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A galvanometer has a resistance of 40 \(\Omega\) 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 o…

Q.24 A galvanometer has a resistance of 40 \(\Omega\) 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?

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Answer: (a) A shunt of about 0.404 Ω in parallel with the galvanometer. (b) A resistance of 1210 Ω in series with the galvanometer.

\(G = 40\ \Omega\), \(I_g = 4\) mA \(= 4\times10^{-3}\) A.

(a) Ammeter, \(I = 0.4\) A: a low resistance shunt \(S\) is connected in parallel. The p.d. across G and S is the same: \(I_g G = (I - I_g)S\).

\[S = \frac{I_g G}{I - I_g} = \frac{4\times10^{-3}\times40}{0.4 - 0.004} = \frac{0.16}{0.396} \approx 0.404\ \Omega\]

(b) Voltmeter, \(V = 5\) V: a high resistance \(R\) is connected in series: \(V = I_g(G + R)\).

\[R = \frac{V}{I_g} - G = \frac{5}{4\times10^{-3}} - 40 = 1250 - 40 = 1210\ \Omega\]
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