A voltmeter has a range `O - V` with a series resistance `R`. With a series resistance `2R`, the range is `O - V'`. The correct relation between `V` and `V'`is

Explanation

Here, galvanometer resistance `G = 3Omega` and full scale deflection current`I_(g) = 2 mA`
(a) To convert the galvanometer into an ammeter of range 0.3 ampere. a resistance of value .S. is connected in parallel with it such that the current through G should not be more than `I_(g)=0.3 A` and`(I-I_(g)` should pass through S.
`(I-I_(g) S = I_(g),G`
therefore `S=(I_(g)G)/(I-I_(g))=(2xx10^(3)xx30)/(0.3-2xx10^(-3))=0.2Omega`Hence, to convert the galvanometer into an ammeter of the desired range a shunt resistance (a small valued resistance) of `0.2Omega` is connected parallel to the meter. This shunt resistance gives us a low resistance instrument with a deflection current `l_(3)` - 0.3 ampere, while the current through the `galvanometer is 2 mA`
(b) To convert the galvanometer into a voltmeter of range 0.2 volt, a resistance R is. connected in series with it such that`V=I_(g)(R+G)i.e.0.2=2xx10^(-3)(30+R)`Or,R=100-30=70Ohms
Thus, to convert the galvanometer into a ! voltmeter of the desired range, a high resistance (R_(s)` is connected in series with the galvanometer. The equivalent meter resistance is `R_(eg) = 30 + 70 = 100 ohm` In this case, most of the voltage appears across the series resistor. The current through the voltmeter is 2 mA