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.

Explanation

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`