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(D) Given: Resistance of the device $G = 1000\,\Omega$,full-scale current $I_g = 100\,mA = 0.1\,A$,and the desired range $I = 1\,A$.To convert a galva

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$111$

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(D) Given: Resistance of the device $G = 1000\,\Omega$,full-scale current $I_g = 100\,mA = 0.1\,A$,and the desired range $I = 1\,A$.To convert a galvanometer (or voltmeter) into an ammeter,a shunt resistance $S$ is connected in parallel.The formula for shunt resistance is $S = \frac{I_g \cdot G}{I - I_g}$.Substituting the values: $S = \frac{0.1 	imes 1000}{1 - 0.1}$.$S = \frac{100}{0.9} = \frac{1000}{9} \approx 111.11\,\Omega$.Rounding to the nearest integer,we get $111\,\Omega$.

$A$ voltmeter of resistance $1000\,\Omega$ gives full scale deflection when a current of $100\,mA$ flows through it. The shunt resistance required across it to enable it to be used as an ammeter reading $1\,A$ at full scale deflection is ............... $\Omega$.

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