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A galvanometer of resistane G can measure 1 A current. If a shunt S is used to convert it into an ammeter to measure 10 Acurrent . The ratio ofG/Sis |
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Answer» ` 1/9` I = 10 A Fromfigure, `I_(g)G = ( I - I_(g))S ` ` G/S = (I-I_(g))/(I_(g))` Substitutingthe GIVENVALUES, we GET ` G/S= (10 A -1 A)/( 1 A) = 9/1 `
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