divisions. The galvanometer be designed to read (i) `6` A per division and (ii) `1 V` per division? |
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Answer» Full-scale voltage `= 150` divisions`//2` divisions per `m V` `= 75 m V` Full-scale current`= 150` divisions`//10` divisions per `m A` `= 15 mA` So resistance of galvanometer is `G = ("Full-scale voltage")/("Full-scale current"1) = (75 xx 10^(-3))/(15 xx 10^(-3)) 5 Omega` (i) Range of ammeter is `I = 150 xx 6 = 900 A`. So, `S = (I_(G)G)/(I - I_(G)) = (15 xx 10^(-3) xx 5)/((900 - 15 xx 10^(-3))) = 8.3 xx 10^(-5) Omega` (ii) Range of voltmeter is `V = 150 xx 1 = 150 V`. So, `R = (V)/(I_(G)) - G = (150)/(15 xx 10^(-3)) - 5 = 9995 Omega`. |
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