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A galvanometer has a current range of 15 mA and a voltage range of 750 mV. To convert this galvanometer into an ammeter of range 25 A, the required shunt is

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$0.8 Ω$

$0.93 Ω$

$0.03 Ω$

$2 Ω$

answer is A.

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Detailed Solution

$G = \frac{V}{I_{g}} \Rightarrow \frac{750 \times 10^{- 3}}{15 \times 10^{- 3}} = 50 ∴ G = 50 Ω$

Shunt, $S = \frac{G}{(\frac{I}{I_{g}} - 1)} \Rightarrow S = \frac{50}{(\frac{25}{15 \times 10^{- 3}} - 1)}$

$∴ S = \frac{50}{\frac{5 \times 10^{3}}{3} - 1} \Rightarrow S = \frac{50 \times 3}{4997} = 0 .03 Ω$

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