A galvanometer h aving 30 divisions has current sensitivity of `20muA`/division. . It has a resistances of 25` Omega`
(i) How will you convert it into an ammeter measuring upto 1 ampere.
(ii) How will you convert this ammeter into a voltmeter upto 1 volt.

Explanation

The full scale deflection current
`i_(g)=30xx(20xx10^(-6))=6xx10^(-4)A`.
If S is the required value of the shunt connected in parallel with the galvanometer, then
`i_(g)=S/(S+G)irArr6xx10^(-4)=S/(S+25)xxl`
After solving, we get `S=150/9994Omega-0.01150Omega`
The resistance of the ammeter
`R_(A)=(SG)/(S+G)=(0.0150xx25)/(0.0150+25)=0.0150Omega`
To convert this ammeter into the voltmeter, we can use
`V=i_(g)(R_(A)+R_(0))` Here `V=1V,i_(g)=1A`
`therefore1=1(0.0150+R_(0))` or `R_(0)=0.985Omega`