A galvanometer coil has a resistance of 10 O | vedclass

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(C) Given:Resistance of galvanometer $(G)$ = $10 \ \Omega$Full-scale deflection current $(I_g)$ = $3 \ 	ext{mA} = 3 	imes 10^{-3} \ 	ext{A}$Range o

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$3 	imes 10^{-3}$

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(C) Given:Resistance of galvanometer $(G)$ = $10 \ \Omega$Full-scale deflection current $(I_g)$ = $3 \ 	ext{mA} = 3 	imes 10^{-3} \ 	ext{A}$Range of ammeter $(I)$ = $10 \ 	ext{A}$The formula for shunt resistance $(S)$ required to convert a galvanometer into an ammeter is given by:$S = \frac{I_g \cdot G}{I - I_g}$Substituting the values:$S = \frac{3 	imes 10^{-3} 	imes 10}{10 - 3 	imes 10^{-3}}$Since $3 	imes 10^{-3} = 0.003 \ 	ext{A}$,the denominator is $10 - 0.003 = 9.997 \ 	ext{A}$.$S = \frac{0.03}{9.997} \approx \frac{0.03}{10} = 0.003 \ \Omega$Therefore,$S = 3 	imes 10^{-3} \ \Omega$.

$A$ galvanometer coil has a resistance of $10 \ \Omega$ and the meter shows full-scale deflection for $3 \ \text{mA}$. The value of the shunt required to convert this meter into an ammeter of range $0$ to $10 \ \text{A}$ is . . . . . . $\Omega$.

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