A galvanometer reads `5*0V` at full scale deflection and is graded according to its resistance per volt at full scale deflection as `5000OmegaV^-1`. (i) How will you convert it into a voltmeter that reads `20V` at full scale deflection? (ii) Will it still be graded `5000OmegaV^-1`? (iii) Will you prefer this voltmeter to one that is graded `2000OmegaV^-1`?

Explanation

Here, resistance per volt=`5000OmegaV^-1`. It means for `1` volt pot. difference, the resistance of galvanometer is `5000Omega`. So current for full scale deflection,
`I_g=1/5000A=0*2xx10^-3A`
(i) In order to convert it into a voltmeter and range 0 to `20V`, let a resistance R be connected in series with it. Then on applying an extra potential difference `=20V-5V=15V`, the current potential it for full scale deflection is again `0*2xx10^-3A` i.e.,
`I_g=0*2xx10^-3A`.
`:. RI_g=15` or `R=15/I_g=(15)/(0*2xx10^-3)=75000Omega`
Thus, to convert a given voltmeter (of range 0 to 5V) into a voltmeter (of range 0 to 20V) a resistance of `75000Omega` should be connected in series. With the given meter.
(ii) Original resistance of voltmeter
`=5000OmegaV^-1xx5V=25000Omega`
Total resistance after conversion
`=25000+75000=100,000Omega`
Resistance per volt of new meter
`=(100,000)/(20)=5000OmegaV^-1`
i.e. it has the same grading as before.
(iii) If greater is the resistance per volt of a meter, the lesser will be the current drawn from the circuit by it. Due to it, it works better. That is why this meter (graded `5000OmegaV^-1`) is more accurate that the one graded as `2000OmegaV^-1`.