A galvanometer of resistance 10 ohm and maximum current of `2muA` is converted into voltmeter of range 100mV and when converted into ammeter then range is 1mA. When these voltmeter and ammeter are connected by a (ideal) battery is series with a resistance of `R=1000 Omega`, then

Options

A. measure value of R is between `979 Omega` and `996 Omega`
B. resistance of voltmeter `10^(5) Omega`
C. shunt resistance is `20 m Omega`
D. If the ideal battery is replaced by non ideal battery with internal resistance of `5Omega` then R will be gt`1000 Omega`

Answer

A, C

Explanation

`v=100xx10^(-3)v`
`v=l_(g)(R_(g)+R)`
`(10^(-1))/(2xx10^(-6))=R_(g)+R`
`5xx10^(4) =R_(v)`

`I=Ixx10^(-3)`

`I_(g)R_(g)=(I-I_(g))S`
`S=(2xx10^(-6)xx10)/(10^(-3)-2xx10^(-6))`
`S=2xx10^(-5)xx10^(3)`
`implies 2xx10^(-2)`
`implies 20 m Omega `
`R_(A)=(20xx10^(-3)xx10)/10=20xx10^(-3)`