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 = 30 xx(20 xx 10^(-6))`
`= 6 xx 10^(-4)A.`
It S is the required value of the shunt connected in parallel with the galvanometer, then
`i_g = (S)/(S+G) i or 6 xx 10^(-4) = (S)/(S +25) xx 1`
After solving , we get `S = (150)/(9994) Omega = 0.0150 Omega`
The resistance of the ammter `R_A = (SG)/(S+G) = (0.0150 xx 25)/(0.0150 + 25) = 0.0150 Omega`
To convert this ammeter into the voltmeter , we can use `V = i_(g) (R_A + R_0)`
Here `V = 1V, i_g = 1A`
`therefore 1 = 1(0.0150 + R_0) or R_0 = 0.985 Omega`