Two identical moving coil galvanometers have `10Omega` resistance and full scale deflection at `2muA` current. One of them is converted into a voltmeter of 100mA full scale reading and the other into an Ammeter of 1mA full scale current using appropriate resisters. These are then used to measure the voltage and current in the Ohm’s law experiment with `R=1000 Omega` resistor by using an ideal cell. Which of the following statement(s) is(are) correct?

Options

A. The measured value of R will be `978 Omega lt R lt 982 Omega`.
B. The resistance of the Voltmeter will be `100 k Omega`.
C. The resistance of Ammeter will be `0.02 Omega` (round off to `2^(nd)` decimal place).
D. If the ideal cell is replaced by a cell having internal resistance of `5 Omega` then the measured value of R will be more than `1000 Omega`.

Answer

A, C

Explanation

Let `R_(v)` be the resistance connected in series with the coil `(10 Omega)` of galvanometer to convert it to measure up to 100mV or 0.1V. We are given that a current of `2mu` A is sufficient for full scale deflection.

`0.1= (2 xx10^(-6)) (R_(V) +10) rArr R_(V) = 49990 Omega` Hence option (b) is wrong.
Let `R_(A)` be the resistance connected in parallel to the coil of galvanometer to convert galvanometer into ammeter

Potential difference across `10Omega` resistance and `R_(A)` will be same.
`(2 xx 10^(-6)) xx 10= (10^(-3) -2 xx 10^(-6))R_(A)`
`rArr (2 xx 10^(-6)) xx 10 = (10^(-3)-2 xx 10^(-6))R_(A)`
`rArr R_(A) = 0.02 Omega`
Hence option (c) is correct
Resistance of the voltmeter is `50000Omega`. Circuit for Ohm.s law can be drawn as follows:

`y50000 =(x-y)1000`
`:. 51y=x`
Reading `=(y50000)/(x)=980`
Hence option (a) is correct
Measured value of resistance cannot be more than `1000Omega`, hence option (d) is wrong.