A galvanometer coil has a resistance of `12Omega` and the meter shows full scale deflection for a current of `3mA`. How will you convert the meter into a voltmeter of range 0 to 18V?

Explanation

Let R be the resistance connected in series with the galvanometer to convert it into a voltmeter of range ` 0 " to " 18 V` .

If G is the resistance of the galvanometer and `I_(g)` is the current flowing the network, then:
` V = I_(g) (R + G) ` ....(i)
Here `G = 12 Omega, I_(g) = 3 mA = 3 xx 10^(-3) A, V = 18 V`
From (i) , `R = V/I_(g) - G`
` :. " " R = (18)/(3 xx 10^(-3)) - 12 `
` = (6,000 - 12 ) Omega`
` rArr" " R = 5,988 Omega`