A galvanometer coil has a resistance of `12Omega` and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V ?

Explanation

1. To convert galvanometer to ammeter low value resistance (shunt) should connect with it so relation between S and G is given by,
`thereforeGI_(G)=S(I-I_(G))`
where G= resistance of galvanometer
`I_(G)` = current capacity of galvanometer
For `I_(1)=10mA,G=G`
`S_(1)+S_(2)+S_(3)=(I_(G)G)/(I_(1)-I_(G))`
`S_(1)+S_(2)+S_(3)=((1xx10)mA)/((10-1)mA)=10/9" "...(1)`
For `I_(2)=100mA,G=10+S_(1)`
`S_(2)+S_(3)=(I_(G)(G+S_(1)))/((I_(2)-I_(G)))`
= `(1mA(10+S_(1)))/((100-1)mA)=(10+S_(1))/99" "...(2)`
For `I_(3)=1A,G=10+S_(1)+S_(2)`
`S_(3)=(I_(G)(G+S_(1)+S_(2)))/((I_(3)-I_(G)))=(1mA(10+S_(1)+S_(2)))/((1000-1)mA)`
`thereforeS_(3)=(10+S_(1)+S_(2))/999" "...(3)`
2. Putting the value of equation (2) in equation (1),
`S_(1)+(10+S_(1))/99=10/9`
`thereforeS_(1)+S_(1)/99=10/9-10/99`
`therefore100S_(1)=110-10` multiplying both sides by 99
`thereforeS_(1)=100/100=1Omega" "...(4)`
3. Putting `S_(1)=1Omega` in equation (2),
`S_(2)+S_(3)=(10+1)/99=11/99=1/9" "...(5)`
4. The value of equation (3) in equation (5),
`thereforeS_(2)+(10+S_(1)+S_(2))/999=1/9`
`thereforeS_(2)+S_(2)/999=1/9-S_(1)/999-10/999`
`therefore1000S_(2)=111-1-10`
[`because` Multiplying both sides by 999]
`thereforeS_(2)=100/1000=0.1Omega`
5. From equation (2),
`S_(2)+S_(3)=(10+S_(1))/99`
`thereforeS_(3)=(10+1)/99-1/10" "[becauseS_(1)=1Omega]`
`thereforeS_(3)=1/9-1/10=1/99=" "...0.0111`
`thereforeS_(3)~~0.01Omega`