A galvanometer of resistance G has a voltage | vedclass

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(D) The correct option is $(D)$.To convert a galvanometer into a voltmeter,we need to connect a large resistor $R$ in series with the galvanometer.Let

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$G \cdot \left[\frac{V}{V_g}-1\right]$

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(D) The correct option is $(D)$.To convert a galvanometer into a voltmeter,we need to connect a large resistor $R$ in series with the galvanometer.Let $I_g$ be the full-scale deflection current of the galvanometer.Then,the voltage range of the galvanometer is $V_g = I_g G$.When a resistor $R$ is connected in series,the total resistance becomes $G + R$.The new voltage range $V$ is given by $V = I_g(G + R)$.Dividing the two equations: $\frac{V}{V_g} = \frac{I_g(G + R)}{I_g G} = \frac{G + R}{G}$.$\frac{V}{V_g} = 1 + \frac{R}{G}$.$\frac{R}{G} = \frac{V}{V_g} - 1$.Therefore,$R = G \left( \frac{V}{V_g} - 1 \right)$.

$A$ galvanometer of resistance $G$ has a voltage range $V_g$. The resistance required to convert it to read voltage up to $V$ is

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