A galvanometer with a scale divided into `100` equal divisions has a current sensitivity of `10` divisions per mA and a voltage sensitivity of `2` divisions per mV. What adoptions are required to read (i) `5A` for full scale and (ii) 1 division per volt?

Explanation

(a) .
`I_g G = (I - I_g) S`
`0.1 I G = (I - 01 I)S`
`S = (G)/(9) = (99)/(9) = 11 Omega`
Resistance of ammeter
`R_A = (GS)/(G + S) = (99 xx 11)/(99 + 11) = 9.9 Omega`
(b) Given : `G = 30 Omega, I_g = 2mA = 2 xx10^-3 A`
(i) To convert galvanometer into ammter, connect a small resistance (shunt `S`) in parallel to it.
`I_g G = (I - I_g) S`
`2 xx 10^-3 xx 30 = (0.3 - 2 xx 10^-3)S`
`S = 60 xx 10^-3 ~= 0.3 S`
`S = (60 xx 10^-3)/(0.3) = 0.2 Omega`
(ii) To convert galvanometer into voltmeter, connect a large resistance (R ) in series to it.

`V = I_g (G + R)`
`0.3 = 2 xx 10^-3(30 + R)`
`30 + R = 150 rArr R = 120 Omega`
( c) `I_g = 10 mA, V_g = 50 mV`
`V_g = I_g G rArr G = (V_g)/(I_g) = (50)/(10) = 5 Omega`
(i) `I_g G = (I - I_g) S`
`10 xx 10^-3 xx 5 = (5 - 10 xx 10^-3)S`
`50 xx 10^-3 ~= 5 S`
`S = 10 xx 10^-3 = 0.01 Omega`
(ii) `V = I_g (G + R)`
`100 = 10 xx 10^-3(5 + R)`
`R =10^4 - 5 = 10000 - 5 = 9995 Omega`
(d)
`V_1 = (2000)/((2000 + 9000)) xx 220 = (2)/(11) xx 220 = 40 V `
Reading of voltmeter `= 40 V`.