A galvanometer having `30` divisions has a current sensitivity of `20muA//di vision`. It has a resistance of `25Omega`. How will you convert it into an ammeter upto 1 ampere? How will you convert this ammeter into a voltmeter up to 1 volt?

Explanation

The full scale deflection current
`i_(g3)xx(20xx10^(-6))=6xx10^(-4)A` .
If `S` is the required value of the shunt connected in parallel with galvanometer , then
`i_(g)=(S)/(S+G)iimplies6xx10^(-4)=(S)/(S+25)xx1`
After solving, we get `S=(150)/(9994)Omega=0.0150Omega`
The resistance of the ammeter
`R_(A)=(SG)/(S+G)=(0.0150xx25)/(0.0150+25)=0.0150Omega`
To conver this ammeter into the voltmeter, we can use
`V=i_(g)(R_(A)+R_(0)) `Here `V=1V,i_(g)=1A`
`:. 1 =1(0.0150+R_(0))` or `R_(0)=0.98Omega`