There are 21 marks (zero to 20) on the dial of a galvanometer, that is there are 20 divisions. On passing `10muA` current through it, it shows a deflection of 1 division. Its resistance is `20Omega`.
(a) How can it be converted into an ammeter which can measure 1 A current ?
(b) How can we change it into a voltmeter to measure a pd of 1 V ? Also find the effective resistance of both of the above mentioned meters.

Explanation

1. When a current of `10muA` passes through the galvanometer, its pointer shows a deflection of 1 division. There are 20 divisions in this galvanometer.
`therefore` The maximum current which can be measured by it (current capacity)
`I_(G)=10xx10^(-6)xx20=200xx10^(-6)A`.
2. For ammeter, the required shunt to be joined in parallel to galvanometer is,
`S=(GI_(G))/(I-I_(G))`
= `(20xx200xx10^(-6))/(10000xx10^(-4)xx10^(-4))`
= `(20xx2xx10^(-4))/(10000xx10^(-4)xx10^(-4))`
= `40/9998~~0.004Omega`
`I_(G)=200xx10^(-6)A`
= `2xx10^(-4)A`
`G=20Omega`
I = 1 A
= `10000xx10^(-4)A`
3. Thus to convert this galvanometer into an ammeter which can measure 1A current, a shunt of `0.004Omega` should be joined.
The effective resistance of this ammeter will be
`G.=(GS)/(G+S)=(20xx0.004)/(20+0.004)~~0.004Omega`.
4. In order to convert the galvanometer into a voltmeter, the required series resistance is
`R_(S)=V/I_(G)-G`
= `1/(2xx10^(-4))-20`
= `0.5xx10^(4)-20`
= 5000 - 20
= `4980Omega`
Here, V = 1 volt
`I_(G)=2xx10^(-4)A`
`G=20Omega`
5. In order to convert this galvanometer into a voltmeter which can measure 1 volt, a series resistance of 4920 should be joined with it.
The effective resistance of this voltmeter will be `R._(S)=R_(S)+G=4980+20=5000Omega`.
(`thereforeR_(S)` and G are in series)