Tha scale of a galvanometer is divided into `150` equal divisions. The galvanometer be designed to read (i) `6` A per division and (ii) `1 V` per division?

Options

A. `S=8.3xx10^(-5)Omega, R=9995Omega`
B. `S=8.3xx10^(-2)Omega,r=995Omega`
C. `S=4.3xx10^(-5)Omega,R=995Omega`
D. `R=8.3xx10^(-5)Omega,S=995Omega`

Answer

1

Explanation

Here, `I_(g)=((1mA)/(10 d iv))(150di v)=15mA`
`V_(g)=((1mA)/(2di v))(150di v)=75mV`
`G=(V_(g))/(I_(g))=(75mV)/(15mA)=5 Omega`
`(a) I (` current to be measured `)`
`=((6A)/(di v))(150di v)=900A`
As `n=(I)/(I_(g))=(900A)/(15mA)=(900A)/(15xx10^(-3)A)=6xx10^(4)`
`S=(G)/((n-1))=(5Omega)/((6xx10^(4)-1))~~(5Omega)/(6xx10^(4))=8.3xx10^(-5)Omega`
`(b)V (` voltage to be measured `)`
`=((1V)/(di v))(150d i v)=150V`
As `n=(V)/(V_(g))=(150V)/(75mV)=(150V)/(75xx10^(-3)V)=2xx10^(3)`
`R=G(n-1)=5Omega(2xx10^(3)-1)`
`=5 Omega(2000-1)=9995Omega`