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.

Answer

A, B, C

Explanation

For ammeter `99I_g=(I-I_g)`
or `I=100 I_g` ………..i

`I_g` is the ful scale deflection current of the galvanometer and `l` the range of ammeter,.
for the circuit in fig 1 given in the question
`(12V)/(2r+(99xx1)/(99+1)=3Aimpliesr=1.01Omega`
For voltmeter, range
`V=I_g(99+100)`
`V=200l_g`
Also resistance of the voltmeter
`=99+101=200Omega`

in fig 2 resistance across the terminals of the battery
`R_1=r+(200xx2)/(200+2)=2.99Omega`
`V=7.96xx5/4=99.5V`
using eq. ii `I_g=9.95/200=0.05A`
Using eqn i range of the ammeter ltbr. `I=100 I_g=5A`