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A galvanometer of resistance of 10 Omega...

A galvanometer of resistance of `10 Omega` gives fullscale defiection for a current of 4 mA. How can it be converted into an ammeter of range `0-5 A` ?

Text Solution

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We are given `:G = 10 Omega ,lg = 4 m A = 4 xx 10 ^(-3) A`
We know that, Resistance required `S = ( I _(R))/(I - I _(R)) G`
`therefore S = ( 4 xx 10 ^(-3))/( 5 xx 4 xx 0 ^(-3)) xx 10 = 8 xx 10 ^(-3) Omega `
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