Current senstivity of moving coil galvanometer is `5 "div"//mA` and its voltage senstivity (angular deflection per unit voltage applied) is `20 "div"//V`. The resistance of the galvanometer is

Options

A. `250 Omega`
B. `25 Omega`
C. `40 Omega`
D. `500 Omega`

Answer

A

Explanation

Current sensitivity of a moving coil galvanometer is the deflection `(theta)` per unit current (I) flowing through it, i.e.
`I_(S) = theta/I = (NAB)/k` ……….(i)
Where, N= numberof turns in the coil,
A=Area of each turn of coil,
B = magnetic field
k= restoring torque per unit twist of the fibre stirp.
Similarly, voltage sensitivity is the deflection per unit voltage. i.e.,
`V_(s) = theta/V = (NAB)/(k)(I/V) = (NAB)/(kR_(G)`............(ii)
where, `R_(G)` is the resistance of the galvanometer.
From eq(s) (i) and (ii), we get
`R_(a) = I_(S)/V_(S)` ...............(ii)
Here, `I_(s) = 5"div"/mA = 5 xx 10^(-3)"div"/A` and `V_(B)`=20 "div"/V
Substituting the given values in Eq. (iii), we get
`R_(a) = (5 xx 10(3))/20 = 250`
`therefore` The resistance of the galvanometer is `250 Omega`