A moving coil galvanometer of resistance $20 Ω$ gives a full scale deflection when a current of 1 mA is passed through it. It is to be converted into an ammeter reading 20 A on full scale. But the shunt of $0.005 Ω$ only is available. What resistance should be connected in series with the galvanometer coil ? (in $Ω$ ).

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answer is 79.995.

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Detailed Solution

$n = \frac{I}{I_{g}} = \frac{20}{10^{- 3}} = 2 \times 10^{4}; S = \frac{G + R}{n - 1}$

$\Rightarrow G + R = (n - 1) S$

$\Rightarrow 20 + R = 5 \times 10^{- 3} [2 \times 10^{4} - 1]$

$R = 79.995 Ω$

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