Two moving coil metres `M_(1) and M_(2)` have the following particular `R_(1) =10Omega, N_(1) =30, A_(1) = 3.6 xx 10^(-3) m^(2), B_(1) =0.25 T,`
`R_(2) = 14Omega , N_(2) =42, A_(2) = 1.8 xx 10^(-3) m^(2), B_(2) =0.50 T`
The spring constants are identical for the two metres. What is the ratio of current sensitivity and voltage sensitivity of `M_(2)" to" M_(1)?`

Options

A. 1.4, 1
B. 1.4, 0
C. 2.8, 2
D. 2.8, 0

Answer

A

Explanation

For meter `M_(1),R_(1)=10Omega, N_(1)=30,`
`A_(1)=3.6xx10^(-3)m^(2), b_(1)=0.25 T, k_(1)=k`
For meter `M_(2), R_(2)=14 Omega, N_(2)=42,`
`a_(2)=1.8xx10^(-3)m^(2), B_(2)=0.50T, k_(2)=k`
So, current sensitivity `I_(s)=(NBA)/(k)`
`(I_(s_(2)))/(I_(s_(1)))=(N_(2)B_(2)A_(1)//k_(2))/(N_(1)B_(1)A_(1)//k_(1))=(42xx0.50xx1.8xx10^(-3)//k)/(30xx0.25xx3.6xx10^(-3)//k)=1.4`
and voltage sensitivity `V_(s)=(N_(2)BA)/(kR)`
Now, `(V_(s_(2)))/(V_(s_(1)))=(N_(2)B_(2)A_(2)//(k_(2)R_(2)))/(N_(1)B_(1)A_(1)//(k_(1)R_(1)))=(N_(2)B_(2)A_(2)R_(1)k_(1))/(N_(1)B_(1)A_(1)R_(2)k_(2))`
`=(42xx0.50xx(1.8xx10^(-3))xx10xxk)/(30xx0.25xx(3.6xx10^(-3))xx14xxk)=1.`