# A Millivoltmeter Of 25 Millivolt Range Is To Be Converted Into An Ammeter Of 25-ampere Range. The Value (In Ohm) Of Necessary Shunt Will Be: (A) 1 (B) 0.01 (C) 0.05 (D) 0.001

The correct answer is (D) 0.001

$$i_{g} = \frac{25 mV}{G} ampere$$

Where G is the resistance of the meter. The value of the shunt required for converting it into ammeter of range 25 A will be

$$S = \frac{i_{g}G}{i – i_{g}}$$ $$\Rightarrow \frac{25 mV}{25}$$ $$\Rightarrow 0.001 \Omega$$

Hence, proved

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