Calculate the value of resistance needed to convert a moving-coil galv

1.	Calculate the value of resistance needed to convert a moving-coil galvanometer of 60 Ω which gives a full scale deflection for a current of 50 mA into (i) an ammeter of range 0 – 5 A (ii) a voltmeter of range 0 – 50 V.
Answer» Data : G = 60 Ω, I_g = 50 mA = 5 × 10^-2 A, I = 5 A, V = 50 V (i) S = \(\cfrac{GI_g}{I-I_g}\) = \(\cfrac{60\times5\times10^{-2}}{5-5\times10^{-2}}\) = \(\cfrac3{4.95}\) = 0.6061 Ω A resistance of 0.6061 Ω should be connected in parallel to the coil of the galvanometer to measure current up to 5 A. (ii) R_S = \(\cfrac{V}{I_g}\) – G = \(\cfrac{50}{5\times10^{-2}}\) – 60 = 1000 – 60 = 940 Ω A resistance of 940 Ω should be connected in series with the coil of the galvanometer to measure voltage up to 50 V.

Calculate the value of resistance needed to convert a moving-coil galvanometer of 60 Ω which gives a full scale deflection for a current of 50 mA into (i) an ammeter of range 0 – 5 A (ii) a voltmeter of range 0 – 50 V.

Answer»

Data : G = 60 Ω,

I_g = 50 mA = 5 × 10^-2 A,

I = 5 A,

V = 50 V

(i) S = \(\cfrac{GI_g}{I-I_g}\)

= \(\cfrac{60\times5\times10^{-2}}{5-5\times10^{-2}}\) = \(\cfrac3{4.95}\) = 0.6061 Ω

A resistance of 0.6061 Ω should be connected in parallel to the coil of the galvanometer to measure current up to 5 A.

(ii) R_S = \(\cfrac{V}{I_g}\) – G

= \(\cfrac{50}{5\times10^{-2}}\) – 60

= 1000 – 60 = 940 Ω

A resistance of 940 Ω should be connected in series with the coil of the galvanometer to measure voltage up to 50 V.