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A galvanometer has a resistance of 100Omega. A current of 10^(-3)A pass through the galvanometer. How can it be converted into (a) ammeter of range 10 A and (b) voltmeter of range 10v. |
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Answer» Solution :`G=100Omega,i_(g)=10^(-3)A` a) `i=10A,n=i/(i_(g))=10^(4)` `S=G/((n-1))=100/((10^(4)-1))=100/999Omega` b) `V_(g)=i_(g)G=10^(-3)xx100=10^(-1)V` `V=10VrArrn=V/(V_(g))=10/(10^(-1))=100` `thereforeR=G(n-1)=100(100-1)=9900Omega` |
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