A galvanometer (coil resistance `99 Omega`.) is converted into an ammeter using a shunt of `1 Omega` and connected as shown in figure (a). The ammeter reads `3 A`. The same galvanometer is convened into a voltmeter by connecting a resistance of `101 Omega` in series. This voltmeter is connected at, shown in figure (b). Its reading is found to be `4/5` of the full scale reading. Find :
(a) internal resistance r of the cell
(b) range of the ammeter and voltmeter
(c) full scale deflection current of the galvanometer.

Explanation

For ammeter
`99I_(8)=(I-I_(8))I`
or `I=100I_(8)`……`(1)`
`I_(8)` is the full scale deflection current of the galvanometer and `I` is the range of ammeter
For the circuit in the adjacent figure,
`(12)/(2+r+(99xx1)/(99+1))=3A`
`rArr r=1.01Omega`
For voltmeter, range
`V=I_(8)(99+101)=200I_(8)`.......`(2)`
Also resistance of the voltmeter `=99+101=200Omega`
In the adjacent figure, resistance across the terminals of the battery
`R_(1)=r+(200xx2)/(202)=2.99Omega`
`:. ` Current drawn from the battery, `I_(1)=(12)/(2.99)=4.01A`
`:. ` Voltmeter reading
`(4)/(5)V=12-I_(8)r=12-4.01xx1.01`
`V=7.96xx(5)/(4)=9.95V`
Using Eq. `(2)`, `I_(8)=(9.95)/(200)=0.05A`
Using Eqs. `(1)`, range of the ammeter
`I=100I_(8)=5A`.