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A galvanometer has a resistance of 100 Omega. A current of 10^(-3) A is permissible through the galvanometer. How can it be converted into a) an ammeter of range 10 A and b) a voltmeter of range 10 V. |
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Answer» <P> Solution :G = 100 `OMEGA p, i_1 = 10^(-3)A `a) `i_2 = 10 A , n = (i_2)/(i_1) = 10^4` `S = (G)/((n-1)) = (100)/((10^4 -1)) = 100/999 Omega` b) `V_1 = i_1 G = 10^(-3) XX 100 = 10^(-1) V` `V_2 = 10 V rArr n = (V_2)/(V_1) = (10)/(10^(-1)) = 100` `therefore R = G (n-1) = 100 (100 -1) = 9900 Omega` |
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