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Q.A galvanometer has a resistance of 40Ω and a current of 4 mA is needed for full scale deflection. What is the resistance and how is it to be connected to convert the galvanometer

(a) into an ammeter of 0.4 A range and
(b) into a voltmeter of 5 V range?
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 3mImportance★★★★★
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A small shunt in parallel converts the galvanometer into an ammeter; a large resistance in series converts it into a voltmeter — both sized using IgGI_g G relations.

Given: galvanometer resistance G=40 ΩG = 40\ \Omega, full-scale deflection current Ig=4 mA=4×10−3 AI_g = 4\ \text{mA} = 4\times10^{-3}\ \text{A}.

(a) Ammeter of range I=0.4I = 0.4 A:

A shunt SS is connected in parallel with the galvanometer so that most of the current bypasses the coil. At full-scale deflection, the shunt carries (I−Ig)(I - I_g) while the galvanometer carries IgI_g, and both share the same potential difference:

IgG=(I−Ig)S  ⟹  S=IgGI−IgI_g G = (I - I_g)S \implies S = \frac{I_g G}{I - I_g}

S=(4×10−3)(40)0.4−0.004=0.160.396≈0.404 ΩS = \frac{(4\times10^{-3})(40)}{0.4 - 0.004} = \frac{0.16}{0.396} \approx 0.404\ \Omega

(b) Voltmeter of range V=5V = 5 V:

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