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(B) Let $i_{g}$ be the full-scale deflection current of the galvanometer.For the first case,the range is $0-1\, V$ with resistance $R_{1}$ in series:$

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$R_{1}+G$

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(B) Let $i_{g}$ be the full-scale deflection current of the galvanometer.For the first case,the range is $0-1\, V$ with resistance $R_{1}$ in series:$1 = i_{g}(G + R_{1}) \quad \dots(1)$For the second case,the range is $0-2\, V$ with an additional resistance $R_{2}$ in series:$2 = i_{g}(G + R_{1} + R_{2}) \quad \dots(2)$Dividing equation $(2)$ by equation $(1)$:$\frac{2}{1} = \frac{i_{g}(G + R_{1} + R_{2})}{i_{g}(G + R_{1})}$$2 = \frac{G + R_{1} + R_{2}}{G + R_{1}}$$2(G + R_{1}) = G + R_{1} + R_{2}$$2G + 2R_{1} = G + R_{1} + R_{2}$$R_{2} = G + R_{1}$Thus,the additional resistance required is $G + R_{1}$.

$A$ galvanometer of resistance $G$ is converted into a voltmeter of range $0-1\, V$ by connecting a resistance $R_{1}$ in series with it. The additional resistance that should be connected in series with $R_{1}$ to increase the range of the voltmeter to $0-2\, V$ will be

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