The scale of galvanometer is divided into `150` equal divisions. The galvanometer has a current sensitivity of `10"divisions"//mA` and the voltage sensitivity of `2"divisions"//mV`. How the galvanometer be designed to read (i) 6A per division and (ii) 1 V per division?

Answer

(a) `8*3xx10^-5Omega` in parallel
(b) `9995 Omega` in series

Explanation

Galvanometer resistance is
`=(I_s)/(V_s)=(nBA//k)/(nBA//kG)=G`
So, `G=(I_s)/(V_s)=10/2=5Omega`
Current for full scale deflection
`I_g=(150)/(10)mA=15mA=15xx10^-3A`
(i) The total current to be read,
`I=6xx150=900A`
As, `S=(I_gG)/(I-I_g)=(15xx10^-3xx5)/(900-15xx10^-3)`
`=(15xx5xx10^-3)/(900)`
`=8*3xx10^-5Omega` in parallel
(ii) The total voltage to be read,
`V=1xx150=150V`
As `R=(V)/(I_g)-G=(150)/(15xx10^-3)-5`
`=10^4-5=9995Omega` in series